This is a hard problem. Note: IMO 2026 hasn’t happened; this resembles a known problem. Actually this resembles ISL / a known problem: sequence where each next term is smallest integer > a_n sharing a factor with all previous terms. Claim: eventually periodic with linear growth.

Key observation: gcd(a_{n+1}, a_i) > 1 for ALL i ≤ n. So the set of primes dividing a_{n+1} must “hit” every previous term. As n grows, the constraint accumulates. Let me think about structure.

Let P_n = set of primes p such that p | a_i for all… no. Condition: a_{n+1} must share a prime with each a_i, i ≤ n.

Intuition: eventually every term is divisible by some fixed prime p? If some prime p divides all a_i for i ≥ N… Actually consider: if 2 divides a_n for infinitely many n? Hmm.

Classic approach: Show that eventually all terms share a common prime p, and then the sequence becomes: a_{n+1} = smallest multiple of … hmm not quite; a_{n+1} just needs gcd>1 with each earlier term; if all earlier terms from some point are divisible by p, but earlier finitely many terms a_1..a_N may need other primes.

Suppose eventually all a_n divisible by p (say p=2). Then condition for a_{n+1}: even and gcd with each of a_1,…,a_N >1 (the odd early ones… actually all terms; even terms automatically satisfied). So a_{n+1} = smallest even number > a_n with gcd>1 with each of the finitely many “exceptional” terms b_1,…,b_k (those not divisible by 2). The set S = {m : 2|m, gcd(m,b_j)>1 ∀j} is a union of residue classes mod M where M = 2·∏(primes dividing b_j’s, radical). Indeed membership in S depends only on m mod M where M = 2·rad(b_1···b_k). Then a_{n+1} = next element of S after a_n, and S is periodic mod M, nonempty (M itself… gcd(M,b_j) > 1 yes, M ∈ S). So the sequence eventually walks through S in order: once a_n ∈ S and all constraints are just “be in S”, we get a_{n+1} = successor in S. S periodic with period M and density d = |S∩[1,M]|·… Then a_{n+T} = a_n + M with T = number of elements of S in an interval of length M. That gives periodicity for n large: a_{n+T} = a_n + L for n ≥ N0. But problem wants for every positive integer n! Hmm, “for every positive integer n” — with a_{n+T}=a_n+L for all n ≥ 1? That seems too strong unless… Wait, maybe they allow choosing T large so that… a_{n+T}=a_n+L must hold also for small n. Since eventually periodic: exists N0, T0, L0 with a_{n+T0} = a_n + L0 for n ≥ N0. For n < N0 it may fail. Hmm but the statement says for every positive integer n. That forces exact periodicity from the start? That can’t be derived from eventual… Actually it can: if we prove eventual periodicity, we can’t extend backwards generally. But maybe the problem intends exists T, L such that for all n ≥ some point? Actual IMO 2025 Problem 4? Wait — this is actually IMO 2025 Problem 4? No, IMO 2025 P4 was about divisors: a_{n+1} = sum of three largest divisors. Hmm.

Hmm, but maybe the statement as given (“for every positive integer n”) is fine because we can choose T ≥ N0 and note… no, a_{1+T} = a_1 + L needs a_1 in the periodic pattern. Not necessarily.

Wait—maybe indeed the eventual behavior forces alignment: replace T by T’ ≥ N0? Condition needs a_{n+T} = a_n + L for n=1: a_{1+T} vs a_1. Not controllable.

But maybe actually we can prove exact periodicity: the walk is deterministic given the set of constraints, and constraints only from b_j’s and the ever-present… Hmm, but the early terms before entering S break the pattern.

Perhaps the intended reading: prove there exist T and L such that the relation holds for all sufficiently large n — but the problem explicitly says every positive integer n. Since it’s a “prove” problem from a (fictional/future) IMO, I’ll trust the statement but I suspect the truth is: the relation holds for all n eventually, and one can then also… Hmm. Alternatively maybe the structure forces the sequence to be “in pattern” from the very beginning? No: a_1, a_2 arbitrary start: a_1 given only. Sequence determined by a_1. E.g., a_1 = 2: a_2 = 4, a_3=6, then a_4 must share factor with 2,4,6: 8. Everything even: a_n = 2n. Then T=1, L=2 works for all n. a_1=3: a_2 = 6, a_3 = 8? gcd(8,3)=1 no. a_3 = 9: gcd(9,6)=3 ok. a_3=9? Check smallest >6 with gcd>1 with 3 and 6: 7 no, 8 gcd(8,3)=1, 9 yes. a_4 >9 sharing with 3,6,9: 12. a_5: >12 share with 3,6,9,12: 15. 13,14: gcd(14,3)=1. 15 yes. a_6: 18 (16 fails vs 3, 17 no, 18). a_7: 21, 20 gcd(20,3)… 20 vs 9: gcd 1. 21: yes. So multiples of 3: 3,6,9,12,15,18,21,… a_n = 3n, T=1, L=3, holds all n. Hmm interesting. Maybe generally the “junk” early terms also eventually get absorbed: possibly the eventual periodic pattern retroactively matches? Not necessarily but with T chosen as multiple covering… For “every n” we need the whole sequence periodic-ish. Counterexample attempt: a_1 = 6? a_2 = 8, a_3=10, … all even: a_n even multiples? a_1=6, a_2: smallest>6 gcd>1 with 6: 8. a_3: >8 gcd>1 with 6 and 8: 10. Then even forever: a_n = 2n+4, T=1,L=2 holds all n. a_1=15: a_2: 18 (16 gcd1 with15,17), a_3: 20 (gcd(20,15)=5, gcd(20,18)=2) yes 20? check 19 no. gcd(20,15)=5>1, gcd(20,18)=2>1. a_3=20. a_4: >20, gcd>1 with 15,18,20: 21? gcd(21,20)=1. 22? gcd(22,15)=1. 24: gcd(24,15)=3,24&18=6,24&20=4. a_4=24. a_5: 25? vs18 gcd1. 26 vs15 gcd 1. 27 vs 20 gcd 1. 28 vs15 gcd1? gcd(28,15)=1. 30: gcd with all >1. a_5=30. a_6: 32? vs15 no. 33 vs20 no. 34 vs15 no. 35 vs18 no. 36: gcd(36,15)=3, 36&20=4. a_6=36. a_7: 38 vs15 no; 39 vs20 no; 40: gcd(40,15)=5, 40&18=2. a_7=40. Hmm terms: 15,18,20,24,30,36,40,… a_8: >40 gcd>1 with {15,18,20,24,30,36,40}: 42: gcd(42,15)=3,42&20=2,42&40=2. yes 41 no. a_8=42. a_9: 44 vs15 no; 45 vs 20: gcd5! 45&18=9, 45&24=3, 45&40=5, 45&42=3, 45&36=9, 45&15=15. So 45? Check 43,44: 44 vs 15 gcd 1. So a_9=45. a_10: 46 vs15 no,47,48: 48&15=3, 48&45=3, all even ok, 48&25? fine. a_10=48. a_11: 50: 50&15=5, 50&18=2, 50&21? terms:15,18,20,24,30,36,40,42,45,48. 50 vs each: 45:5, 42:2 => yes 49? 49 vs 15 gcd1. a_11=50. a_12: 51 vs 20 gcd1; 52 vs 15 gcd1? gcd(52,15)=1. 54: 54&15=3, &20=2, &25… &40=2,&45=9,&50=2 ok. 53 prime. a_12=54. Pattern: numbers divisible by 2 or… membership: need gcd>1 with 15 (i.e., div by 3 or 5) AND gcd>1 with 20 (div by 2 or 5) AND with all others… Terms so far all divisible by (3 or 5) and (2 or 5): 45 is 9·5. Eventually S = {m: (3|m or 5|m) and (2|m or 5|m) and …}. Every term ≥18 even or div by 5… condition with 45: div by 3 or 5. With each even term: any even m fine. So S candidates: m such that for every term a_i: gcd>1. If m even and divisible by 15 always works. The set stabilizes: constraints only from terms whose prime sets are “new”. Anyway.

So the general proof:

Step 1: There is a prime p and index N such that p | a_n for all n ≥ N. Step 2: Let B = {a_i, i< N with p ∤ a_i} (finite). Let M = p·rad(∏ a_i over i<N). Define S = { m > 1 : p|m and gcd(m, a_i) >1 for all i < N }. Then S is a union of residue classes mod M. For n ≥ N, a_{n+1} = min element of S greater than a_n? Need: for n ≥ N, condition “gcd(a_{n+1}, a_i)>1 for i ≤ n” equivalent to a_{n+1} ∈ S and >a_n — because for N ≤ i ≤ n, p | a_i and p | a_{n+1} handles those. And a_{n+1} must be div by p? No wait — a_{n+1} need not be divisible by p; S defined via p|m is sufficient but the actual a_{n+1} is the min over all m with gcd condition, which might not be div by p. But step 1 says p | a_n for all n ≥ N anyway. Then indeed for n ≥ N: a_{n+1} = min{ m > a_n : gcd(m, a_i) > 1 ∀ i ≤ n }. Since p | a_{n+1} (by Step 1) and the conditions for i in [N, n] are then automatic, and conversely… Hmm careful: a_{n+1} is defined as min over all m satisfying gcd condition, and we know (from step 1) this min happens to be divisible by p. Define S’ = { m : gcd(m,a_i)>1 ∀ i < N and (p | m or gcd(m,a_i)>1 for i in [N,n])}. Messy. Better: define S = { m : gcd(m, a_i) > 1 for all i ≥ 1 } — depends on all terms, infinite. But if p | all a_n, n ≥ N, then S = { m : gcd(m,a_i)>1 for i < N, and gcd(m, a_i)>1 for i≥N }. For m divisible by p, second part automatic. For m not divisible by p, hard. Instead, we don’t need S to capture all; we need: a_{n+1} = min element > a_n of some fixed periodic set, for n large.

Alternative cleaner: Step 1 stronger: show that eventually the set of “active constraints” stabilizes. Define R_n = rad(a_n) (squarefree kernel? set of primes). Constraint from a_i is: m must be divisible by some prime in P_i = primes(a_i). A constraint P_i is redundant if some earlier/other P_j ⊆ P_i. Claim: only finitely many distinct minimal constraints appear, i.e., the family of minimal sets among {P_i} stabilizes. Since each P_i is a finite set of primes, and… minimal sets could keep shrinking? A new term with P_{n+1} ⊊ nothing… e.g. new prime singleton {q}: then all future terms divisible by q. Hmm.

Standard solution outline (this is IMO 2025 Problem 3? No… Actually I recall this: it’s IMO 2025 Problem 3? IMO 2025 P3 was “bonza functions”. Hmm, this could be from ISL 2024 N?). Let me just construct the proof myself.

Let P_n = set of primes dividing a_n. The condition: P_{n+1} ∩ P_i ≠ ∅ ∀ i ≤ n.

Key Lemma A: The sequence has upper density type bound: a_{n+1} ≤ a_n + C eventually? Since if p | a_i for some fixed small prime… Let’s find a prime p dividing infinitely many a_n. Some prime among P_1 must divide infinitely many a_n? Each a_n (n≥2) shares a prime with a_1, and P_1 finite, so yes: ∃ p ∈ P_1 dividing a_n for infinitely many n. Fix such p (could take p smallest such? maybe choose p to be minimal prime with p | a_n infinitely often… good idea: let p be the SMALLEST prime dividing infinitely many terms).

Claim 1: liminf (a_{n+1} - a_n) ≤ … Given infinitely many terms divisible by p: whenever a_n ≡ ? Hmm. Note that any multiple of p·(product of needed primes) works as candidate. Let’s think growth: define Q = p · ∏_{q ∈ P_1} q … Candidate m = multiple of a_1 · p? gcd(m,a_i)>1 needs primes hitting each a_i, unknown.

Better approach via “eventually all terms divisible by p”:

Claim: Let p be a prime dividing infinitely many terms. Suppose infinitely many terms are NOT divisible by p… derive structure. Hmm.

Known solution idea (I recall similar problem: “Choose a_{n+1} smallest > a_n with gcd(a_{n+1}, a_n·…)” ) Let me think concretely.

Since every a_{n+1} shares prime with a_1, primes dividing terms from P_1 partition indices; some p ∈ P_1 divides infinitely many terms. Among all primes dividing infinitely many terms, let p be the smallest.

Main Lemma: only finitely many terms are not divisible by p? Suppose infinitely many terms not divisible by p. Each such term a_n shares a prime with every a_i div by p… hmm need more leverage: counting/density argument. Terms divisible by p that are > x: at least roughly? Consider gaps: if a_n divisible by p and also by all needed primes…

Density argument: The number m = p·k for suitable k… Let’s bound a_n linearly. Let q_1 < q_2 < … be primes dividing infinitely many terms; actually let’s define D = set of primes dividing infinitely many terms; finite? Each prime in D… Could be infinite (e.g., all even numbers: every prime divides infinitely many terms? a_n = 2n: yes D is all primes, infinite). Hmm, but p = 2 smallest.

Let me recall: this is ISL 2024 N? Or it is actually “IMO 2026 P6” per the user (future). I need to construct full rigorous proof. Let me do it carefully.

Setup: p := smallest prime dividing infinitely many terms. Let F = set of primes q < p. Each q ∈ F divides only finitely many terms. So ∃ N1 such that for n ≥ N1, a_n has no prime factor < p. Also infinitely many n with p | a_n.

Step A: Show a_{n+1} - a_n is bounded. For n with p|a_n? Consider the number m = next multiple of p·r where r = product of one prime from each “hard” constraint… constraints from a_i with p ∤ a_i. Problem: number of such constraints could grow.

Step B (key): Show that only finitely many terms are coprime… i.e., ∃ N2: p | a_n ∀ n ≥ N2.

Proof attempt: Suppose infinitely many terms not divisible by p. All terms ≥ N1 have least prime factor ≥ p. A term a not divisible by p has least prime factor ≥ p’ = next prime after p… at least > p. Hmm use counting: terms divisible by p are ≥ p apart… all terms are increasing; consider window [x, 2x]. Terms in there: each pair of terms shares… no, only later shares with earlier — but actually the condition is symmetric-ish: for i < j, gcd(a_j, a_i) > 1. So ALL pairs of terms share a common prime! That’s a strong pairwise condition. Nice: the whole sequence is pairwise non-coprime.

Now: Suppose b = a_m not divisible by p, with m ≥ N1 (so all prime factors of b are > p, and finitely many of them, say set Q_b). Every term divisible by… every term shares a prime with b. Infinitely many terms divisible by p; they also must share prime with b, i.e., divisible by p·q for some q ∈ Q_b > p. Fine, no contradiction yet.

Growth/greedy argument: Show a_{n+1} - a_n bounded: Let W = product of p and one prime chosen from each a_i among the finitely many terms with small prime factors… but constraints from terms not divisible by p and appearing arbitrarily late are the issue. However! Any multiple of rad-friendly number: choose m divisible by p and by every prime factor of b for each such b? If infinitely many b’s with disjoint prime sets… but pairwise gcd>1: all the b’s pairwise share primes too.

Let’s consider the set B = { n ≥ N1 : p ∤ a_n }. If B infinite: terms {a_n}_{n∈B} pairwise share primes, all their prime factors > p.

Density counting: In [1, x], number of terms of sequence ≥ ? and ≤ ? Each a_{n+1} ≤ smallest number > a_n divisible by ALL primes dividing a_1…a_n? Too big. Need a smarter candidate: m ≡ 0 mod p, and m ≡ … Let me get a uniform candidate: Let v = p·∏{i < N} (least prime factor of a_i)?? For terms with index ≥ N… if all terms with n ≥ N were divisible by p, then candidate any multiple of p·rad(a_1···a{N-1}) works: gap bounded. So Step B is the crux; after Step B, gaps bounded and periodicity follows.

Proving Step B: Suppose for contradiction p ∤ a_n for infinitely many n. Let p be smallest prime dividing infinitely many terms; also every prime < p divides finitely many. Consider n ∈ B large. Since a_n not div by p and shares factor with all previous, and a_{n} is minimal… Use minimality: a_{n} < any candidate. Candidate for a_n at step n-1: the smallest multiple of P := p·∏(primes covering the finitely many terms not divisible by p among a_1..a_{n-1})? But that set may grow (that’s what we’re contradicting). Hmm circular. Let’s think again.

Alternative known approach: consider the smallest prime p dividing infinitely many a_n. Show: eventually, consecutive multiples-of-p appear, and any term not divisible by p must be < next multiple of p that satisfies constraints, forcing it to be divisible by a smaller-prime pattern… Let me attempt:

Let n ≥ N1 (no prime factors < p). Claim: a_{n+1} < a_n + Cp for constant C? Suppose at stage n, let m* = least multiple of p exceeding a_n that satisfies gcd(m, a_i) > 1 for all i ≤ n. Then a_{n+1} ≤ m. Bound m: constraints from a_i divisible by p auto-satisfied. Others: a_i with p∤a_i, i ≤ n — call them exceptional set E_n, may be growing. To satisfy them need m divisible by primes hitting each. If E_n grows unboundedly, no bound. Hmm.

Second idea: pairwise intersecting family of prime sets. The family {P_n} is pairwise intersecting. Helly-type/sunflower: For pairwise intersecting family of finite sets… If infinitely many terms with p ∤, their prime sets pairwise intersect and each intersects all P of p-divisible terms.

Maybe use minimality differently: For n ∈ B (a_n not divisible by p), compare a_n with alternative candidate m = a multiple of p… The minimality says every integer in (a_{n-1}, a_n) fails some constraint. In particular multiples of p in that interval fail constraints from exceptional terms. Since a_n itself not divisible by p, and p | infinitely many terms…

Let me think about what the answer structure is (recall actual solutions to this problem — this is in fact ISL 2024 N7? Hmm “gcd(a_{n+1}, a_i) > 1” greedy — I believe I’ve seen this exact problem: it’s ISL 2024 N4 or so, or maybe USAMO?). I think I recall a similar problem: “EGMO 2023”? Not sure. Construct proof independently.

Let me define things more carefully and try to prove: (Lemma 1) The set of primes dividing at least one term but only finitely many terms… hmm could be infinite (primes dividing a_n=2n each appear infinitely often actually every prime divides infinitely many multiples… a_n=2n: prime 3 divides a_n when 3|n: infinitely often. So D=all primes.)

Try: Lemma 1: gaps a_{n+1}-a_n are bounded. Proof: Let p ∈ P_1. Consider m = multiple of a_1·p? Hmm gcd(m, a_i) needs >1 for all i. Take m = k·∏_{i≤n} lpf? no.

Hmm, is there an a priori bound like a_{n+1} ≤ 2a_n or a_n ≤ C n? Candidate: m = product of a_n and something? m must be > a_n but candidates like lcm huge. But note a_{n+1} ≤ any valid candidate; valid candidate: 2·a_1·a_2? no need > a_n. Candidate: a_n + rad? Hmm: Let r_n = rad(a_1 a_2 ⋯ a_n)?? multiple of that > a_n works: a_{n+1} ≤ r_n·⌈(a_n+1)/r_n⌉ ≤ a_n + r_n. r_n grows. Weak.

Alternative: strong candidate: m = a_j·k for a_j an earlier term? gcd(a_j k, a_i) ≥ gcd(a_j, a_i) > 1 for all i ≤ n if j ≤ n! Yes!! Any multiple of any single term a_j is a valid candidate (satisfies all constraints i ≤ n as long as j ≤ n… wait need gcd(m, a_i) > 1 for i ≤ n; gcd(a_j·k, a_i) ≥ gcd(a_j, a_i) > 1 for i ≠ j… and for i = j it’s a_j > 1). So a_{n+1} ≤ smallest multiple of a_1 greater than a_n ≤ a_n + a_1. Gaps bounded by a_1! Excellent. So a_{n+1} - a_n ≤ a_1 for all n. More generally a_{n+1} - a_n ≤ min_j a_j = a_1.

Great, Lemma 1: a_{n+1} ≤ a_n + a_1. So density positive: in any interval of length a_1 there’s at least one term beyond a_1.

Now Lemma 2: smallest prime p dividing infinitely many terms; primes < p divide finitely many; choose N1 with: for n ≥ N1, lpf(a_n) ≥ p, and additionally all terms with a prime factor < p are among a_1..a_{N1}.

Lemma 3: p | a_n for all sufficiently large n. Proof attempt with bounded gaps: Suppose not: infinitely many n with p ∤ a_n. All prime factors of such a_n are > p (n ≥ N1). Hmm, want contradiction with minimality/greedy: Since p divides infinitely many terms, consider the term a_m, huge, p | a_m. Numbers of form m’ = a_m + tp? no.

Counting argument: terms have gaps ≤ a_1, so at least x/a_1 terms up to x (asymptotically). Terms not divisible by p have lpf > p; but that’s not restrictive density-wise… e.g., numbers with lpf > p have density ∏_{q ≤ p}(1−1/q) > 1/a_1 possibly. Hmm need different idea.

Idea: use pairwise intersecting + greedy minimality. Let n ∈ B (p ∤ a_n). Consider the largest multiple of p… Actually consider candidate at step n−1: m = smallest multiple of p·s exceeding a_{n−1}, where s = some fixed product handling exceptional constraints… exceptional terms are exactly B-terms (plus early ones) — circular again but maybe induction: assume B infinite; the B-terms pairwise share primes, all > p.

Let q = smallest prime dividing infinitely many B-terms (exists: P of first B-term finite, later B-terms each share prime with it… wait B-terms pairwise intersect, so some prime of the first B-term divides infinitely many B-terms). q > p. Hmm, then consider terms divisible by q: infinitely many. Also p divides infinitely many.

Maybe the right claim: Let p smallest prime dividing infinitely many terms. Claim: all but finitely many terms divisible by p. Suppose not. Let q = smallest prime dividing infinitely many of the non-p terms; q > p.

Now consider large x and interval I = [x, x + K] with K = a_1·(something). Terms in I: ≥ K/a_1. Terms are: divisible by p, or divisible by q (or in finite junk), or divisible by (other primes…). Hmm not immediate.

Alternative approach for Lemma 3 via minimality: Take n huge with p ∤ a_n. Let P’ = product of p and q? Try to show a “multiple of pq or of p·(primes)” beats a_n. Since a_n was chosen minimal, every multiple of p in (a_{n−1}, a_n) fails some constraint: fails only constraints from terms coprime to it: i.e., ∃ earlier B-term (or early junk) b with gcd(mult of p, b) = 1.

Let’s set J = the finite set of terms with a prime factor ≤ p??? Hmm.

Perhaps stronger claim: p·q’ | for suitable… Let me look at it via the structure “eventual behavior is: S = set of m with gcd(m, a_i) > 1 for all i in some finite ‘core’”.

Alternative plan avoiding Lemma 3: Let’s define C = set of m > 1 such that gcd(m, a_i) > 1 for ALL i ≥ 1 (compatible with entire sequence). Claim (i): all but finitely many terms belong to C. Claim (ii): C is eventually periodic (union of residue classes modulo some M) — need C determined by finitely many constraints. Constraint from a_i: divisible by some prime in P_i. C = ∩_i {m : P(m) ∩ P_i ≠ ∅}. If the family {P_i} has infinitely many minimal elements this needn’t be periodic. But maybe Claim: finitely many minimal constraints. Since a_1 has finitely many primes, constraints P_i ⊇ … hmm no, minimal constraints under ⊇: P_i is “stronger” if subset. Infinitely many pairwise-incomparable P_i possible in principle (e.g., {2,3},{2,5},{3,5}… pairwise intersecting, incomparable, infinite family exists: all P_i = pairs from {2,3,5}: only 3 distinct though. Infinite antichain of pairwise intersecting finite sets: {2, r} for primes r — pairwise intersecting (share 2), infinite antichain. Then C = {m : 2·r… m div by 2 or by r, for all r} = multiples of 2 union… m must hit every {2,r}: either 2|m, or r|m for all r (impossible) ⇒ C = even numbers ∪ finite… = even numbers essentially: still periodic! In general, C = {m: hits all P_i}. If m ∈ C and m not divisible by any prime that divides infinitely many… Let’s define: since gaps bounded, terms up to x ≥ x/a_1.

Claim (ii)‘: C ∩ [large] = C’ ∩ [large] where C’ = {m : m hits all P_i, i ≤ N} ∩ {m : divisible by some prime in D… } hmm. Let’s directly prove: ∃ finite N such that for m sufficiently large, m ∈ C ⟺ gcd(m, a_i) > 1 ∀ i ≤ N. Direction ⇐ needs: hitting first N constraints ⇒ hitting all. Not obviously.

I think the cleanest known route is indeed: prove eventually all terms divisible by a single prime p. Let me try to prove Lemma 3 properly.

Setup recap: gaps ≤ a_1 =: g. p = smallest prime dividing infinitely many terms. Every prime < p divides only terms in finite initial junk; let N1 with lpf(a_n) ≥ p for n ≥ N1. Suppose B = {n ≥ N1 : p ∤ a_n} infinite.

Note: for n ∈ B, lpf(a_n) > p, so a_n has all prime factors ≥ p⁺ (next prime).

Key: consider a specific term a_n, n ∈ B, n huge. At the step choosing a_n, the candidate m0 = smallest multiple of L exceeding a_{n−1}, where L := p·rad(∏{i ∈ E} a_i) with E = set of indices i < n with p ∤ a_i. Then m0 satisfies all constraints and m0 ≤ a{n−1} + L. Minimality: a_n ≤ m0. Not obviously contradiction.

Try counting in interval of length L·something: Within any interval of length L there is a valid candidate (multiple of L). So between consecutive B-terms? Hmm.

Different angle: think about why non-p terms must eventually stop: A non-p term a_n must have gcd > 1 with all previous, using only primes > p. Future terms must all hit P_n too. The “cost”: multiples of p that also hit P_n are sparser. Greedy prefers smaller numbers. Intuition: a non-p term is chosen only if it beats every multiple of p candidate, meaning a_n < next “good” multiple of p. Good multiples of p (hitting all exceptional constraints) have density; each new B-term makes good multiples sparser for the future, but B-term itself must be a number with lpf ≥ q ≥ p⁺ hitting all previous exceptional constraints…

Density comparison: Let E = current exceptional prime-set family (from B-terms so far + junk). A future B-term must hit all constraints using primes > p (its own primes > p). A future p-term must be div by p and hit exceptional constraints (using any primes). The B-term must hit every P_i including p-terms’ P_i! p-terms a_i, i ≥ N1: their prime sets contain p and maybe others; if some p-term a_i = p^k exactly (prime power), then every later term must be divisible by p!! So: if any term is a power of p, then all subsequent terms divisible by p, done-ish. Similarly, the real obstruction: p-terms typically have other primes so B-terms hit those.

Hmm, so aim: show some term is (eventually forced) ≡ power of p? Unlikely direct. But maybe: show some term a_i has P_i = {p} ∪ (primes each > everything relevant)…

Alternative known solution sketch (I now genuinely recall this problem is from ISL 2024 (N6?) “gcd greedy sequence periodicity” — the solution I vaguely recall uses exactly: 1) gaps bounded by a_1; 2) let p smallest prime dividing infinitely many terms; show all large terms divisible by p by a clever argument; 3) then the exceptional constraints are finite, S periodic, done.) Let me work out step 2’s clever argument.

Suppose infinitely many B-terms. Consider consecutive terms a_n (p-term) around a B-term. Since gaps ≤ g = a_1: in interval (a_{n}, a_n + g] there’s a term. Consider t large, and the multiple of p… Let’s think mod p: terms not divisible by p occupy residues mod p; there are infinitely many p-terms too.

Try: q := lpf of some fixed B-term b* (q > p, b* large so no small primes). Every subsequent term hits P(b*)… primes all > p.

Hmm think about candidate construction: Let n−1 ≥ some index; define m = smallest number > a_{n−1} of form p·(multiple of R) where R = rad(prod of exceptional terms up to n−1). a_n ≤ a_{n−1} + pR. If a_n is a B-term: a_n has lpf > p AND hits every previous constraint AND a_n ≤ every multiple-of-p valid candidate in the interval.

Sharper: every valid candidate in (a_{n−1}, a_n) — there are none by minimality. So all multiples of p in (a_{n−1}, a_n) are invalid: each fails some exceptional constraint (constraints from p-terms are auto-OK for multiples of p? gcd(mult of p, p-term) ≥ p, yes OK). So each multiple of p in the gap misses some exceptional prime-set. Meanwhile a_n itself, coprime to p, hits ALL constraints, including all p-terms’ prime sets. There are ~a_n/p… within gap ≤ g, at most g/p multiples of p — few. Not a contradiction locally. Need global/asymptotic argument.

Global density: Let’s count terms in [1, x]: ≥ x/g − C. Every term ≥ N1 has lpf ≥ p. Numbers with lpf ≥ p have density δ_p = ∏_{q<p} (1 − 1/q). So need x/g ≤ δ_p x roughly: gives 1/g ≤ δ_p: constraint but not contradiction.

Let’s use exceptional constraints to shrink density: If B infinite, then B-terms force ALL later terms (both kinds) to be divisible by primes > p hitting each B-term. Take k B-terms b_1,…,b_k with prime sets Q_1,…,Q_k (all primes > p; pairwise intersecting). All sufficiently late terms lie in T = {m : lpf(m) ≥ p, m hits every Q_j}. Density of T: m must, for each j, be divisible by some prime in Q_j (primes > p, so each condition has density ≤ ∑{r ∈ Q_j} 1/r ≤ ω(b_j)/p⁺… but ω(b_j) can be large ~ log b_j / log p. Hmm ∑ 1/r over r | b_j, r > p: at most (1/p)·log_p(b_j)… roughly (log b_j)/(p log p). Not < 1/g necessarily since b_j large ⇒ many primes possible. But wait: can also bound: number of prime factors of b_j is ≤ log b_j / log p, and each ≥ p, so ∑ 1/r ≤ ?? primes distinct so ∑{r|b_j, r>p} 1/r ≤ ∑{i≥0} 1/(p+i-th prime) hmm bounded by H = ∑ over primes r>p up to … For the density of “hits Q_j” ≤ ∑{r∈Q_j} 1/r. If Q_j ⊆ primes > p, worst case Q_j = all primes in (p, y]: sum ≈ log(log y/ log p) — can exceed any bound?? No wait density of hitting Q_j is ≤ min(1, ∑ 1/r) — could be close to 1. Hmm, so single constraints don’t shrink density enough. But we get to intersect MANY constraints. Do densities multiply down to < 1/g? If Q_j’s share primes (pairwise intersecting), independence fails. E.g., all Q_j ∋ q: then all constraints satisfied by multiples of q: density 1/q ≥ … fine that’s still allowed: then q divides all late terms?? Not quite—hitting Q_j doesn’t require q.

Hmm wait, if that happens (some prime q > p in ⋂ Q_j for infinitely many B-terms) it’s consistent; but then consider minimality between p and q: p divides infinitely many, q > p…

Let me think about small cases to understand why non-p terms die out. Suppose p = 2 divides infinitely many terms, and suppose odd terms appear infinitely often. Odd terms pairwise share odd primes, and share odd primes with… no wait odd term must share factor with EVERY previous term including even ones: shares odd prime with each even term. And even terms must share factor with all previous odd terms: even term divisible by odd primes hitting each odd term. Now greedy: even candidates are dense (every other number), but must hit odd-term constraints. Odd terms: sparse-ish. Why can’t we have all terms divisible by 3, alternating 2·3k and 3·odd? E.g., all terms multiples of 3: then constraints all satisfied among themselves plus junk. Sequence would be: a_{n+1} = next multiple of 3 (if all constraints just “div by 3”)… then greedy picks a_n+… next multiple of 3 always valid, but is it minimal? Suppose current terms all multiples of 3, junk none, a_1 = 3: we computed a_n = 3n? Wait earlier with a_1=3 we got 3,6,9,12,…: but check a_2: smallest >3 with gcd>1 with 3: 6? What about 4? gcd(4,3)=1. 5 no. 6 yes. a_3: 8? gcd(8,3)=1… but gcd(8,6)=2, fails with a_1=3. 9. a_4: 10: gcd(10,3)=1. 12. Yes all multiples of 3, even though 2 divides infinitely many terms (6,12,18,…). Smallest prime dividing infinitely many terms is 2! And infinitely many terms NOT divisible by 2 (9, 15, 21,…). Uh-oh: Lemma 3 as stated is FALSE. In example a_n = 3n: 2 divides a_n for even n — infinitely many; 2 ∤ a_n infinitely often. So “smallest prime dividing infinitely many” isn’t the right p. Instead p = 3 divides ALL. So correct target: ∃ prime p dividing all sufficiently large terms — in this example p=3. Need to choose p correctly, not merely smallest with infinite multiplicity.

So Lemma 3’: ∃ prime p and N: p | a_n ∀ n ≥ N. Hmm is that even true in general?? Could the eventual set S be like {m : m ≡ hits {2,3}} i.e., multiples of 2 or 3, with no single common prime? Suppose constraints stabilize to: every term divisible by 2 or 3, and the greedy walks through numbers divisible by 2 or 3: but pairwise gcd condition: a term ≡ 2 mod 6 (div by 2 not 3) and a term ≡ 3 mod 6 (div by 3 not 2) have gcd… 2·odd and 3·odd: gcd could be 1! e.g. 4 and 9. So pairwise intersecting must hold; the family of ALL terms pairwise share primes. Can we have a pairwise-intersecting infinite family with no common prime and periodic structure? E.g., all terms divisible by at least two of {2,3,5}: pairwise any two terms share at least one of 2,3,5 (two 2-subsets of a 3-set intersect). Sets like: multiples of 6, 10, 15. S = {m : m div by 6, 10, or 15}. Is such a steady state reachable and stable? Greedy on S: a_{n+1} = next element of S? Check: next element m of S after a_n: does m satisfy gcd>1 with all previous (which are in S plus junk)? Two elements of S: 6a and 10b: gcd ≥ 2. 6a,15b: ≥3. 10a,15b: ≥5. Yes pairwise fine! So S is “self-compatible”: any two elements of S have gcd > 1. So an eventual regime with no universal prime is conceivable! But is a_{n+1} = next element of S? Need: no smaller number m’ ∉ S with gcd>1 with all previous terms. m’ must hit every previous term. Possibly m’ with primes {2,3} only div by… m’=2·3·k? that’s in S (div by 6). m’ = 2·7·… must share with terms ≡ 15·odd (div by 3,5 patterns): a previous term = 15 exactly ⇒ m’ div by 3 or 5. And previous term = 10: div 2 or 5; term 6: 2 or 3. m’ hits all: e.g. m’ div by 2 and 3 ⇒ in S. Or div by 3 and 5 ⇒ in S. Or 2,5 ⇒ S. Or… could hit using different primes per term? If terms include 6,10,15 themselves then any m’ must be div by (2∨3)∧(2∨5)∧(3∨5) ⇒ at least two of {2,3,5} ⇒ m’ ∈ S (if also every other previous term is hit… other previous terms are in S, m’ has two of 2,3,5, S elements have two of 2,3,5, two 2-subsets of {2,3,5} intersect ⇒ gcd>1 automatically!). So actually S = {m : at least two of 2,3,5 divide m} is a stable orbit: if 6, 10, 15 all appear as terms, subsequent terms = successive elements of S. E.g., start a_1 = 6? Then a_2 = 8 (gcd 2), even forever: a_n even: 6,8,10,12,… all even. Hmm because greedy prefers density-1/2 evens. To get S(two of {2,3,5}) as the orbit we’d need starting terms preventing evens: a_1 = 15: computed earlier: 15,18,20,24,30,36,40,42,45,48,50,54: div by: 15:{3,5};18:{2,3};20:{2,5};24:{2,3};30;36;40:{2,5};42:{2,3,7};45:{3,5};48;50:{2,5};54:{2,3}. Indeed all have ≥ two of {2,3,5}? 15 ✓, 18 ✓(2,3), 20 ✓(2,5), 24 ✓, 30 ✓, 36 ✓, 40 ✓, 42 ✓, 45 ✓(3,5), 48 ✓, 50 ✓, 54 ✓. And S elements between: after 15: S-elements: 18? 16 has only {2}, but wait 16 needs gcd with 15 only: fails. 18 ✓. After 18: 20 ✓ (need two of? m must hit 15 and 18: (3∨5)∧(2∨3): 20: 5,2 ✓). Candidates 19 no. After 20: 21: hits 15(3),18(3),20? gcd(21,20)=1 ✗. 22: 15? ✗. 24 ✓. Looks like the orbit IS S = numbers div by at least two of 2,3,5 — let me verify 45→48: 46: vs15 ✗; 47; 48 ✓. And 50: 49 vs 18 ✗. 50 ✓. 54: 51=3·17: vs 20 ✗; 52: vs 15? gcd=1 ✗; 53; 54 ✓. Next: 56? {2,7}: vs 15 ✗. 57? vs 20 ✗. 60 ✓. Also 55={5,11}: vs 18 ✗. So 60. S in [1,60]: multiples of 6,10,15: 6,10,12,15,18,20,24,30,36,40,42,45,48,50,54,60. From 18 on the terms match S exactly! So indeed no single prime divides all large terms in this example. Great—so Lemma 3’ false too. The truth: eventual orbit = S where S is defined by finitely many constraints and is “self-compatible” (pairwise gcd > 1 within S). Period M = 30 here, T = |S ∩ (0,30]| = 8: terms from 18: 18,20,24,30,36,40,42,45,48,… a_{n+8} = a_n + 30 for n ≥ 3 (15 = 45 − 30 and 15 is a term (a_1), 18 = 48−30 ✓, 20 ✓… actually even n≥1: 15,18,20,24,30,36,40,42 | 45,48,50,54,60,66,70,72? Let’s not verify; also the problem claims for EVERY n, interesting — in this example it seems to hold from n=1! Because a_1 = 15 ∈ S and the orbit was S-greedy from the start? But a_2 = 18 = next S element after 15? S after 15: 18 ✓. So yes whole sequence = S-walk. Coincidence? The problem statement demands existence of T, L valid for all n ≥ 1. Hmm, but consider junk early terms not in final S: e.g. a_1 = 2: orbit = evens from start: fine. Try to design a_1 so early terms ∉ eventual S: a_1 = 21 = 3·7: a_2: 24 (22 vs21 ✗, 23), a_3: 27? gcd(27,24)=3 ✓ gcd 21=3 ✓; 25? vs21 ✗ vs24 ✗; 26 vs21 ✗; 27 ✓. a_4: 28? vs27 gcd 1 ✗. 30 ✓ (28 ✗,29). a_5: 33 (31,32 vs21 ✗… 32 vs 27 ✗ anyway; 33: vs21=3,24=3,27=3,30=3 ✓). a_6: 36 ✓ (34 vs21✗,35 vs 24 ✗? gcd(35,24)=1 ✗). a_7: 39? vs 28? not a term. terms: 21,24,27,30,33,36. 38 vs21 ✗ 37; 39: gcd with all: 3|39 and all terms div by 3! ✓. Interesting: all terms so far divisible by 3 ⇒ sequence = multiples of 3 forever? a_8: 42, then 45, 48, … any next multiple of 3 works and is minimal? Need check numbers between: e.g., after 39: 40: vs 21 gcd 1 ✗. 41. 42 ✓. After 42: 43,44 (vs21: gcd(44,21)=1 ✗), 45. So a_n = 3n + 18 for n ≥ 1? a_1=21=3·7 ✓ T=1, L=3 works for all n ≥1 again! Because 21 ≡ 0 mod 3. Hmm.

When could early terms break the pattern? Eventual orbit S with period M, and early terms a_1..a_k before entering S-walk. For “every n” we need a_{n+T} = a_n + L also for small n, i.e., a_n + L is a term for every n, with fixed lag T. If T is the eventual period count and L = M·(T/T0)… For small n we need a_n + L to also be a term exactly T positions later. If all early terms happen to lie in S and be consecutive S-elements… In general?? Suppose a_1 = odd prime power like 25: a_1=25: a_2: 30 (26,27,28 vs25 ✗… gcd(30,25)=5 ✓; 30 is first >25 with gcd(·,25)>1: multiples of 5: 30 ✓). a_3: 32? vs25 ✗. 34 ✗ vs 25. 35: vs30=5 ✓ vs25 ✓ ⇒ 35. a_4: 36? vs25,35 ✗. 40: vs25=5,30,35=5 ✓; 38 vs25✗; 39 vs 25 ✗? gcd(39,25)=1 ✗; 40 ✓. a_5: 42? vs 25 ✗. 44 ✗. 45: ✓(5 with 25, 30… 45&40=5, 45&35=5, 45&42? not term) ⇒ 45. Everything div by 5: a_n = 5n + 20, all n: a_1 = 25 ✓. T=1,L=5 all n. Because a_1 = 25 ≡ 0 mod 5 and orbit = multiples of 5 = 5ℕ from 25 onward, and 25 itself on the lattice. Seems robust: the orbit S always contains a_1?? Since every element of orbit must have gcd > 1 with a_1 anyway, and a_1 ∈ … S is defined by constraints from all terms; a_1 satisfies gcd(a_1, a_i) > 1 for all i (pairwise property). So EVERY term is compatible with all constraints, i.e., all terms lie in the “compatible set” C = {m : gcd(m, a_i) > 1 ∀i}. And the eventual orbit is exactly: a_{n+1} = next element of C after a_n?? Hmm is that true from the start? a_{n+1} = smallest m > a_n with gcd(m, a_i)>1 for i ≤ n. Since constraints only accumulate, a_{n+1} ≥ next element of C_n ⊇ C. All terms ∈ C. If a_{n+1} > (next element of C after a_n) =: c, then c fails some constraint i ≤ n? c ∈ C means c compatible with all a_i including future ones. So c satisfies constraints i ≤ n too! And c > a_n, c < a_{n+1} contradicts minimality unless c = a_{n+1}. Wait: a_{n+1} is minimal m > a_n satisfying constraints i ≤ n; c satisfies those and c > a_n so a_{n+1} ≤ c. And a_{n+1} ∈ C? a_{n+1} is a term, and all terms are in C (since gcd(a_j, a_i) > 1 for ALL i,j: for i < j by construction; for i > j by construction applied at step i… gcd(a_i, a_j) > 1 when j < i by definition; symmetric so all pairs ✓; also gcd(a_j,a_j)>1). YES: every term lies in C, where C = {m > 1 : gcd(m, a_i) > 1 ∀ i ≥ 1}. And a_{n+1} ∈ C, a_{n+1} > a_n, so next element of C after a_n satisfies… c ≤ a_{n+1} and c ≥ a_{n+1} by minimality (c is > a_n, satisfies constraints 1..n, so a_{n+1} ≤ c; a_{n+1} ∈ C and > a_n so c ≤ a_{n+1}). Hence a_{n+1} = min{c ∈ C : c > a_n} FOR ALL n ≥ 1!!! Beautiful. So the whole sequence from a_1 is precisely successive elements of C greater than a_1… wait need also every element of C in (a_n, a_{n+1}) — none, shown. But also elements of C below a_1 or between… irrelevant. Also is a_1 itself relevant: sequence = all elements of C that are ≥ a_1?? Not exactly: sequence = a_1 followed by successive C-elements; and a_1 ∈ C. So sequence = enumeration of C ∩ [a_1, ∞) in increasing order.

So the whole problem reduces to: prove C is eventually periodic (union of arithmetic progressions mod some M, up to finite discrepancies)? Wait but for the FULL claim (all n) we need C ∩ [a_1, ∞) to be EXACTLY periodic from a_1 on? Hmm, no: we need ∃T, L: the (n+T)-th element of C≥a_1 equals n-th element + L for all n ≥ 1. If C is eventually periodic mod M (i.e., ∃X: for m ≥ X, m ∈ C ⟺ m + M ∈ C), then for elements beyond X it’s a shift, but early elements (between a_1 and X) need not satisfy. Hmm! But we can choose L = M·k and T = number of C-elements in [a_n, a_n + Mk)… which is constant only in periodic region. So we DO need exact periodicity on [a_1, ∞): for all m ≥ a_1: m ∈ C ⟺ m + M ∈ C. Is that plausible? Direction ⇒: m ∈ C, show m + M ∈ C: if M divisible by rad(stuff)… If M = rad(∏ over a “core” set of terms whose constraints imply all others)·? Let’s think: Claim: if m ∈ C then m + M ∈ C where M := any common multiple of “relevant primes”… Take M with property: for every i, and every m ∈ C… we need gcd(m + M, a_i) > 1 ∀i. Hmm if p | m and p | a_i and p | M then p | m + M ✓. So if we choose M divisible by every prime that divides both some element of C and some term… those primes could be infinitely many (e.g., C ⊇ evens, prime 7 divides some even element and some term). But we only need SOME prime witnessing each constraint to survive. Choose M = product of all primes ≤ W for suitable W? Let’s define: Claim A: ∃ finite set of primes 𝒫 and finite set of indices I such that for all sufficiently large… hmm but we need from a_1, not sufficiently large.

Let’s reconsider: define for each i, constraint K_i = P_i (primes of a_i). C = {m: ∀i, P(m) ∩ P_i ≠ ∅}. Claim B: there are finitely many “essential” constraints: ∃ finite I s.t. ∀ m ≥ 2? (all m, not just large): [∀ i ∈ I: gcd(m, a_i) > 1] ⇒ m ∈ C. I.e., the intersection of hitting-conditions stabilizes. Since constraints are “upward monotone in prime set”… The family {P_i} ordered by ⊆: constraint from a_i implied by constraints in I if every set hitting all {P_j}_{j∈I} hits P_i. Sufficient: ∃ j ∈ I: P_j ⊆ P_i. Is it true that every P_i contains some minimal P_j and there are finitely many minimal ones? By Dickson-type lemma? Infinite antichains of finite subsets of primes exist in general ({2,r}: antichain, infinite). Could the sequence’s prime sets contain such an antichain {P_i} = {2, r_i}? Terms 2^{a}·r_i{b}. Pairwise ok (share 2). But hitting condition: every term must hit every {2, r_i}: satisfied since all div by 2. C would = even numbers meeting other constraints. But is greedy consistent with infinitely many terms of form 2^a r^b with r large and NOT… they’re even, fine. Then constraint {2, r_i} is implied by constraint {2, r_1}? No: m hitting {2,r_1} could be r_1·odd, then must also hit {2, r_i}: r_1·odd·… not implied. So C = {m : (2|m ∨ r_1|m) ∧ (2|m ∨ r_2|m) ∧ …} = evens ∪ {odd multiples of all r_i} = evens (if infinitely many r_i) — as an infinite intersection it’s periodic-ish only “in the limit”: m odd multiple of r_1 might satisfy first million constraints but ∉ C. But C itself = evens ∪ (finitely many exceptional odd… no: odd m ∈ C requires r_i | m ∀i infinite ⇒ impossible) ⇒ C = evens ∩ (other constraints…) — exactly periodic! Interesting: infinite antichain still gave periodic C. Because infinite conjunction of (2|m ∨ r_i|m) equals (2|m) exactly for m fixed… any fixed m has finitely many prime factors, so if m ∈ C and m not divisible by 2, then m hits each {2,r_i} via r_i ⇒ all r_i | m impossible. General principle: C = ∩ hitting conditions; is C always a periodic set (exactly, union of residue classes mod some M)? Hmm: C = {m : P(m) is a “transversal” hitting all P_i}. Let 𝒯 = collection of finite prime sets Q (minimal ones) such that Q hits all P_i. C = {m : P(m) ⊇ some Q ∈ 𝒯_min}. Minimal transversals: could be infinitely many? Each minimal transversal is a minimal hitting set of the family {P_i}. If infinitely many minimal transversals exist, C = union over Q of multiples of ∏Q — infinite union of APs, maybe not periodic but that’s fine?? We need periodic for the final count argument. Hmm, but actually maybe only finitely many minimal transversals can be relevant below any bound… For exact statement “∃T,L ∀n”, we need: exists M such that ∀ m ≥ a_1: m ∈ C ⟺ m + M ∈ C. Suppose minimal transversals Q_1, Q_2, … (each finite set of primes). m ∈ C iff ∃k: ∏Q_k | m. This is a union of APs with moduli d_k = ∏Q_k. If infinitely many minimal transversals with d_k → ∞, the set ∪ d_kℤ is typically NOT eventually periodic (e.g., squarefree… hmm, actually a union of infinitely many APs 0 mod d_k can fail periodicity). So we must show finitely many minimal transversals suffice/exist, USING the greedy structure (bounded gaps!). Recall gaps ≤ a_1 ⇒ C has bounded gaps ≥… C ⊇ terms, so C has gaps ≤ a_1 beyond a_1. Hmm C could still have infinitely many minimal transversals? Let’s use: Claim C: Let d = ∏ Q for the minimal transversal Q₀ minimizing the product? All multiples of d are in C? Yes: m div by all of Q₀ hits everything ⇒ dℤ ⊆ C (for m>1). So C ⊇ dℤ.

Now the key finiteness: Claim D: the number of essential constraints is finite. Essential constraint: P_i such that removing it changes C∩[a_1,∞)? Suppose infinitely many i with P_i not implied by others… Let’s use bounded gaps + minimality more cleverly. Recall sequence = increasing enumeration of C from a_1. So terms and C beyond a_1 coincide! C ∩ [a_1, ∞) = {a_n}. Wow, right: every term ∈ C and every element of C > a_n and ≤ a_{n+1} equals a_{n+1}. And elements of C in (a_1… ) so C ∩ [a_1, ∞) = {a_1, a_2, …}.

So now the problem is a beautiful self-consistency: C = {m : gcd(m, c) > 1 ∀ c ∈ C ∩ [a_1,∞)}… i.e., letting A = C ∩ [a_1, ∞): A = {m ≥ a_1 : gcd(m, a) > 1 ∀a ∈ A}. A is a maximal-ish pairwise non-coprime set. Need: A is exactly periodic: ∃M ∀m ≥ a_1: m ∈ A ⟺ m+M ∈ A. Then T = #(A ∩ [m, m+M)) constant? Careful: T = number of A-elements in any window of length M is constant by exact periodicity, and a_{n+T} = a_n + M… yes with L = M.

So reduces to: Lemma: Let A ⊆ [N, ∞) satisfy A = {m ≥ N : gcd(m, a) > 1 ∀ a ∈ A} (with A nonempty). Then A is periodic: ∃M: m ∈ A ⟺ m + M ∈ A for m ≥ N. Hmm also A automatically contains all multiples of a for a ∈ A (≥N) hence bounded gaps.

Proof attempt: Define 𝒬 = set of minimal transversals of {P(a) : a ∈ A}. A = {m ≥ N : ∏Q | m·rad…} precisely m ∈ A iff P(m) contains a transversal iff ∃Q ∈ 𝒬: ∏Q | m. Want: finitely many minimal transversals?? Hmm think: minimal transversal Q: hits every P(a), minimal. Note every P(a) for a∈A is itself a transversal-containing set (P(a) hits all P(a’)). So each P(a) ⊇ some minimal transversal Q_a. Also each Q∈𝒬 ⊆ ∪… each element of Q is “needed”: ∀q∈Q ∃a: P(a)∩Q = {q}.

Suppose Q1, Q2 minimal transversals: any m1 div by ∏Q1… is in A (if ≥N), and m2 div by ∏Q2 in A, and gcd(m1,m2)>1 required?? Since both in A and A pairwise non-coprime: gcd(∏Q1·k1, ∏Q2·k2) > 1 for ALL k1,k2 ⇒ Q1 ∩ Q2 ≠ ∅ (choose k avoiding extra primes: gcd(∏Q1, ∏Q2)>1 needed: take m1 = ∏Q1^s ≥ N with same radical: m1 ∈ A ✓; similarly m2; gcd > 1 ⇒ Q1∩Q2 ≠ ∅). So 𝒬 is pairwise intersecting family of minimal sets… still could be infinite ({2,r}? Q1={2,r1}, Q2={2,r2}: pairwise ok. Both minimal transversals of the SAME family: Q1 minimal transversal means removing r1: {2} not transversal ⇒ ∃a ∈ A odd… a odd with r1|a? a∈A must be hit by Q1 via 2 or r1; minimality of Q1 requires some a with P(a)∩Q1={r1}: a odd, r1|a. Similarly odd a’ with r2|a’. And also 2 necessary: some a even with… P(a)∩Q1 = {2}. Now the odd a (r1|a, 2∤a) must be hit by Q2={2,r2}: so r2|a. So a divisible by r1, r2, …, all r_i for every Q_i = {2, r_i}: forces finitely many. Generalizes? Conjecture: 𝒬 finite always. Proof: Suppose Q_1, Q_2, … infinitely many distinct minimal transversals. Each pair intersects. Hmm, use: for each Q_i and each q ∈ Q_i, ∃ a ∈ A with P(a) ∩ Q_i = {q} (“private witness”). Take witness a_1 for some q_1 ∈ Q_1 with q_1 ∉ Q_j for some…? Since Q_j distinct minimal transversals, ∃ q1 ∈ Q_1 Q_2 (if Q1 ⊄ Q2… minimal transversals are an antichain, so Q_1 ⊄ Q_2 and vice versa). Witness a with P(a) ∩ Q_1 = {q_1}. Q_2 hits P(a): via some q’ ∈ Q_2 ∩ P(a), q’ ≠ q_1? q’ could equal q1? q1 ∉ Q_2 so q’ ≠ q1, and q’ ∉ Q_1 (since P(a) ∩ Q_1 = {q1}). Hmm, so P(a) contains elements from every Q_j, all different from Q_1 except q1… For infinitely many Q_j: P(a) must hit all Q_j; P(a) finite ⇒ some prime q* ∈ P(a) lies in infinitely many Q_j. König/compactness flavor: Build an infinite structure. Let 𝒬 infinite. All Q ∈ 𝒬 hit P(a₀) for fixed a₀ ∈ A: some prime p₁ ∈ P(a₀) in infinitely many Q ∈ 𝒬₁ ⊆ 𝒬. Each Q ∈ 𝒬₁ ∋ p₁, minimal ⇒ witness a_Q with P(a_Q) ∩ Q = {p₁}?? no wait minimality gives witness for each element. Since Q’s in 𝒬₁ are distinct and all contain p₁, they differ elsewhere: |Q| ≥ 2 for infinitely many… take q_Q ∈ Q {p₁} distinct-ish; witness b_Q ∈ A with P(b_Q) ∩ Q = {q_Q} ⇒ p₁ ∤ b_Q. Now b_Q must be hit by every OTHER Q’ ∈ 𝒬₁: via primes ≠ p₁. Hmm, getting complicated; use a cleaner approach:

Cleaner: A has bounded gaps (A ⊇ multiples of d₀ := ∏Q₀ ≥ N). Sunflower lemma? Alternatively bound |Q| for Q ∈ 𝒬 and bound primes in Q: Claim: every prime q appearing in any minimal transversal is ≤ some bound. If Q ∋ q huge: minimality ⇒ ∃ a ∈ A with P(a) ∩ Q = {q}… hmm a could be huge, fine. But consider m = ∏Q ·k ∈ A: fine. Try instead: Claim E: every minimal transversal Q satisfies ∏_{q∈Q} q ≤ some bound M₀, and primes bounded ⇒ finitely many minimal transversals. How to bound: Let a* = min A = a_1 (=N). Every Q hits P(a)?? No wait Q is transversal: hits every P(a), a ∈ A. Hmm that bounds ONE element of Q (≤ max prime of a). Other elements unbounded a priori? Minimality: q ∈ Q has witness a_q ∈ A with P(a_q) ∩ Q = {q}. Hmm.

Use greedy/minimum structure of A more: A = C∩[N,∞) where C defined by A itself. Extra property: everything ≥ N compatible with A is IN A (maximality!). So if m ≥ N and gcd(m, a) > 1 ∀ a ∈ A, then m ∈ A. Use maximality to show structure: Let R = set of primes q such that q | a for ALL… hmm.

Let’s consider 𝒬 (minimal transversals). A = {m ≥ N: ∃Q ∈ 𝒬, Q ⊆ P(m)}. Pairwise intersecting antichain 𝒬. Also: every Q ∈ 𝒬, the number m_Q = (∏Q)^k ≥ N is in A. Apply transversal property to a = m_Q: every Q’ hits P(m_Q) = Q: that’s just pairwise intersecting again. Minimality witnesses: for q ∈ Q, ∃ a ∈ A, P(a) ∩ Q = {q}; P(a) ⊇ some Q_a ∈ 𝒬 with Q_a ∩ Q ⊆ … Q_a ∩ Q ≠ ∅ and Q_a ⊆ P(a) so Q_a ∩ Q ⊆ P(a) ∩ Q = {q} ⇒ q ∈ Q_a. So: for every Q ∈ 𝒬 and q ∈ Q, there is Q’ ∈ 𝒬 with Q’ ∩ Q = {q}. Nice clean combinatorial condition: 𝒬 is an intersecting antichain such that ∀Q∀q∈Q ∃Q’: Q∩Q’ = {q}. Question: must 𝒬 be finite?! Suppose 𝒬 infinite. Hmm, known result: a maximal intersecting family generated… Counterexample attempt: 𝒬 = {{1,2},{1,3},{1,4},…} ∪ … condition: for Q={1,k}, q=k: need Q’ with Q∩Q’={k}: Q’ ∋ k, 1∉Q’. Add sets {2,3,…}? Let’s try 𝒬 = all 2-subsets of ℕ? Not intersecting ({1,2},{3,4}). Try 𝒬 = {{1,x}: x≥2} ∪ {{2,3,4,…}}? infinite set not allowed (finite sets only, since transversals minimal ⊆ P(a) finite? A minimal transversal needn’t be finite a priori! Hmm—but every P(a) contains a minimal… every m ∈ A: P(m) finite contains a transversal (P(m) itself is one, hits all P(a) since gcd… wait does P(m) hit all P(a)? m ∈ A, a ∈ A ⇒ gcd > 1 ✓). Within finite P(m) choose minimal sub-transversal: finite ✓. And A = {m: P(m) ⊇ some finite minimal transversal}. So only finite minimal transversals matter; 𝒬 := finite minimal transversals. But maybe infinitely many. The witness argument above: witness a with P(a)∩Q = {q}: does such a ∈ A exist for minimal transversal? Minimality of Q: Q{q} is not a transversal ⇒ ∃a ∈ A: P(a) ∩ (Q{q}) = ∅. And Q hits P(a) ⇒ P(a)∩Q = {q}. ✓. So 𝒬 (finite minimal transversals that are ⊆ P(a)… all of them) satisfies: intersecting? shown via powers-of-∏Q ∈ A trick provided ∏Q^k ∈ A: m = ∏Q^k, P(m) = Q ⊇ Q ∈ 𝒬 ⇒ m∈A ✓. antichain ✓ (minimality). private-intersection property ✓. Also each Q must hit P(a) for the SPECIFIC witnesses… it’s all encoded.

Now: is every intersecting antichain with private-intersection property finite?? Try to build infinite example: Let Q_n = {p} ∪ {r_n}? private for r_n: need Q’ with Q’∩Q_n = {r_n}: Q’ ∌ p. Q’ hits all Q_m = {p, r_m}: ⇒ r_m ∈ Q’ ∀m: infinite ⇒ impossible. So that family fails, consistent with earlier. Try: 𝒬 = 2-subsets of {a,b,c} plus more? Finite. Try “triangle-ish” infinite: Q_n = {x_n, y_n, common structure}? Intersecting antichain infinite with private property: e.g., 𝒬 = { {i, j}’ }… Consider 𝒬 = all pairs {2i, 2i+1}?? not intersecting. Hmm consider 𝒬 = lines of an infinite projective plane! Lines pairwise intersect in one point, antichain, and for line L and point q ∈ L there’s another line through q meeting L only at q ✓. Infinite projective planes exist (free planes). So combinatorially 𝒬 can be infinite!! But additional constraints from our setting: number-theoretic realizability + maximality of A + bounded gaps? Also every P(a), a ∈ A must be hit by every Q… and A = all m with P(m) ⊇ some Q. The family of P(a)‘s = all finite supersets of members of 𝒬 (realizable with m ≥ N: any finite prime set containing some Q arises as P(m) for suitable m ≥ N ✓ all in A). Transversal condition: each Q ∈ 𝒬 must hit every such superset ⟺ hit every Q’ (supersets add primes… Q must hit P(a) for all a: P(a) ⊇ Q’ some Q’; Q ∩ Q’ ≠ ∅ suffices ✓). And 𝒬 = ALL finite minimal transversals: closure condition: any finite minimal transversal of the family must be in 𝒬 — that’s the maximality-driven condition: 𝒬 must equal the set of minimal transversals of the hypergraph {P(a)} which reduces to: minimal transversals of 𝒬↑… Let 𝒬 be an infinite projective plane’s lines with points = primes: is a “new” transversal possible: a finite set hitting all lines that doesn’t contain a line? In infinite projective plane, a finite hitting set of all lines: every point is on infinitely many lines… a finite set S hitting all lines, S not ⊇ line. Hmm, in an infinite projective plane can a finite set of points meet every line? Take two points x,y ∈ S; lines through neither… For finite S not containing a line, is there a line avoiding S? Probably yes in free/infinite planes… If every finite transversal contains a line, then 𝒬 IS the set of minimal transversals and we’d have a consistent infinite 𝒬 ⇒ A = {m : P(m) contains a line}. Would such an A actually arise from the greedy sequence though?! A must also equal C∩[N,∞) with bounded gaps etc. — all satisfiable? But wait: A must be a set of integers = {m ≥ N : P(m) ⊇ some line}. Realizability as the compatible set: C(A) = {m: gcd(m,a)>1 ∀a∈A} = {m : P(m) transversal of {P(a)}} = {m: P(m) ⊇ some line (if all finite transversals contain lines)} = A ✓ self-consistent!! Then the greedy sequence starting at a_1 = min element would enumerate A, and if A is not eventually periodic, the THEOREM WOULD BE FALSE. So there must be an additional constraint ruling this out — the greedy genesis: A isn’t arbitrary self-consistent set; it’s generated by a greedy process starting from a single a_1. Key: the sequence is built greedily and constraints appear over time; C is determined at the end, but the greedy picks minimal elements at each finite stage: a_{n+1} = min{m > a_n : hits P_1..P_n}. We showed this equals min element of C — wait, we showed a_{n+1} ∈ C? Let me re-verify: a_{n+1} satisfies gcd with all i ≤ n by definition and with i > n by later steps’ condition (gcd(a_i, a_{n+1}) for i > n+1 by definition at step i). ✓. And any c ∈ C with a_n < c satisfies step-n constraints ⇒ a_{n+1} ≤ c. So a_{n+1} = min C ∩ (a_n, ∞). ✓✓. So A := {a_n} = C ∩ [a_1, ∞) PROVIDED also every c ∈ C with c ≥ a_1: c ≥ a_1 = … c ∈ (a_n, a_{n+1}] ⇒ c = a_{n+1} ✓ (or c = a_1). So A = C ∩ [a_1, ∞). ✓.

Now C = {m : gcd(m, a_n) > 1 ∀n} — constraints only from A. So A self-consistent set as analyzed. Additional structure from greediness beyond self-consistency: Which self-consistent sets arise? A must satisfy: A = C(A) ∩ [min A, ∞). The projective-plane example might satisfy that as a set, but can it be generated—doesn’t matter! If it satisfies A = C(A) ∩ [min A, ∞), then taking a_1 = min A, the greedy sequence equals enumeration of A, giving counterexample if A non-periodic. Therefore either (a) such number-theoretic projective realization is impossible, or (b) it’s automatically periodic, or (c) I’ve erred. Since the theorem is presumably true, the resolution: for a set of INTEGERS, P(m) for m ≤ x has ≤ log₂x elements and primes have sizes; realizing “all finite transversals contain a line” with primes and the maximality within [N,∞)… Wait, actually maybe such A fails “bounded gaps”? A has bounded gaps automatically? A ⊇ multiples of ∏Q₀ for a fixed line Q₀ ⇒ gaps ≤ ∏Q₀ ✓ bounded. Hmm. But does A = C(A) ∩ [N, ∞) hold for plane example: need: every m ≥ N with P(m) transversal contains line — we assumed every FINITE transversal of the line-family contains a line. Is that possible for an infinite projective plane? In classical infinite projective planes (e.g., over ℚ), a finite set of points meeting every line: no! Because there are infinitely many lines avoiding any finite set of points (through a generic point). Lines through a point x ∉ S… take any point x not in S ∪ (lines… ), hmm: given finite S, pick point x ∉ S. Lines through x: at least 3, actually infinitely many in infinite plane; each line through x meets S? The lines through x partition remaining points; S finite meets only finitely many of the infinitely many lines through x ⇒ some line through x misses S ⇒ S not transversal. Hence NO finite transversals other than… wait but lines themselves are infinite in such planes! Minimal transversals must be finite in our setting (⊆ P(m)). If plane’s lines are infinite sets, then no m has P(m) ⊇ line ⇒ A empty. So need infinite plane with FINITE lines — impossible! (An infinite projective plane has infinite lines? A projective plane where lines are finite: order n plane is finite. Yes: in a projective plane, if one line is finite then the plane is finite (standard).) More generally, our requirement: 𝒬 infinite family of FINITE sets, pairwise intersecting, antichain, private-intersection property, AND every finite transversal of 𝒬 contains a member of 𝒬 (else A ≠ C(A)∩…: hmm wait we need 𝒬 to be exactly the minimal transversals of the hypergraph of all P(a) = supersets — equivalently every finite minimal transversal of 𝒬 itself… note each Q ∈ 𝒬 IS a transversal (intersecting) and minimal? private property ⇒ minimal as transversal ✓. The issue: other minimal finite transversals S ∉ 𝒬 would mean m with P(m) = S: gcd(m, a) > 1 ∀ a ∈ A ⇒ m ∈ C(A), and if m ≥ N, m ∈ A ⇒ P(m) ⊇ some Q: S ⊇ Q, S minimal transversal & Q transversal ⇒ S = Q ∈ 𝒬 contradiction — so consistency FORCES: every finite transversal of 𝒬 contains a member of 𝒬.)

So crux combinatorial lemma: Let 𝒬 be a family of finite sets (of primes), pairwise intersecting, such that every finite transversal of 𝒬 contains a member of 𝒬. Then 𝒬… must have a finite “core”: actually then the minimal transversals = 𝒬 and claim: 𝒬 finite. Suppose 𝒬 infinite. Build a finite transversal not containing any member: Since all Q ∈ 𝒬 intersect fixed finite Q₀ ∈ 𝒬: pigeonhole: partition 𝒬 by which element of Q₀ they contain… Let me attempt: q ∈ Q₀. Let 𝒬_q = {Q ∋ q}. ∪q 𝒬_q = 𝒬. Some 𝒬{q1} infinite. Try transversal S: start greedy: we want finite S hitting all Q, with no Q ⊆ S. Hmm, like constructing a maximal intersecting… Use the tree/induction on |Q₀|: Standard result: “Helly property”? Let me think: Let 𝒬 pairwise-intersecting family of finite sets with all sets of size ≤ k?? sizes unbounded possibly. Hmm. But sizes: Q ∈ 𝒬 hits Q₀ and more… no bound evident. However! Number theory gives extra: nothing bounds |Q| directly. Hmm.

Alternative: maybe don’t need 𝒬 finite; need A eventually… we need EXACT periodicity of A. A = {m ≥ N: P(m) ⊇ some Q ∈ 𝒬} = [N,∞) ∩ ∪{Q∈𝒬} d_Q·ℤ{sqfree-support}, d_Q = ∏Q. If 𝒬 infinite this union of APs: is it periodic? Not generally. So likely 𝒬 finite is the way; prove the combinatorial lemma:

Lemma F: If 𝒬 is an infinite pairwise-intersecting antichain of finite sets, then there exists a finite transversal T of 𝒬 containing no member of 𝒬. If true ⇒ contradiction ⇒ 𝒬 finite ⇒ A periodic with M = lcm/product related: m ∈ A ⟺ ∃Q: ∏Q | …: P(m) ⊇ Q ⟺ ∏{q∈Q} q | m. So A = [N,∞) ∩ ∪_{Q∈𝒬} (∏Q)ℤ: exactly periodic with period M = lcm(∏Q) = ∏(all primes in ∪𝒬)? M = lcm of the d_Q: finite ✓. m ∈ A ⟺ m ≥ N ∧ (d_Q | m for some Q) ⟺ periodic mod M ✓ EXACTLY periodic. Then T := #{elements of A in any interval (x, x+M]} — constant = #{residues covered}·(M/…)= number of residues r mod M with r ≡ 0 mod some d_Q: T̃. And a{n+T̃} = a_n + M for ALL n ≥ 1: since A ∩ (a_n, a_n + M] contains exactly T̃ elements and a_n + M ∈ A ✓ and count ⇒ the T̃-th element after a_n is exactly a_n + M ✓. So T = T̃, L = M. Also need T ≥ 1: A nonempty infinite ✓.

Now prove Lemma F. Suppose 𝒬 infinite antichain, pairwise intersecting, finite sets. Want finite transversal containing no member. Hmm. Consider a minimal counterexample structure. Since pairwise intersecting with finite sets: Let Q₀ ∈ 𝒬. Every Q hits Q₀. For q ∈ Q₀ let 𝒬_q as above. Idea: build T by taking q1 ∈ Q₀ that’s in infinitely many members, then restrict to those NOT containing q1? Hmm we need to hit everything. Alternative approach: choose T = Q₀’ a slight perturbation: T = (Q₀ {q}) ∪ {stuff hitting 𝒬 with Q∩Q₀={q}}…

Let me think again. Actually maybe easier: prove directly that 𝒬 finite using integers/greedy? Or prove periodicity differently. Alternatively realize: since 𝒬 = minimal transversals and each Q ∈ 𝒬 is ⊆ P(a) for… each Q ∈ 𝒬 is realized: m with P(m) = Q lies in A? m = ∏Q^k: is m ∈ A? Yes shown. OK here’s another thought: use the “private” property + König: Suppose 𝒬 infinite. Members are finite; by pairwise intersection with Q₀, some q₁ ∈ Q₀ in infinitely many members: 𝒬₁ = 𝒬_{q₁} infinite. Does some member avoid q₁? If NOT, all members contain q₁ ⇒ {q₁} is a transversal, contains no member iff {q₁} ∉ 𝒬; if {q₁} ∈ 𝒬 then antichain ⇒ 𝒬 = {{q₁}} finite. Else {q₁} is finite transversal with no member inside (members have size ≥ 2 distinct from {q₁}? member ⊆ {q₁} would be {q₁} ∉ 𝒬 ✓) ⇒ done (Lemma F conclusion). So assume ∃Q₁ ∈𝒬, q₁ ∉ Q₁. All infinitely many members of 𝒬₁ ∋ q₁ also hit finite Q₁: some q₂ ∈ Q₁, 𝒬₂ = {Q ∈ 𝒬₁: q₂ ∈ Q} infinite. Note q₂ ≠ q₁. Continue: T_k = {q₁,…,q_k}: if T_k is a transversal: does it contain a member? If not: done. If yes: some member Q ⊆ T_k… hmm then need to handle. If T_k not transversal: ∃ Q_k ∈ 𝒬 with Q_k ∩ T_k = ∅; members of 𝒬_k (infinite, all ⊇ T_k? no—𝒬_k = {Q: {q1..qk} ⊆ Q} infinite) each hit Q_k ⇒ q_{k+1} ∈ Q_k with 𝒬_{k+1} infinite, q_{k+1} ∉ T_k. So as long as no T_k is transversal, we grow, and infinitely many members contain T_k for every k — but members are finite sets… yet 𝒬_k infinite members each ⊇ T_k with |T_k| = k → ∞: sizes of members in 𝒬_k ≥ k: fine, sizes unbounded allowed. Hmm this doesn’t terminate by itself. But wait: members containing T_k: they’re supersets of T_k; antichain condition: no member contains another. If some member equals… Consider member Q* ∈ 𝒬 with Q* ∩ … hmm. Let’s add the private-intersection property into the mix (haven’t used): ∀Q ∈ 𝒬, ∀q ∈ Q, ∃Q’ ∈ 𝒬: Q ∩ Q’ = {q}. Members of 𝒬_k contain T_k = {q1,…,q_k}; take Q ∈ 𝒬_k, and its private partner for q₁: Q’ with Q∩Q’ = {q₁} ⇒ q₂,…,q_k ∉ Q’. Hmm.

Alternative cleaner combinatorial theorem: A pairwise intersecting family of finite sets that is an antichain and equals the set of ALL its minimal transversals (“self-dual clutter”) — question: must a self-transversal (self-blocking) clutter on infinite ground set with finite edges be finite?? Known: A clutter equals its blocker iff … For finite clutters, b(b(𝒬)) = 𝒬; self-blocking finite clutters exist (e.g., triangle {12,13,23}, also {1}…). Infinite self-blocking clutter with finite edges? Hmm. If it exists number-theoretically we’d contradict the IMO statement, so it must not exist, i.e., theorem: every self-blocking clutter with finite edges is finite. Hmm wait, careful: blocker of 𝒬 = minimal transversals; we need 𝒬 ⊆ b(𝒬) (each edge is a minimal transversal: from pairwise intersecting + private property… hmm actually pairwise intersecting gives edge is transversal; minimality of edge as transversal: need for q ∈ Q some Q’ with Q’∩(Q) = ∅; Q’∩Q={q} private ✓) and b(𝒬) ⊆ 𝒬 (every minimal finite transversal is an edge — the consistency condition; but also could there be minimal transversals that are infinite? For blocker theory with infinite families, a minimal transversal might be infinite or might not exist… For our purposes we only care about finite ones: consistency forces finite minimal transversals ∈ 𝒬. Also every edge… fine.)

So target Lemma F’: Let 𝒬 be a clutter (antichain) of finite nonempty sets, pairwise intersecting, such that every finite transversal contains an edge. Then 𝒬 is finite.

Proof attempt: Suppose 𝒬 infinite. Strategy: construct a finite transversal T containing no edge. Use compactness/greedy: Order… Take any edge Q₀ = {x₁,…,x_r}. Every edge meets Q₀. Define for S ⊆ Q₀ the (sub)family… Induction on r = |Q₀| maybe: r = 1: Q₀ = {x}: every edge contains… every edge meets {x} ⇒ contains x ⇒ antichain ⇒ 𝒬 = {{x}} finite ✓. General: For each x ∈ Q₀, consider 𝒬^x = edges NOT containing x. If some 𝒬^x = ∅ ⇒ all edges ∋ x ⇒ 𝒬 = {{x}}? no: all edges contain x, antichain okay with many edges, but then T = {x} is a transversal; contains an edge iff {x} ∈ 𝒬 ⇒ 𝒬={{x}} finite; else T contains no edge ⇒ contradiction with hypothesis ⇒ so consistency forces… So either finite or every x has 𝒬^x ≠ ∅.

Hmm, think of the structure of maximal intersecting families… Alternative: model as graph/hypergraph and use König’s lemma to build INFINITE… but transversal must be finite.

Try: Claim: exists finite transversal T with T ⊆ union of … Let me look at it via “critical” edges. Since every finite transversal contains an edge, the blocker (finite part) = 𝒬 ⊆ itself. Known theorem (Edmonds–Fulkerson): a clutter equals its blocker iff it’s a “square” clutter… in finite case the only self-blocking clutters?? Actually finite self-blocking clutters: b(𝒬)=𝒬. Examples: odd projective-plane-like? {{1}}: b = {{1}} ✓. Triangle: edges {12,13,23}: minimal transversals: {1,2},{1,3},{2,3}? transversal must hit all three pairs: {1,2} hits 12,13(1),23(2) ✓ minimal ✓. Also {1,2,3}? not minimal. So triangle self-blocking ✓ finite. Lines of finite projective plane: self-blocking ✓. So finite ones exist aplenty; need: no infinite one (with finite edges). Proof of Lemma F’: Suppose 𝒬 infinite self-blocking-ish. Pick edge Q₀. Each edge meets Q₀ ⇒ some x ∈ Q₀ with infinitely many edges containing… Let me try a minimal-degree / compactness argument:

Since edges are finite and pairwise intersecting and 𝒬 infinite: Claim ∃ finite T transversal, no edge inside. Construct greedily a MINIMAL transversal that’s infinite?? If we can show existence of an infinite minimal-ish transversal… but hypothesis only restricts finite transversals. Hmm hypothesis: every finite transversal contains an edge. Suppose 𝒬 infinite; want contradiction.

Take edge Q₀, |Q₀| = r minimal among edges. Consider T₀ = Q₀: it’s a finite transversal containing edge (itself) - fine. Perturb: replace element x of Q₀: T = (Q₀{x}) ∪ Y where Y = finite set hitting all edges that miss Q₀{x}: edges Q with Q∩Q₀ = {x} (must contain x!) since every edge hits Q₀. So ℰ_x := {Q : Q∩Q₀={x}} — all contain x. So Y = {x} works trivially; need avoid x to make progress… Edges in ℰ_x all contain x and pairwise intersect; also Q∩Q₀={x} means Q ∖ {x} misses Q₀. To hit ℰ_x without x: need finite Y hitting all Q∖{x}, Q ∈ ℰ_x — possible iff finitely-generated… If ℰ_x infinite with “spread” tails, impossible finitely?? e.g. ℰ_x = {{x, r_i}}: hitting without x impossible. Then every transversal contains x… IF ℰ_x has no finite transversal avoiding x, then x ∈ every finite transversal. Define K = {x : x in every finite transversal of 𝒬} hmm interesting: K = intersection of all finite transversals… K itself finite (⊆ Q₀ any edge? K ⊆ every transversal… K ⊆ Q₀ since Q₀ is a transversal ✓). K could be empty. Every edge is a transversal ⇒ K ⊆ ⋂𝒬 edges. If K ≠ ∅: x ∈ K in every edge: then pairwise intersection trivial… and antichain: edges ∋ x. Private property for edge Q ∋ x, element q ≠ x∈ Q: Q’ with Q∩Q’={q} ⇒ x ∉ Q’ contradicting x ∈ every edge — unless Q = {x}. So if K ≠ ∅ ⇒ all edges = {x} ⇒ finite ✓. So assume K = ∅: for every x, ∃ finite transversal avoiding x?? wait K defined via all finite transversals; K = ∅ means ∀x ∃ finite transversal T_x ∌ x. Hmm not obviously; K=∅ ⟸ shown above must hold in infinite case… let me redo: If 𝒬 infinite, suppose x ∈ every edge… private property kills (as above) unless {x} ∈ 𝒬 ⇒ 𝒬 finite. So no common element among edges. But K (common to all finite transversals): K ⊆ ⋂edges = ∅ ✓. So ∀x ∈ Q₀ ∃ finite transversal T avoiding x ⇒ T hits ℰ_x avoiding x ⇒ Y_x := T works! So define T* = (⋃_{x∈Q₀} stuff)… Let’s now build transversal with no edge: Hmm. For each x ∈ Q₀, Y_x finite hits every edge Q with Q ∩ Q₀ = {x}, x ∉ Y_x. Consider T = Q₀ ∪ ⋃Y_x minus…? We want NO edge inside T — hard since T big. Different tack: use minimal transversals: every finite transversal contains an edge ⇒ minimal finite transversals ARE edges. Build a finite transversal T that is minimal and different from all edges?? Since 𝒬 infinite, but transversal minimal ⇒ ∈ 𝒬: no direct contradiction. Need: construct finite transversal T whose minimal shrink can’t be an edge, e.g., T avoiding some element of every edge…? An edge Q ⊆ T: prevent by ensuring T misses at least one point of each edge: T transversal + T “co-transversal” (independent-ish: complement is also transversal?? complement infinite). Condition: T finite, hits every edge, and complement hits every edge (i.e., no edge ⊆ T). Since edges are finite and there are infinitely many, and T finite: edges not ⊆ T: automatic for all but finitely many edges? Edges ⊆ T: only finitely many possible (T finite ⇒ ≤ 2^|T| subsets). So: take ANY finite transversal T; the edges inside T are among subsets of T. Remove them: for each edge Q ⊆ T, we want to destroy: delete a point? deleting may break transversality; instead we can add?? adding doesn’t remove edges. Hmm: choose T minimal transversal ⇒ T ∈ 𝒬 = edge. So T itself is an edge—can’t avoid via minimality. Use two disjoint-ish transversals? Any transversal T and edge Q ⊆ T with Q ≠ T… Let’s do: take minimal transversal T₁ ∈ 𝒬 (an edge). Since 𝒬 infinite, take another edge Q₂ ⊄… all edges intersect T₁. Consider T = (T₁ {x}) ∪ Y_x where Y_x is finite transversal-part avoiding x as above: T hits: edges meeting T₁{x} ✓; edges with Q∩T₁={x}: hit by Y_x ✓ (Y_x from T_x := finite transversal avoiding x: T_x hits ALL edges; set Y_x = T_x). So T’ := (T₁{x}) ∪ T_x is a finite transversal avoiding… contains x? T_x ∌ x, T₁{x} ∌ x ⇒ x ∉ T’. Now shrink T’ to minimal transversal T’’ ⊆ T’: T’’ ∈ 𝒬, x ∉ T’’. Fine—edges without x exist anyway; no contradiction yet.

Need a smarter invariant. Perhaps use sizes: let r = min edge size; hmm.

Let me step back: maybe use the NUMBER THEORY more instead of pure combinatorics; the ground set is primes and A consists of integers ≥ N with bounded gaps? We haven’t used: A has an enumeration with gaps ≤ a_1! Gaps: consecutive elements of A differ ≤ a_1 = min A =: g. Now A = ∪{Q∈𝒬} multiples of d_Q (∩[N,∞), N = g). Gap bound: every interval of length g beyond g contains a multiple of some d_Q. If 𝒬 finite, fine. If 𝒬 infinite: for intervals to be covered, small moduli must do the job: only d_Q ≤ g can cover ALL intervals? Not exactly: infinitely many large moduli could conspire. But: consider interval (x, x+g]: must contain m ∈ A: m div by d_Q ≤ … m ≥ x arbitrary large but d_Q could be large too. Hmm, still, DENSITY: A ⊆ ∪ d_Q ℤ; density of A ≥ 1/g. ∑ over Q with d_Q > D of 1/d_Q could be small… Let 𝒬_small = {Q: d_Q ≤ D}: finite set! (moduli bounded, squarefree, distinct edges ⇒ distinct d_Q ✓ finitely many). Choose D = large. Union bound: density(A) ≤ ∑{d_Q ≤ D} 1/d_Q + ∑{d_Q > D} 1/d_Q. If ∑{Q} 1/d_Q converges… does it? Not obviously. Hmm. Edges distinct squarefree numbers d_Q; ∑ 1/d over ALL squarefree diverges. But edges pairwise intersecting etc. Suppose we show tail ∑{d_Q > D} 1/d_Q → 0 (i.e., ∑ 1/d_Q < ∞), then A is “almost” the finite union A₀ = ∪{small} d_Qℤ, within density ε. Then use maximality: every m ≥ N in A₀… wait A ⊇ each d_Qℤ ∩ [N,∞) exactly, A = ∪ over ALL Q. Periodicity fails only due to large-modulus multiples m ∈ A with P(m) not ⊇ any small edge. To finish we’d need: only finitely many edges — hmm, alternatively show directly: every edge Q has d_Q ≤ some bound ⇒ finitely many edges. Suppose edge Q with d_Q huge. m = d_Q ∈ A (if ≥ N; take m = d_Q·(power)= just d_Q^2 whatever ∈ A). Consider interval around… use greedy gap bound at m: predecessor/successor within g: the interval (m − g, m) ∩ A? Not obviously conflicting.

Use instead: private witnesses are integers: for q ∈ Q, ∃ a ∈ A with P(a) ∩ Q = {q}. And ALL elements of A pairwise non-coprime… I think the right path: bound the primes appearing in edges. Claim G: every prime q dividing some edge… Let q ∈ Q ∈ 𝒬 with private partner Q’ (Q∩Q’ = {q}). Both edges. Hmm consider integers u = ∏(Q)^∗ and v = ∏(Q’)^∗ ∈ A: gcd(u,v) = q-power: fine.

New idea: use the MINIMALITY of greedy at small scales: A ∩ [N, N + something] is fixed finite; edges must all intersect… every edge is a minimal transversal, so must hit P(a_1): Q ∩ P(a_1) ≠ ∅ ∀ edges ✓ (a_1 ∈ A). Similarly hit P(a) for every a. Edges hit every P(a_n)!! Each edge is a transversal of the family of ALL terms’ prime sets. So ∏-small edges… Take edge Q: every a ∈ A divisible by some prime of Q ⇒ A ⊆ ∪{q∈Q} qℤ. COMBINED with density ≥ 1/g: ∑{q ∈ Q} 1/q ≥ 1/g for EVERY edge Q!! (upper density of A ≤ ∑ 1/q.) Nice constraint: every edge has ∑{q∈Q} 1/q ≥ 1/g. So edges can’t consist of only huge primes: min prime of Q: at least one q ∈ Q with q ≤ |Q|g… but |Q| unbounded? Also NUMBER of small primes is limited: primes ≤ some bound… Let 𝒫₀ = set of primes q ≤ some threshold. Hmm, edges satisfy: ∑{q∈Q}1/q ≥ 1/g. Since ∑ 1/q over q ∈ (Y, ∞) primes diverges slowly… an edge could be many medium primes. But also pairwise intersecting + antichain + private…

Let’s also get better structural: For each prime q define A_q = {a ∈ A : q | a}. Edge Q: A = ∪{q∈Q} A_q. Consider the finite set R = {q prime: upper density of A_q ≥ 1/(2g|Q₀|… )} hmm. Define δ(q) = upper density of A_q ≤ 1/q. For every edge Q: ∑{q∈Q} δ(q) ≥ 1/g (density union bound: actually need d̄(∪) ≤ ∑ d̄ ✓, and lower density of A ≥ 1/g — gaps ≤ g ⇒ lower density ≥ 1/g ✓). So each edge contains q with δ(q) ≥ 1/(g·|Q|)… sizes unbounded ⇒ weak. Hmm but ∑{q∈Q} δ(q) ≥ 1/g with δ(q) ≤ 1/q means edges are “heavy”. Define H = {q : q ≤ 2g… } no: define heavy primes W_ε = {q: δ(q) ≥ ε}: finite? δ(q) ≤ 1/q ⇒ W_ε ⊆ {q ≤ 1/ε} finite ✓✓. Every edge Q: ∑{q ∈ Q} δ(q) ≥ 1/g; contribution from q ∉ W_ε with… hmm ∑{q∈Q, q>1/ε} δ(q) ≤ ∑{q > 1/ε, q ∈ Q} 1/q ≤ ∑{primes q > 1/ε} 1/q over q ∈ Q: can be ≥ 1/g if Q huge. Damn, ∑ 1/q diverges ⇒ no finite tail bound. Need δ(q) bounds better than 1/q for primes actually… δ(A_q) ≤ 1/q always. Tail over a single edge: Q finite but arbitrarily large; ∑{q∈Q, q>T} 1/q ≤ ∑{T<q≤X}… unbounded as Q grows. So an edge with primes = all primes in (T, T’] with ∑1/q ≥ 1/g conceivable size-wise. Antichain + private + integer-realizability might kill it. Realizability of witnesses: private partner Q’ for q ∈ Q: Q’ ∩ Q = {q}: Q’ also heavy: ∑{q’∈Q’} 1/q’ ≥ 1/g with all q’ ∉ Q except q.

Hmm, maybe the finiteness claim should be proven with a cleverer argument using EXACT structure: A = ∪{Q ∈ 𝒬} d_Q ℤ ∩ [N,∞) and gaps ≤ g. Take x with x ≡ … Chinese remainder to build a long gap unless finitely many edges dominate: Given finite subfamily 𝒬₀={Q: d_Q ≤ D}, D ≥ ?: if 𝒬₀ is a transversal-complete… Suppose ∃ edge Q* with Q* ∩ … hmm. Suppose infinitely many edges. Consider gap construction: find interval (x, x+g] with no multiples of any d_Q: need for each Q: x ≢ −1,…,−g mod d_Q i.e. avoid g residues mod d_Q. By CRT choose x mod ∏ over finite subfamily; infinite family is the problem: interval (x, x+g] might contain multiples of huge d_Q’s — but a multiple of huge squarefree d > x+g can’t be in (x, x+g] if d > x + g. So: choose x as follows: handle edges with d_Q ≤ some bound via CRT — need d_Q > g for avoidance possible (g residues out of d_Q; if d_Q ≤ g impossible: multiples of d_Q appear in every interval of length g!). AH: if some edge has d_Q ≤ g, that’s the “small” edge:then multiples of d_Q are ≤ g apart… that’s fine and consistent. So: If ALL edges have d_Q > g: then pick D and x ≤ some bound with x avoiding residues for all d_Q ≤ D (possible: each modulus excludes g of d_Q > g residues, CRT product nonempty ✓, choose x in [1, ∏d_Q] can’t control vs “d_Q > x+g” for the rest… The remaining edges d_Q ∈ (D, x+g]: not handled! Because x could be as large as ∏{d_Q≤D} d_Q, and mid-size edges (D, x+g] may hit the interval. Since infinitely many edges, can’t finish. Hmm, unless we also avoid those: infinitely many constraints, each excluding g/d_Q fraction: total excluded density ≤ ∑_{d_Q > D} g/d_Q: if ∑ 1/d_Q converges, choose D with tail < 1/(2g…) then a positive density of x survive ⇒ gap of length g exists in A ⇒ contradiction with gaps ≤ g!! ⇒ So: either some d_Q ≤ g… wait need to handle edges with d_Q ≤ g: they guarantee coverage. Let me restructure the final argument: Let 𝒬_s = {Q : d_Q ≤ G} for a threshold… Actually goal is not “some small edge exists” but “finitely many edges total”. Hmm wait maybe we don’t need finitely many! Re-examine what’s needed for the theorem: need exact periodicity of A on [N, ∞): ∃M ∀m ≥ N: m∈A ⟺ m+M∈A. A = ∪_Q d_Qℤ (∩[N,∞)). If we take M = lcm of d_Q over Q ∈ 𝒬_fin (finite subfamily): then m ∈ A via small edge ⇒ m + M ∈ A ✓; but m ∈ A only via huge edge: m + M ∈ A? need d | m and d | M: no. So exact periodicity needs essentially all-but-none… If ∃ m ∈ A with P(m) ⊇ only huge edges… such m exist whenever huge edges exist (m = d_Q^k: P = Q ⊉ any other edge (antichain!) ⇒ m ∈ A only via Q). So exact periodicity forces… M must be divisible by d_Q for ALL Q ⇒ if infinitely many edges, no finite M works?? m = d_Q^k ∈ A, need m + M ∈ A: possible via a different edge though! Not immediate contradiction, but morally we should just prove: finitely many edges.

Let me now try to prove finiteness cleanly with the density/CRT argument: Claim: ∑_{Q ∈ 𝒬} 1/d_Q < ∞? Hmm need that for the gap argument. Alternative: Show directly: ONLY finitely many edges, via: every edge Q must contain a prime from the finite set 𝒫* = primes dividing d for every… hmm.

Density argument refined: lower density of A ≥ 1/g. For each edge Q, define its “own” elements: m ∈ A with P(m) ⊇ Q but ⊉ any other edge — nonempty. Not density-relevant.

Let’s use the transversal property against a cleverly chosen single element a ∈ A: every edge hits P(a) for every a ∈ A. So edges ⊆ “hit everything” ⇒ ⋂ over a of (∪{p | a} p): Every edge contains a prime factor of a for each a. Take a = d{Q₁}^k: every edge hits Q₁ (knew that). Take several elements with few prime factors: elements of A of the form q^j (prime powers in A): if q^j ∈ A then EVERY edge contains q!! When is a prime power in A? q^j ∈ A iff q^j ≥ N and gcd(q^j, a) > 1 ∀a ∈ A iff q divides every element of A iff {q} is a transversal iff {q} ∈ 𝒬 (then 𝒬 = {{q}} by antichain!). So either 𝒬 = {{q}} — the “single prime” case, A = multiples of q ≥ N, periodic ✓ — or A contains no prime powers. Cute but side.

OK here’s another classical idea: since gaps ≤ g, look at g consecutive elements… Actually maybe bound |Q|: every edge Q is a MINIMAL transversal: for each q ∈ Q, witness a_q ∈ A with P(a_q) ∩ Q = {q}. The witnesses {a_q}{q∈Q} pairwise non-coprime (all in A): gcd(a_q, a{q’}) > 1 via primes outside Q (P(a_q)∩Q={q}, P(a_q’)∩Q={q’}, common prime ∉… common prime in P(a_q)∩P(a_q’): could be anything not in Q except… if common prime r: r ∈ P(a_q) ⇒ r ∈ Q ⇒ r = q and r = q’ ⇒ r ∉ Q). So witnesses pairwise share primes OUTSIDE Q. Hmm.

Density again, but sharper using union over edge Q AND second edge: A ⊆ ∪{q ∈ Q} qℤ for every edge. So A ⊆ ⋂{Q edge} ∪{q∈Q} qℤ =: full constraint. Lower density 1/g. If infinitely many edges: the density of ⋂… = density of {m: P(m) transversal} = density of ∪Q d_Qℤ ≤ ∑ 1/d_Q hmm same thing: Actually A EQUALS ∪{Q∈𝒬} d_Qℤ ∩ [N,∞) — wait is that right? m ∈ A iff P(m) ⊇ some edge iff… m ∈ A iff P(m) is a transversal of 𝒬-generated hypergraph; minimal transversals (finite) = 𝒬; P(m) finite transversal ⇒ contains minimal transversal ⇒ hmm: does every finite transversal contain a MINIMAL transversal? Yes: finite set, remove elements while transversal ✓ minimal finite transversal, which is an edge ∈ 𝒬 by consistency. And conversely P(m) ⊇ edge ⇒ transversal ✓ (edges pairwise intersect… edge hits every edge? Edge Q hits edge Q’ ✓ pairwise intersecting; but transversal must hit every P(a), a ∈ A: P(a) ⊇ some edge… Q hits P(a)? Q ∩ P(a) ⊇ Q ∩ (edge in P(a)) ≠ ∅ ✓). So A = [N,∞) ∩ ∪{Q∈𝒬} d_Q ℤ ✓ EXACTLY. Lower density ≥ 1/g (gaps: gaps of A are ≤ g? gap bound from greedy: a_{n+1} ≤ a_n + a_1 = a_n + N… a_1 = N = min A = g. ✓).

Now: Lemma H: if ∪{Q} d_Qℤ has gaps ≤ g beyond N with d_Q distinct squarefree (𝒬 antichain: d’s not dividing each other actually: Q ⊄ Q’ ⇒ d_Q ∤ d{Q’} ✓✓ antichain w.r.t. divisibility!), then 𝒬 finite?? Is that TRUE as pure number theory? Suppose 𝒬 = {Q₀} ∪ {huge edges}: gaps ≤ d_{Q₀} =: covered if d_{Q₀} ≤ g ✓ any extra huge edges don’t hurt gaps! So Lemma H false as stated: ∪ could be multiples of 6 ∪ multiples of huge squarefree numbers: gaps fine. But then is A = that union realizable as our A? A must be self-consistent: 𝒬 = {{2,3}, HugeQ}: pairwise intersecting ⇒ HugeQ ∩ {2,3} ≠ ∅ say 2 ∈ HugeQ ⇒ antichain ok. Private property for edge {2,3}, q=3: need edge Q’ with Q’∩{2,3} = {3}: HugeQ ∋ 2 ∌3? then Q’∩{2,3}∋2 no. So need another huge edge containing 3 not 2; and consistency: every finite minimal transversal is an edge: transversals of {{2,3},H₂,H₃,…}: e.g. {2} ∪ (hit H₃’s without 2…) etc. This forces lots of structure. So pure gap/density argument insufficient; must use blocker-consistency deeply. So back to combinatorial Lemma F’. Let me search memory: “A clutter such that every finite transversal contains an edge” on infinite ground set, edges finite, pairwise intersecting — prove edge set finite… Hmm I recall the actual solution of this ISL/IMO problem probably goes: eventually the sequence is union of APs — with solution steps: (1) a_{n+1} ≤ a_n + a_1; (2) consider primes ≤ a_1 (“small primes”); since density ≥ 1/a_1, hmm, each m ∈ A: does it need a small prime factor? If all prime factors of every element > g = a_1… elements ≥ N are ≥ N; A ⊆ ∪{q∈Q} qℤ for edge Q… Claim: every edge contains a prime < c(g)?? From ∑{q ∈ Q} 1/q ≥ 1/g we can NOT conclude bounded smallest prime. Hmm but actually we can get better than union bound? A has lower density 1/g. A ⊆ ∪_{q∈Q} qℤ. Also intersections… nothing.

Alternatively bound |Q ∩ [2, T]|≥1 with T = exp(O(g))? ∑{q>T} 1/q over q ∈ Q ≤ ∑{T < q ≤ d_Q} 1/q ≈ ln(ln d_Q / ln T): unbounded. OK so edges may consist of only huge primes in principle…but every edge hits P(a_1): a_1 ≤ g! primes of a_1 ≤ g. EVERY EDGE CONTAINS A PRIME FACTOR OF a_1, in particular a prime < g. More: every edge hits P(a) for EVERY a ∈ A. So edges all intersect the fixed finite set P(a_1) =: {p_1 < … < p_s}, p_i ≤ g.

Now, pigeonhole edges by their intersection pattern with P(a_1) and iterate with witnesses: Suppose 𝒬 infinite. ∃ nonempty S₁ ⊆ P(a_1), infinitely many edges Q with Q ∩ P(a_1) = S₁… Consider the “second-level”: is there an element a ∈ A with P(a) ∩ S₁ = ∅? If for every a ∈ A, P(a) ∩ S₁ ≠ ∅, then… hmm S₁ is a transversal of {P(a)}! Then S₁ contains an edge… an edge Q̂ ⊆ S₁ ⊆ P(a_1). Edges Q in our infinite family: Q ∩ P(a₁) = S₁ ⊇ Q̂: antichain ⇒ Q = Q̂ only one ⇒ contradiction with infinitude of that class unless… Q ⊇ Q̂ contradicts antichain (Q ≠ Q̂ ⇒ Q ⊋ Q̂ ✗). So each class S with infinitely many edges is NOT a transversal: ∃ a_S ∈ A with P(a_S) ∩ S = ∅. But every edge Q in class S hits P(a_S): via primes outside S ∪… Q ∩ P(a_S) ≠ ∅ with P(a_S) ∩ P(a_1)… So now edges in class S₁ hit the finite set P(a_{S₁}) ∖ S₁-part: refine: infinitely many edges Q with Q ∩ (P(a_1) ∪ P(a_{S₁})) = S₂ ⊋ S₁ (S₂ ∩ P(a_{S₁}) ≠ ∅ strictly grows). Iterate: get S₁ ⊊ S₂ ⊊ S₃ ⊊ ⋯ each S_k = Q ∩ F_k for infinitely many edges Q, F_k finite sets growing. This builds edges of unbounded size — no contradiction yet (sizes unbounded allowed?). Hmm, but wait: at each stage, class S_k (infinitely many edges Q with Q∩F_k = S_k): S_k not a transversal (else edge ⊆ S_k ⊆ each such Q ✗antichain) ⇒ ∃a_k ∈ A, P(a_k) ∩ S_k = ∅ ⇒ refine. The process never terminates ⇒ infinite ascending chain S₁ ⊊ S₂ ⊊ …, with all S_k ⊆ … and for each k, ∃ edge Q_k ⊇ S_k (with Q_k ∩ F_k = S_k). So edges of size → ∞: consistent so far. Need extra ingredient to kill: maybe DENSITY: the elements a_k ∈ A with P(a_k) ∩ S_k = ∅: as S_k grows, a_k avoids many primes… but also A ⊆ ∪{q∈Q_j} qℤ ∀j: a_k div by some prime of each Q_j outside… hmm hmm. Let’s consider the limit set S∞ = ∪S_k: infinite set of primes q₁ = elements added… At stage k we have: infinitely many edges Q with Q ∩ F_k = S_k. Also a_k ∈ A with P(a_k) ∩ S_k = ∅; a_k is divisible by some prime in EVERY edge; The edges of class S_k hit P(a_k) outside F_k… And a_k’s are integers in A: pairwise gcd > 1.

DENSITY IDEA concretized: A ⊆ ∩k (∪{q ∈ Q_k} qℤ) where Q_k are our chain edges ⊇ S_k. Hmm not obviously small.

Try yet another angle — maybe use the WITNESS/private property to derive that S_k chain impossible: The edge Q_k has Q_k ∩ F_k = S_k, and F_k ⊇ S_{k}. Private property applied to Q₁… meh.

Let me instead consider the following cleaner claim: Claim I: the set 𝒫_A := {q prime : q | a for infinitely many… } no. Claim J: only finitely many primes appear in edges. Suppose prime q ∈ edge Q. Then ∃ witness a with P(a) ∩ Q = {q} ⇒ hmm q | a. Every edge hits every P(a). Let me count density with inclusion of TWO structures: lower density(A) ≥ 1/g and A ⊆ ∪{q∈Q} qℤ ⇒ ∃q ∈ Q: upper density(A ∩ qℤ) ≥ 1/(g|Q|). Define for prime q: σ(q) := upper density of A_q(≤ 1/q). For edge Q: ∑{q∈Q} σ(q) ≥ 1/g ⇒ ∑{q ∈ Q} min(1/q, σ(q)) ≥ 1/g. Now KEY: bound ∑{all primes} σ(q)?? = expected number of prime divisors — unbounded. Bound instead: for primes q > G₀ with σ(q) > 0: q divides some element…

Honestly, let me look at this from the actual known solution perspective. I believe this problem is ISL 2024 N7 or similar; recalling AoPS discussions… “a_{n+1} smallest with gcd condition to all previous; show eventually a_{n+T} = a_n + L.” I genuinely think the intended solution: (1) a_{n+1} − a_n ≤ a_1. (2) Let d = the minimal integer such that infinitely many terms divisible by… Actually here’s a neat approach: Let D = liminf-type: consider the set 𝒮 of positive integers d ≥ 2 such that all sufficiently large terms are divisible by some prime … hmm.

Alternative approach: strong induction on a_1 (=g)? Since gaps ≤ g and everything ≥ … Consider the finitely many residues…

Let me try yet another angle on finiteness of 𝒬 exploiting that EVERY edge must hit EVERY P(a), a ∈ A — that’s much stronger than pairwise intersecting among edges; and elements a realize all supersets of edges: For edge Q and any other edge Q’, a = (d_{Q’})^k ∈ A: Q hits Q’ — again pairwise. But also: mixed elements: a = product over various… any m ≥ N with P(m) ⊇ edge is in A. So the constraint on transversals: hit all supersets of edges ⟺ hit all edges. Fine, so really structure is: clutter 𝒬, pairwise intersecting, every finite transversal contains an edge (“𝒬 is transversal-complete”). Realizability over primes is automatic (any clutter of finite sets can be labeled by primes!) — EXCEPT constraints: g = a_1 = min A and gaps ≤ g comes from a_1 itself: a_1 has P(a_1) ⊇ some edge; g = a_1 ≥ d_{Q(a_1)} ≥ d_min. And gap bound uses a_1: gaps ≤ a_1. Any transversal-complete infinite clutter would give counterexample by setting a_1 = d_Q (min). SO the mathematical truth must be: Lemma F’’: every pairwise-intersecting clutter of finite sets with “every finite transversal contains an edge” is FINITE. This must be a true combinatorial lemma; prove it!

Proof attempt of F’’: Suppose 𝒬 transversal-complete, infinite. Consider edge Q₀. For each x ∈ Q₀ define 𝒬(x) = {Q ∈ 𝒬: Q ∩ Q₀ = {x}}? We know each edge meets Q₀. Build a finite transversal T with no edge inside, by finite induction on |Q₀| =: r with a minimal-size edge Q₀ (r minimal over edges). Case r = 1 done earlier.

Inductive idea: contract/delete: clutter minors! In finite blocker theory: b(𝒬) = 𝒬 characterizations… For minors: 𝒬 / x (contract) and 𝒬 x (delete) correspond to blocker operations: b(𝒬 ∖ x) = b(𝒬)/x etc. If 𝒬 is transversal-complete (self-blocking), consider x in some edge: Deletion 𝒬∖x := edges not containing x, made into clutter… hmm might lose pairwise intersecting. Contraction 𝒬/x := minimal sets of {Q ∖ {x}: Q ∈ 𝒬}. Induct on min edge size?

Alternative concrete approach: Let Q₀ = {x_1, …, x_r} minimal-size edge. For each i, private edge Q_i with Q_i ∩ Q₀ = {x_i}. Now consider T = ∪_{i=1}^r (Q_i ∖ {x_i})?? T hits?: given edge Q: Q hits every Q_i; if Q ∩ Q_i ⊆ {x_i} for all i, then x_i ∈ Q ∀i ⇒ Q ⊇ Q₀ ⇒ Q = Q₀ (antichain) — but Q₀ ∩ T: Q₀ ∩ Q_i = {x_i} ⇒ Q₀ ∩ (Q_i ∖ x_i) = ∅ ⇒ Q₀ ∩ T = ∅!! So T hits every edge except exactly Q₀ (any Q ≠ Q₀ has some i with Q ∩ Q_i ∋ y ≠ x_i… wait: shown: if Q misses T then Q ∩ Q_i ⊆ {x_i}∀i and Q∩Q_i ≠ ∅ ⇒ x_i ∈ Q∀i ⇒ Q ⊇ Q₀ ⇒ Q = Q₀ ✓). So T’ := T ∪ {x_1} is a finite transversal (hits Q₀ via x_1 ✓). Transversal-complete ⇒ ∃ edge Q* ⊆ T ∪ {x_1}. Where’s Q: Q hits Q_1: Q∩Q_1 ⊆ (T∪{x_1}) ∩ Q_1 = {x_1} ∪ (T ∩ Q_1): T∩Q_1 ⊇ Q_1∖{x_1}… yes Q_1∖x_1 ⊆ T so fine, no info. Hmm Q ⊆ T∪{x_1}: with T built from finitely many edges — no contradiction yet; Q* is just some edge inside that finite set. We want to derive contradiction with INFINITUDE. Where does infinitude enter?? So far nothing. Right: transversal-completeness alone admits infinite examples?? Let’s test: is there an infinite transversal-complete clutter of finite sets? Suppose 𝒬 = {{x}} ∪ …no. Try to construct: ground {p} ∪ {r_1, r_2, …} ∪ …: want every finite transversal to contain an edge. Take 𝒬 = all sets of form {p_i, p_j… } hmm. Try 𝒬 = {A_n} with A_n = {n-th level…}. Consider 𝒬 = { {i, i+1}: i ∈ ℕ }? not pairwise intersecting. Infinite intersecting antichain: Δ-system {x} ∪ B_n with B_n disjoint: private property for elements of B_n needs edge meeting only there — would have to avoid x but every edge… edges all contain x?? In sunflower with core {x}: pairwise intersection = {x}: private edge for y ∈ B_n: Q’ ∩ ({x}∪B_n) = {y}: Q’ = {x}∪B_m: ∩ = {x}(∪∅) ∌… = {x} ≠ {y} ✗. So sunflowers fail private, hence fail transversal-complete? Check transversal-completeness directly for sunflower: T = {x}: transversal, contains an edge? edges are {x}∪B_n ⊄ {x} ⇒ not transversal-complete ✓ fails (as expected since {x} would be in blocker). General principle: transversal-complete ⟹ private property (blocker ⊆ 𝒬… each edge minimal? hmm let me make sure the needed implications for the proof come from consistency of A directly, which we established: (i) pairwise intersecting; (ii) every finite transversal contains an edge; (iii) antichain; (iv) [derived] private property.)

Try infinite construction: “iterated triangles”/free plane style. Free projective plane generated by finite configuration has infinite lines I think, but maybe some infinite transversal-complete clutter with finite edges exists via different construction? If it exists, IMO problem would be false… but wait — is realizability really unconstrained? The ground set of 𝒬 = primes; edges = prime sets; A = ∪ d_Qℤ ∩ [N, ∞) where N := min A must equal a_1 and also we need… any N works? The sequence starts at arbitrary a_1 > 1; for counterexample need a_1 = min of A and A = C(A) ∩ [a_1, ∞): C(A) = {m: P(m) transversal} = ∪ d_Qℤ (all m whose prime set contains an edge — using transversal-complete for finite prime sets ✓). Then set a_1 := min element ≥ 2 of ∪d_Qℤ = d_min := min d_Q. The greedy sequence from a_1: equals enumeration of C ∩ [a_1, ∞)?? We proved: greedy sequence = enumeration of C(A_seq) where A_seq = its own terms. Self-consistency: if we define sequence as enumeration of A = ∪d_Qℤ ∩ [d_min, ∞), does it satisfy the greedy recurrence? Need: a_{n+1} = min{m > a_n: gcd(m, a_i) > 1 ∀i ≤ n}: RHS ≤ a_{n+1} clear (a_{n+1} ∈ A compatible with all of A ⊇ {a_i}). Could RHS be smaller: some m ∉ A, m compatible with first n terms only: possible!! The greedy sequence isn’t automatically the enumeration of A: greediness sees only finitely many constraints at each step; enumeration-of-A is a valid “limit” but greedy might deviate below: m compatible with a_1..a_n but not with later. Hmm wait but for the actual greedy sequence we PROVED a_{n+1} = min C ∩ (a_n, ∞) where C = compatibles of the ACTUAL sequence. For a purported counterexample we’d need to verify the greedy generates exactly A — not automatic. OK so realizability is genuinely constrained; but for our PROOF direction we don’t care: we have the actual sequence, its A, its 𝒬 satisfying (i)-(iv), plus MORE properties we can derive from greedy dynamics if needed (like gaps ≤ a_1 — follows from A ⊇ d_minℤ anyway ✓… a_1 = min A relation: a_1 ∈ A ⇒ P(a_1) ⊇ some edge ⇒ d_Q ≤ a_1 for that edge; gaps ≤ d_Q ≤ a_1 ✓ consistent).

So the question stands: prove (i)+(ii)+(iii) ⇒ 𝒬 finite, or find additional property from greedy dynamics to force finiteness. Hmm hold on — maybe (i)+(ii)+(iii) infinite example EXISTS; let me try hard to construct: Want: infinite antichain, pairwise intersecting, finite sets, s.t. every finite transversal ⊇ edge. Equivalently in blocker terms: the blocker (all minimal transversals, possibly infinite sets too) — every FINITE minimal transversal is an edge, and every finite transversal contains a finite minimal one ✓(finite shrink) so condition ⟺ finite minimal transversals ⊆ 𝒬 ⟺ (with edges being minimal transversals themselves ⊆) blocker’s finite part = 𝒬.

Construction attempt via “duplicating projective planes to infinity”: Take finite projective planes 𝒫_n of growing order on disjoint point sets, then “join”: edges = lines of 𝒫_1 ∪ … no: lines from different planes are disjoint ✗ not intersecting. Join by adding a common point p to everything: fails private property. Cross-join: edges = L₁ ∪ L₂ (line from plane1 + line from plane2)? Pairwise intersecting ✓ (first components intersect). Antichain ✓. Transversals: hit every L₁∪L₂: e.g., transversal of plane1’s lines alone works: contains a line of plane1… but does it contain an EDGE (which is L₁∪L₂)? NO ✗ — need every finite transversal contains a full edge: T = line ℓ₁ of plane 1: hits all edges via first component ✓; contains no edge (edges are bigger) ✗ ⇒ not transversal-complete. Hmm. So constructing seems hard; likely the lemma is TRUE: pairwise intersecting + transversal-complete ⇒ finite. Let me try to prove via the T-construction above which showed: T = ∪(Q_i ∖ x_i) hits everything except Q₀ exactly. Generalize: for ANY edge Q₀ we can build finite transversal T_{Q₀} that misses Q₀… T∪{x}: contains an edge Q* with… let’s analyze Q* ⊆ T ∪ {x_1}: note Q* ∩ Q₀ ≠ ∅ and Q₀ ∩ (T ∪ {x_1}) = {x_1} ⇒ x_1 ∈ Q*. Also we can do same with x_2: T ∪ {x_2} ⊇ edge Q ∋ x_2, Q ∖ {x_2} ⊆ T. Interesting: for each i: edge R_i with R_i ∩ Q₀ = {x_i} and R_i ⊆ T ∪ {x_i} — so private edges can be found INSIDE the finite set T ∪ Q₀ =: U₁ (finite!). Now iterate/localize: maybe show ALL edges ⊆ some finite set U?! If ground set effectively finite ⇒ 𝒬 finite ✓ (finite ground ⇒ finitely many finite subsets… edges ⊆ U finite ⇒ ≤ 2^|U| edges ✓✓). So goal: every edge ⊆ U for a fixed finite U. Take U := T ∪ Q₀ ∪ …? Suppose edge Q ⊄ U with some element y ∉ U. Hmm need transversal avoiding y to trap…: If for every y there’s a finite transversal avoiding y (shown earlier: no common element in all edges + … wait need transversal avoiding y: y not in every edge… need stronger: ∃ finite transversal ∌ y. Hmm: is that automatic? Take edge Q’ ∌ y (exists unless y ∈ ⋂ edges = ∅ handled): Q’ is itself a transversal avoiding y ✓✓!! EVERY edge is a transversal. So for any y: pick edge avoiding y. OK so strategy: show that the union of ALL edges is finite. Suppose not: infinitely many primes appear across edges. Note each element y appearing in some edge Q: private… Let’s use the “T-trick” more cleverly: T = ∪_i (Q_i∖{x_i}) hits every edge except Q₀. So every edge except Q₀ intersects the FINITE set T. And every edge intersects Q₀ too anyway. Now do this for another base edge Q₁ (≠ Q₀, with private edges): get finite T₁ hitting all edges except Q₁. Then W = T ∪ T₁ hits ALL edges (edge missing T is Q₀ ⊆… Q₀ ∩ T₁: does T₁ hit Q₀? T₁ hits all except Q₁ ⊇ hits Q₀ ✓). So W := T ∪ T₁ finite transversal. Every edge meets W: fine but edges could still stick out. Trap edges: For edge Q, consider transversal-completeness applied to sets like W’ = W ∪ (something) designed so the edge inside must be Q… Use “localization” lemma: Claim K: for every edge Q and every y ∈ Q, ∃ edge Q’ ∋ y with Q’ ⊆ W ∪ {y}?? From T-trick: with base Q₀ and x_i: found R_i ∋ x_i, R_i ⊆ T ∪ {x_i}: localized private edges. So define U₀ = Q₀ ∪ T (finite). All edges hit… Let me iterate the T-trick with base edges INSIDE U₀: every edge within U₀ has its private edges; can choose private edges within…? T-trick construction of T requires choosing private edges Q_i — choose the localized ones R_i ⊆ U₀ when base = Q₀! Then T ⊆ U₀. Hmm circular but suggests a finite closed “core” exists: Define U := Q₀ ∪ T where T from ORIGINAL private edges. We found: ∀i ∃ edge R_i ⊆ U with R_i ∩ Q₀ = {x_i}. Now recurse: the finite family 𝒬_U := {edges ⊆ U}: it’s a finite clutter, pairwise intersecting; is IT transversal-complete (within ground U)? If yes, then any edge Q ⊈ U: Q must intersect… consider transversal S of 𝒬_U inside U, minimal ⇒ S ∈ b(𝒬_U): is S a transversal of the WHOLE 𝒬? If yes ⇒ S ⊇ some edge of 𝒬 ⇒ edge ⊆ U ⇒ S itself… then b(𝒬_U) ⊆ 𝒬_U-ish. Then claim edges outside U can’t exist because: take y ∈ Q ∖ U for edge Q: Q is a transversal; … hmm I want: some transversal of 𝒬 within U that Q misses… if Q ∩ U is NOT a transversal-hitting-necessity… Every edge Q hits every edge R ⊆ U ⇒ Q ∩ U hits 𝒬_U ⇒ Q ∩ U ⊇ … Q∩U is a transversal of 𝒬_U ⇒ contains minimal transversal S ⊆ Q ∩ U of 𝒬_U. IF every minimal transversal of 𝒬_U is a transversal of ALL of 𝒬, then S ⊇ edge of 𝒬 (transversal-completeness, S finite) ⇒ edge E ⊆ S ⊆ Q ⇒ antichain ⇒ Q = E ⊆ U ✓!! That would give: every edge ⊆ U ⇒ finitely many edges ⇒ DONE. So need: construct finite U such that every minimal transversal of 𝒬_U (edges inside U) is a transversal of 𝒬 (hits every edge). Equivalently: for every S ⊆ U hitting all edges-inside-U, S hits all edges. Hmm, equivalently the edges inside U “dominate”: any edge Q of 𝒬 is “blocked”: every S ⊆ U hitting 𝒬_U hits Q.

How to build such U: closure process: Start U₀ = Q₀ ∪ (∪ private edges of Q₀). If some minimal transversal S ⊆ U₀ of 𝒬_{U₀} fails to hit some edge Q_S: then S ∪ … note S ∪ (Q ∖ …): hmm add witness edges to U: U₁ = U₀ ∪ Q_S ∪ …: does this terminate?? Each minimal transversal S of 𝒬_{U₀}: finitely many such S (finite ground); for each bad S, add an edge Q_S missed by S. Then U₁ finite. Recurse: might not terminate. Termination needs a rank/measure decreasing. Hmm.

Alternative: known theorem! “If a clutter of finite sets is pairwise intersecting and τ(𝒬)…” Let me think about König-type: In an intersecting family where every finite transversal contains an edge, consider minimum edge Q₀ size r and think of it as “τ = ν”… Actually here’s a classical result by … hmm “self-blocking clutters are finite”? Let me just try to prove directly with a clever compactness: Suppose 𝒬 infinite. Since all edges hit finite W (transversal): some w₁ ∈ W in infinitely many edges. 𝒬¹ = edges ∋ w₁, infinite. There’s an edge E₁ ∌ w₁ (else w₁ common to all edges: earlier showed ⇒ 𝒬 finite… wait showed via private property: if w₁ ∈ every edge then private edge for other elements impossible unless 𝒬 = {{w₁}}: private property comes from (ii)+(iii): does it? Private property: q ∈ Q ∈ 𝒬: is Q∖{q} non-transversal? Q is a MINIMAL transversal? From (ii): if Q∖{q} were a transversal (finite) it’d contain an edge ⊊ Q ✗ antichain. So yes: ∃ a-edge… ∃ EDGE Q’ with Q’ ∩ (Q∖{q}) = ∅?? no wait: Q∖{q} not a transversal means ∃ EDGE Q’’ with Q’’ ∩ (Q∖q) = ∅ (transversal = hits every edge — since hitting every edge ⟺ hitting every P(a)… within clutter world define transversal := hits every edge ✓). And Q’‘∩Q ≠ ∅ ⇒ Q’’∩Q = {q} ✓ private edges exist ✓.) Good. So w₁ ∉ E₁ (edge). All infinitely many edges of 𝒬¹ hit E₁: some w₂ ∈ E₁, 𝒬² := {Q ∈ 𝒬¹: w₂ ∈ Q} infinite; w₂ ≠ w₁. {w₁, w₂}: is it a transversal? If yes: contains an edge: {w₁},{w₂},{w₁,w₂} ∈ 𝒬: {w₁} can’t (E₁ ∌ w₁… {w₁} edge would intersect E₁ ✗). {w₂} edge: then all edges ∋ w₂… but w₂-private? If {w₂} ∈ 𝒬 then antichain ⇒ no other edge contains w₂ ✗ (𝒬² edges contain w₂ and are ≠ {w₂}? they contain w₁ too so ⊋ {w₂} ✗ antichain violation) ⇒ {w₂} ∉ 𝒬. {w₁,w₂} ∈ 𝒬? Possible! Then edges of 𝒬² ⊇ {w₁,w₂} ⇒ = {w₁,w₂} by antichain: only one ✗ infinite. So {w₁,w₂} not a transversal ⇒ ∃ edge E₂: E₂ ∩ {w₁,w₂} = ∅. Continue: 𝒬² edges hit E₂: w₃ ∈ E₂, 𝒬³ infinite, w₃ ∉ {w₁,w₂}. Inductively: W_k = {w₁..w_k}, 𝒬^k infinite edges ⊇ W_k, W_k not transversal (same argument: if W_k transversal, contains edge E ⊆ W_k; but infinitely many edges ⊇ W_k ⊇ E: antichain ⇒ ≤ 1 such edge… edges of 𝒬^k contain W_k ⊇ E and are edges ⇒ each equals E ✗ infinite) ⇒ ∃E_k disjoint from W_k ⇒ extend. So we get W_k growing forever, with infinitely many edges containing W_k for each k. In particular edges of unbounded size. STILL no contradiction from (i)-(iii) alone… unless: consider the INFINITE set W_∞ = {w₁, w₂, …}. Hmm, each E_k is an edge disjoint from W_k. E_k hits W_∞? E_k must hit edges of 𝒬^{k+5}… E_k ∩ Q ≠ ∅ for Q ⊇ W_m: intersection could be outside W_m. Hmm.

So (i)-(iii) alone may be insufficient?! Wait, maybe infinite transversal-complete intersecting clutters DO exist. Let me attempt construction matching the above pattern: edges 𝒬 = {B_k} ∪ {C_k} where B_k = {w_1,…,w_k, b_k} hmm need pairwise intersecting ✓ via w₁? but need edges avoiding w₁ (E₁)… Design: ground = {w_i} ∪ {e_i}…: Let me define edges: A_k = {w₁,…,w_k, u_k} (u_k fresh), and E_k = … must intersect all A_j: for j ≥ k+1: contain some w_… E_k should avoid W_k = {w₁..w_k} but hit A_j ∀j: hit A_j for j ≤ k: via u_j! So E_k ⊇ {u_1?..}: E_k must hit A_1 = {w₁, u₁}: avoid w₁ ⇒ u₁ ∈ E_k. A_2: w₂ ∉ ⇒ u₂ ∈ E_k (if k ≥ 2). So E_k ⊇ {u₁,…,u_k} ∪ {hit A_j for j > k: w_{k+1}? allowed! E_k ∩ W_k = ∅ but w_{k+1} allowed}: E_k = {u₁,…,u_k, w_{k+1}}: hits A_j, j>k via w_{k+1}? A_j ∋ w_{k+1} iff j ≥ k+1 ✓. E_k ∩ E_m (k<m): u₁ ∈ both ✓. Antichain: A_k vs A_j: u’s distinguish ✓; E_k ⊆ E_m? E_k ∋ w_{k+1}, E_m (m>k) ∋ w_{m+1}, u_{k+1} ∈ E_m, w_{k+1} ∈ E_m? no (E_m ∩ W_m = ∅ ∋ w_{k+1}) ⇒ E_k ⊄ E_m ✓; E_m ⊄ E_k (u_m ∉ E_k… |E_m| > |E_k| ✓). A vs E: w₁ ∈ A_k ∉ E_j ⇒ A_k ⊄ E_j; E_j ⊆ A_k? u₁ ∈ E_j, u₁ ∈ A_k iff k=1: E_j ⊆ A_1 = {w₁,u₁}: E_j ∋ u₁, w₂-free… E_1 = {u₁, w₂} ⊄ A₁ ✓. OK antichain ✓ pairwise intersecting: A_k ∩ E_j: j ≥ k: u_k ∈ E_j? E_j ⊇ {u₁..u_j} ∋ u_k ✓. j < k: w_{j+1} ∈ E_j and w_{j+1} ∈ A_k (k ≥ j+1) ✓. A∩A via w₁ ✓, E∩E via u₁ ✓. Now transversal-completeness: is every finite transversal ⊇ some edge? Test T = {w₁, u₁}: hits A_k ✓ (w₁), hits E_k ✓ (u₁). T ⊇ edge? A₁ = {w₁,u₁} = T ✓✓ oh A₁ IS {w₁,u₁}. Hmm so that T fine. Test T = {w₁, u₁,…}: minimal transversals: must hit all A’s and E’s. Candidates: contains w₁ or not. If w₁ ∈ T: need hit E’s: E_k’s: common element? ⋂E_k = {u₁}? E₁={u₁,w₂}, E₂={u₁,u₂,w₃}: ⋂ = {u₁}. If u₁ ∈ T: T ⊇ {w₁,u₁} = A₁ ✓ good. If u₁ ∉ T: hit E₁ via w₂; hit E₂ via u₂ or w₃; … infinitely many E_k with elements {u₁..u_k, w_{k+1}}: T finite ∌ u₁: for large k, T∩E_k ⊆ {u₂,…} ∪ {w}: T finite: let s = max index of any u_i or w_i in T; E_k for k ≥ s+1: E_k ∩ T ⊆ {u_i: i ≤ s} which are in E_k ✓ possible: if T ∋ u₂: hits all E_k, k ≥ 2 ✓ and E₁ via w₂. So T = {w₁, w₂, u₂}: check: A₁={w₁,u₁}: hit w₁ ✓. A_k ✓. E₁ = {u₁,w₂}: ✓ w₂. E_k (k≥2) ∋ u₂ ✓. Is T ⊇ edge? Edges: A₁ ⊄, A₂ = {w₁,w₂,u₂}: A₂ ⊆ T ✓!! T = A₂ exactly. Hmm lucky. If w₁ ∉ T: hit A_k ∀k: A_k = {w₁,..w_k,u_k}: T finite: hit A_k for k > max-index via w_j some j ≤ … A_k ∩ T ⊆ {w₂..w_j, u_?}: T ∋ w₂ would hit all A_k k ≥ 2 ✓, A₁ = {w₁,u₁} needs u₁ ∈ T. So T = {w₂, u₁, …} hit E_k: u₁ ✓ all. T = {w₂, u₁}: transversal?? A₁ ✓ u₁; A_k, k≥2 ✓ w₂; E₁ = {u₁,w₂} ✓; E_k ✓ u₁. Contains an edge? Edges ⊆ {w₂,u₁}: E₁ = {u₁, w₂} ✓!! = T. Wow. Hmm the family seems suspiciously self-blocking. Try T = {w₂, u₂, u₁}? contains E… E₁ = {u₁,w₂} ⊆ T ✓. Try to find transversal avoiding edges: minimal transversals seem to coincide with edges here?! Let me find ALL minimal transversals: T minimal transversal. Case w₁ ∈ T: subcase u₁ ∈ T ⇒ ⊇ A₁ ⇒ minimal ⇒ = A₁ ✓. Subcase u₁ ∉ T: must hit each E_k within {u₂,…,u_k, w_{k+1}}: cofinally many constraints; T finite ⇒ ∃ u_j ∈ T minimal j ≥ 2 hitting all E_k k ≥ j; E_k for k < j: hit via u_{i≤k} (i≥2) or w_{k+1}: Let j = min{i ≥ 2: u_i ∈ T}: then E_1..E_{j-1}: E_k ∩ T ⊆ {w_{k+1}} ∪ {u_2..u_k}∩T(=∅ for k<j) ⇒ w_{k+1} ∈ T for k = 1..j−1 ⇒ T ⊇ {w₁, w₂,…,w_j, u_j} = A_j ∪ {w₁}… A_j = {w₁..w_j, u_j} ⊆ T ⇒ T = A_j ✓. Case w₁ ∉ T: hit A_k: A₁ ⇒ u₁ ∈ T ⇒ E’s ✓ all hit (u₁ ∈ every E_k ✓). A_k (k ≥ 2): via w_{2..k} or u_k: T finite: let’s see minimal: need hit all A_k: cofinal: some w_j ∈ T (j ≥ 2) hits A_k ∀ k ≥ j; A_k for 2 ≤ k < j: need w_{≤k} or u_k: if j = min{i≥2: w_i ∈ T}: for k < j: A_k ∩ T ⊆ {u_k} ⇒ u_k ∈ T for k = 2..j−1… plus u₁: T ⊇ {u₁,…,u_{j−1}, w_j} = E_{j−1} ✓ ⇒ T = E_{j-1}. OR no w_j ∈ T at all: then hit A_k via u_k ∀k ≥ 2: u_k ∈ T ∀k: infinite ✗. So ALL minimal transversals are edges!! And the clutter is INFINITE, pairwise intersecting, antichain, transversal-complete!!!

Oh no. So (i)-(iii) [+private] admits infinite examples. Then the corresponding A = ∪ d_Qℤ would be a counterexample to the theorem UNLESS the greedy dynamics can’t produce it / it’s not truly self-consistent as a SEQUENCE. Let’s double check whether A = ∪{Q ∈ 𝒬} d_Qℤ with this 𝒬 (mapped to primes) satisfies A = C(A) ∩ [N,∞) — yes by transversal-completeness (m ∈ C(A) ⟺ P(m) hits every P(a): ⟺ P(m) hits every edge? P(a) ⊇ edge; hitting all edges ⟹ hits all P(a) ✓; conversely hitting all P(a) ⊇ hitting all edges since d_Q^t ∈ A ✓). So A self-consistent. But is A the greedy orbit of a_1 = min A? NOT necessarily! The greedy orbit A’ from a_1 satisfies A’ = C(A’) ∩ [a_1, ∞) — a possibly different self-consistent set. The theorem being true means greedy orbits are special among self-consistent sets: greedy can’t produce these infinite-𝒬 sets. Why not? Because greedy makes choices based on finite past: at each step picks min compatible with past — when the “constraint horizon” is finite, smaller numbers get in. In infinite-𝒬 A: consider large m = d_Q for huge edge Q: for m to be EXCLUDED for all m’ < … the greedy would include some m’ ∉ A: e.g. numbers whose prime set hits all CURRENT terms but not future edges. Whether greedy generates such A depends on order of appearance… but the theorem says NO greedy sequence is non-periodic; our reduction says greedy sequence enumerates C(its own terms) — so the real task: show the greedy’s 𝒬 is finite USING dynamics/minimality further. Extra leverage: TIMING: the edges’ witnesses appear at specific times; when edge structure “not yet enforced”, greedy picks small numbers, potentially violating the intended infinite structure. Concretely: consider time n, terms a_1..a_n known ⊆ A. a{n+1} = min m > a_n hitting P(a_i) i ≤ n. Now suppose 𝒬 infinite. There are edges Q with d_Q enormous (≥ some function) and Q “not implied” by early terms: there’s m with P(m) hitting all P(a_i), i ≤ n, but P(m) missing edge Q. When could such m < a_{n+1}… but a_{n+1} = min over those m by DEFINITION — and we showed it lands in A always. So automatically for the true sequence: every m ∈ (a_n, a_{n+1}) fails to hit some P(a_i), i ≤ n. The dynamics ARE consistent with A being weird?! But wait — we PROVED a_{n+1} = min C ∩ (a_n,∞) rigorously. And A = C ∩ [a_1, ∞). Both consistent. So potential counterexample: does there exist an actual integer sequence whose orbit realizes infinite 𝒬? The theorem claims no. So additional number-theoretic/greedy constraints must forbid it. What’s special: the greedy at step n only knows a_1..a_n. For the orbit to equal A (avoid all non-A compatibles-with-past), every m ∉ A must be blocked by some EARLIER term at the time it’s passed: m ∉ A, m > a_n, m < a_{n+1}: ∃i ≤ n: gcd(m, a_i) = 1. I.e., the constraint that excludes m must have APPEARED before m. m ∉ A ⟺ some edge Q_m misses P(m) — need witness term a_i < m (roughly) with P(a_i) ∩ P(m) = ∅; P(a_i) ⊇ … every term’s prime set contains an edge; to have gcd(a_i, m) = 1 need an edge (in fact a full prime set P(a_i)) disjoint from P(m). So requirement: for every non-element m ≥ a_1: ∃ term a < m (a ∈ A, a ≤ some a_n < m… precisely a_i with i ≤ n where a_n < m < a_{n+1}, so a_i ≤ a_n < m: a_i < m ✓) with gcd(a_i, m) = 1. NEW CONDITION (v): ∀m ≥ a_1(=N), m ∉ A ⇒ ∃a ∈ A, a < m, gcd(a, m) = 1. THIS is the greedy’s fingerprint. Now show: (i)-(v) ⇒ 𝒬 finite.

In our infinite example: find violating m: want m ∉ A (P(m) misses some edge) but every a ∈ A with a < m has gcd(a,m) > 1. Take m := product over primes hitting “all edges with small d_Q”: e.g., m = ∏(primes q with q “prominent”) choose m’s prime set = T̃ = a transversal of {edges Q: d_Q ≤ x} but missing some huge edge Q, with m relatively small vs the witnesses of Q: The witnesses that exclude m are terms a coprime to m: P(a) ∩ T̃ = ∅: P(a) ⊇ edge disjoint from T̃: edges disjoint from T̃ have d ≥ (big) since T̃ hits all small edges ⇒ a ≥ d ≥ big. If m < that: contradiction with (v). So (v) forces: for every finite non-transversal T̃ that hits all edges with d_Q ≤ x: hmm let’s formalize: define f(x) = … For any m: m ∉ A with all “blockers” huge ⇒ contradiction if m < min blocker. Blocker a needs P(a) ∩ P(m) = ∅ and P(a) ⊇ some edge Q’ with Q’ ∩ P(m) = ∅ and conversely if edge Q’ disjoint from P(m) exists then a := d_{Q’}^k… but need a < m: smallest a ∈ A with P(a) ∩ P(m) = ∅: at least… a is divisible by primes… a ≥ 2^{ω}, roughly a ≥ product of an edge disjoint… no: P(a) disjoint from P(m), P(a) ⊇ edge E: E∩P(m) = ∅? E ⊆ P(a) disjoint from P(m) ✓ automatically E ∩ P(m) = ∅. And a ≥ d_E ≥ min{d_E : E edge, E ∩ P(m) = ∅}. So (v) ⟺ ∀ m ≥ N with P(m) non-transversal: min{d_E: E ∩ P(m) = ∅} ≤ … ∃ multiple of some such d_E below m — precisely ∃ edge E, E ∩ P(m) = ∅, with a multiple of d_E in [N, m): i.e. d_E ≤ m − 1… roughly d_E < m (need multiple ≥ N and < m: if d_E ≥ N: a = d_E < m ok; if d_E < N: multiples ≥ N… fine d_E ≤ m−1 suffices? need some multiple in [N, m): since d_E ≤ m−1 < m and can take multiple in [N, N + d_E) ⊆ [N, m) iff N + d_E ≤ m hmm roughly). SO CONDITION (v’): for every m ≥ N whose prime set misses some edge, there’s an edge E with E ∩ P(m) = ∅ and d_E ≲ m. Now kill infinite 𝒬: take huge edge Q* (d_{Q} > (N + max(d_E over E ∈ 𝒬_small))·∏small…). Build m: P(m) = T := transversal of 𝒬 ∖ {edges ⊇ … } hmm want T* hitting every edge EXCEPT missing Q, with ∏T small — but that’s exactly the T-trick: T := ∪{x ∈ Q}(R_x ∖ {x}) hits every edge except Q (T ∩ Q* = ∅… wait earlier T-trick: base edge Q₀, T misses exactly Q₀). Yes: with base Q, T hits all edges ≠ Q and T ∩ Q* = ∅. Set m = ∏{q ∈ T} q^{c}: m ∉ A (P(m) = T misses Q: is T ∪… P(m) = T: T transversal? no—misses Q ⇒ m ∉ A ✓). Then (v’) ⇒ ∃ edge E disjoint from T with d_E < m: but the ONLY edge disjoint from T is… edges disjoint from T: any edge E with E∩T = ∅ ⇒ (T-trick property: every edge ≠ Q* meets T) ⇒ E = Q* ⇒ d_{Q} < m ≈ ∏T. So condition: d_{Q} < (∏ T)^{c}: but T is built from private edges R_x of Q, which themselves might be huge!! Size of T comparable to Q’s private edges: no bound. Hmm. So need to choose Q* with SMALL private-complement… circular. Refine: we don’t need T to miss exactly only Q; need: m ∉ A and all edges disjoint from P(m) are huge relative to m. Choose P(m) := W = a finite transversal… no m must be ∉ A: P(m) NOT ⊇ any edge, but P(m) hits every SMALL edge (forcing disjoint edges huge), and ∏P(m) small relative to those huge disjoint edges. Does such P(m) exist when 𝒬 infinite? Take increasing exhaustion 𝒬_k = {Q: d_Q ≤ D_k}. Want finite set X: (a) X hits all edges in 𝒬_k; (b) X ⊉ any edge; (c) some edge disjoint from X exists?? No wait: (b) X ⊉ edge ⟺ m ∉ A ⟺ X misses SOME edge (no! X ⊉ any edge means X isn’t a transversal? A = {P(m) ⊇ edge}: m ∉ A ⟺ P(m) ⊉ every edge. But P(m) could still be a transversal without containing an edge?? Transversal-completeness says finite transversal ⊇ edge! So P(m) ⊉ any edge ⟺ P(m) not transversal ⟺ some edge disjoint… no: transversal means hits all edges; not transversal ⟺ ∃ edge E with E ∩ P(m) = ∅ ✓.) So m ∉ A ⟺ ∃edge disjoint from P(m). Then (v’) requires SOME disjoint edge with d_E < m. To contradict: want X with: X ∩ E = ∅ ⇒ d_E > (∏X)·big. I.e., X hits all edges up to size (∏X)·big. “Self-bounded transversal of an initial segment”: Since 𝒬 infinite with d_Q → ∞ (distinct squarefree… d_Q distinct? edges distinct ⇒ d_Q distinct ✓ (squarefree products of distinct prime sets) so d_Q → ∞ along any infinite subfamily… yes infinitely many edges ⇒ unboundedly large d_Q). Want: finite X, not ⊇ edge, hitting every edge E with d_E ≤ Φ(∏X) where Φ(t) = t·C₀ (C₀ = N+1 whatever). Does it exist whenever 𝒬 infinite? Greedy build: X must hit all small edges; cheapest: X = an edge Q₁ (hits everything!) — but X ⊉ edge ✗. So X = “almost transversal”: hits all edges except some huge ones. Existence: take minimal transversal (= edge) Q₁ and swap one element with private structure: X := (Q₁ ∖ {x}) ∪ Y where Y hits {E: E∩Q₁ = {x}, d_E ≤ Φ} =: family ℰ. If we can pick Y = one prime y ∈ ⋂-ish… choose Y = union over E ∈ ℰ_small of one element ≠ x: X = (Q₁∖{x}) ∪ {choices}: then edges disjoint from X: must have E ∩ Q₁ ⊆ {x}: E ∈ “x-private family”, and E missed Y-choices ⇒ E ∉ ℰ ⇒ d_E > Φ. And X ⊉ edge? Need check: edge inside X possible ✗ problem. Hmm—if X ⊇ some edge then X transversal, m ∈ A, no contradiction; so must ensure no edge within X. But transversal-completeness pushes edges inside finite transversals… X hits all edges except huge ⇒ X ∪ {x} ⊇ transversal ⇒ contains edge… The m-trick requires m ∉ A i.e. no edge ⊆ X while X hits all small edges. Alternatively use m with P(m) = X where X = T from T-trick (T misses exactly Q): guaranteed no edge ⊆ T? Edge E ⊆ T: E∩Q* = ∅ ⇒ E = Q* ✗ (Q∩T = ∅, Q⊄T since… Q* ∩ T = ∅ and Q* nonempty ⇒ Q* ⊄ T unless… E ⊆ T disjoint from Q: but every edge ≠ Q MEETS T ✓ that’s fine — meeting T doesn’t prevent ⊆ T! Edge E ⊆ T: E meets T ✓ trivially; E ≠ Q* (disjoint from Q? E ⊆ T, T∩Q = ∅ ⇒ E∩Q* = ∅ ⇒ E doesn’t hit Q* ⇒ E not an edge (pairwise intersecting!!) ✓✓). GREAT: pairwise intersecting ⇒ no edge inside T (any edge must meet Q, but T ∩ Q = ∅). So m with P(m) = T: m ∉ A ✓, and edges disjoint from P(m): only Q* ⇒ (v’) ⇒ ∃ a ∈ A, a < m, a coprime to m ⇒ P(a) ∩ T = ∅, P(a) ⊇ some edge E, E ∩ T = ∅ ⇒ E = Q* ⇒ a is a multiple of d_{Q}… wait P(a) ⊇ Q just means Q* ⊆ P(a): a divisible by d_{Q} ⇒ a ≥ d_{Q}. So condition: d_{Q} ≤ a < m ≤ (∏T)^{1}·(adjust to ≥ N: m = ∏T·(N?) let’s just say m = ∏T times small factor or power to exceed N; minimal such m ≍ ∏T if ∏T ≥ N). CONTRADICTION IF: d_{Q} > ∏T. So the greedy forces: for EVERY edge Q, d_{Q} ≤ ∏ T(Q) where T(Q) = ∪_{x∈Q}(R_x ∖ {x}) built from private edges. Still circular (T built from other edges). But now bootstrap: define d₁ = min d over edges. Hmm, we need: infinitely many edges ⇒ some edge has d_{Q} > ∏T(Q) for SOME choice of private edges: choose Q = edge with maximal d?? none. Choose cleverly: Actually improve the T-trick: we don’t need private edges; ANY finite set T with: T ∩ Q* = ∅, T hits every edge except Q* and except edges with d > m. Recall (v’) contradiction needs only: all edges disjoint from T have d ≥ m ≈ ∏T, and at least one edge disjoint from T exists (to ensure m ∉ A) — Q*. So goal: find finite T with: (α) ∃ edge disjoint from T; (β) every edge disjoint from T has d > C·∏T (C = const like N·2). If 𝒬 infinite, find such T: Consider W = finite transversal… Take T := a minimal transversal of the finite family 𝒬_D := {edges: d ≤ D}, chosen to not be a transversal of whole 𝒬, with ∏T ≤ B(D) and then need all edges disjoint from T to have d > C·B(D): edges disjoint from T have d > D (T hits 𝒬_D); so need D ≥ C·B(D) where B(D) = min over suitable T of ∏T. Need: minimal transversals of 𝒬_D that are not full transversals, of product ≤ D/C, for some large D. Since 𝒬 infinite: is there such? Suppose for contradiction (of the theorem) 𝒬 infinite. For each D let τ(D) := min{∏T : T hits all edges of 𝒬_D, T contains no… T not ⊇ full-edge? automatically if T disjoint from some edge…}. Hmm, alternatively: minimal transversals of 𝒬_D: since transversal-completeness applies to 𝒬 (not 𝒬_D)… A minimal transversal T of 𝒬_D either contains an edge of 𝒬 — an edge E ⊆ T: E hits all… fine possible — or is disjoint from some edge (of 𝒬∖𝒬_D necessarily, d > D) — or hits all edges without containing one: impossible (transversal-complete). Case analysis: if some minimal transversal T of 𝒬_D with ∏T ≤ D/C is disjoint-type ⇒ contradiction via (v’) ✓. So theorem-contradiction forced unless: ∀D: every “cheap” (∏ ≤ D/C) minimal transversal of 𝒬_D contains a full edge. Note the cheapest transversal of 𝒬_D: any edge E ∈ 𝒬_D works, ∏ = d_E ≥ d_min: so τ(D) ≤ d_min =: d₁ (min over all edges, fixed!). So there EXISTS transversal of 𝒬_D with product ≤ d₁ ≤ D/C for D large. Take T := minimal (by inclusion) sub-transversal of that cheap one: ∏T ≤ d₁. For D > C·d₁: if T disjoint from some edge ⇒ contradiction. Else T contains an edge (transversal-completeness needs T transversal of ALL 𝒬: hmm T hits 𝒬_D only; “T contains an edge or T misses an edge”: T finite: either T is a transversal of 𝒬 ⇒ contains an edge, or ∃ edge disjoint?? NO: T could MEET some edges outside 𝒬_D partially fail: not transversal means ∃edge E: E ∩ T = ∅ — “not hitting” = disjoint ✓ (hitting is binary!). ✓✓ So: T hits all of 𝒬 ⇒ ⊇ edge, OR ∃E with E∩T = ∅ and then d_E > D (since T hits 𝒬_D) > C·d₁ ≥ C·∏T ⇒ contradiction via m-trick ✓. THEREFORE: for every D, every minimal transversal T of 𝒬_D with ∏T ≤ d₁… must contain an edge E₀ ∈ 𝒬 with… E₀ ⊆ T, d_{E₀} ≤ ∏T ≤ d₁ ⇒ d_{E₀} = d₁-ish: E₀ ∈ 𝒬_{d₁}: fine, no contradiction yet!! e.g. T = E (an edge) always. Push further: we need cheap transversals of 𝒬_D that DON’T contain edges. Use the swap trick on cheap edge E₀ (d = d₁): x ∈ E₀: consider (E₀ ∖ x) ∪ Y: must hit ℰ_x^D := {E ∈ 𝒬_D: E ∩ E₀ = {x}}: pick Y = {y_E: E ∈ ℰ_x^D, y_E ∈ E∖{x}}: ∏ can blow up with D. To keep product bounded need to hit ℰ_x^D cheaply without x. If for some x and arbitrarily large D that’s expensive ⇒ means the x-private family is “wide”: many edges pairwise sharing only x… they pairwise intersect: E, E’ ∈ ℰ_x: E∩E’ ∋ x possibly more. If E∩E’ = {x} exactly for many: sunflower-ish with core x. Hmm. This is getting long but feels like the right track. Let me reconsider — maybe there’s a slicker characterization: combine (v’) with structure: define for each edge Q: μ(Q) := min{ ∏T : T finite, T ∩ Q = ∅, T hits every edge except Q…} — no.

ALTERNATIVELY revisit: maybe simpler original-solution path exists. Let me reconsider the problem from scratch with the knowledge A = C∩[a₁,∞): maybe directly prove: the set of minimal elements… Actually here’s possibly THE key simplification: (v) with m := multiples of things. Let me use (v) smartly: For any finite set X of primes with X ⊉ edge (i.e., some edge disjoint from X — via completeness) hmm same as before.

Let me define: for finite prime set X, β(X) := min{a ∈ A: gcd(a, ∏X) = 1} = min{multiple of d_E ≥ N over edges E with E∩X = ∅}… ≥ min such d_E and ≤ max(N + d_E, …): essentially β(X) ≍ min{d_E : E ∩ X = ∅} (up to +N… careful: multiples of d_E that are ≥ N: the smallest is ≤ N + d_E; but also that multiple k·d_E might share primes with X via k! choose k = prime > everything: a = k·d_E coprime to X: a ≤ d_E·(prime > max(N/d_E, X)): ugh bounds get multiplicative constant; safer: if d_E ≥ N: a = d_E itself ∈ A, coprime to X ✓ a = d_E.) Since all but finitely many edges have d_E ≥ N: fine.

restated: ∀m ≥ N: [∃edge disjoint from P(m)] ⇒ ∃a ∈ A, a<m, P(a)∩P(m) = ∅. Contrapositive-ish quantitative: min{d_E: E∩P(m) = ∅} ≤ hmm the a from (v): P(a) ⊇ some edge E’ and P(a)∩P(m) = ∅ ⇒ E’ ∩ P(m) = ∅ and a ≥ d_{E’} hmm no a is multiple of… P(a) ⊇ E’ means d_{E’} | rad(a) so d_{E’} ≤ a ✓. So (v) ⇒ (v’’): ∀ finite prime set X, if some edge is disjoint from X, then min{d_E : E∩X = ∅} < min{m ≥ N: P(m) = X-ish}: take m = smallest integer ≥ N with P(m) ⊆ X… we want m with P(m) = X exactly? P(m) ⊆ X suffices for “edges disjoint from P(m) ⊇ edges disjoint from X” hmm we need the disjoint edge for m: E∩P(m) = ∅: if P(m) ⊆ X then E∩X = ∅ ⇒ E∩P(m) = ∅ ✓. And conclusion: ∃edge E, E∩P(m) = ∅ — but E∩P(m) = ∅ is weaker than E∩X = ∅!! The blocker a of m is coprime to m, not to ∏X. So to force E∩X = ∅ need P(m) = X exactly. Smallest m ≥ N with P(m) = X: m = ∏X·(small adjust) — if ∏X ≥ N: m = ∏X ✓ take X with ∏X ≥ N wlog (can always add a prime or use power: P(m) = X allows m = ∏X^k! m = (∏X)^k ≥ N with k minimal: m ≤ max(N·∏X, ∏X): so m ≤ N·∏X always ✓). CONCLUSION (v*): for every finite prime set X such that ∃edge disjoint from X: min{d_E: E ∩ X = ∅} ≤ N·∏X. NICE clean. [Proof: m = (∏X)^k ∈ [N, N∏X], m ∉ A since edge disjoint ⇒ ∃a<m coprime to m, P(a) ⊇ edge E, E∩X = ∅, d_E ≤ rad a ≤ a < m ≤ N∏X ✓.]

Now prove: (i) pairwise intersecting + (ii) transversal-complete + (iii) antichain + (v) ⇒ 𝒬 finite. Use (v) iteratively to build small transversals: Start X₀ = ∅: some edge disjoint (any) ⇒ ∃E₁: d_{E₁} ≤ N. Then X₁ := E₁? hmm we want eventually X with NO edge disjoint from it but X ⊉… Let’s grow: maintain X_k with ∏X_k bounded and X_k… Strategy “greedy hitting with cost control”: Let X₀ = ∅. While ∃ edge disjoint from X_k: by (v): ∃ edge E with E∩X_k = ∅, d_E ≤ N·∏X_k. Set X_{k+1} := X_k ∪ E: ∏X_{k+1} ≤ N(∏X_k)². Terminates? Only if at some point no edge disjoint from X_k ⇒ X_k transversal ⇒ contains an edge… X_k = union of edges E₁ ∪ E₂ ∪ …: does it terminate?? If 𝒬 infinite maybe never terminates — but each step adds an edge DISJOINT from all previous ones!! E_{k+1} ∩ X_k = ∅ ⊇ E_k: DISJOINT EDGES — CONTRADICTS pairwise intersecting (i) at k=2 already!!! WAIT: X₁ = E₁; if some edge disjoint from X₁ = E₁: but all edges intersect E₁ (pairwise intersecting)! ⇒ NO edge disjoint from E₁ ⇒ E₁… so E₁ (any single edge) already hits all edges. Transversal := hits all edges ✓ every edge is a transversal, fine, that’s consistent: process terminates at X₁ immediately. Of course. The subtlety: “transversal of edges” vs constraints P(a): same thing (shown). So hitting all edges is easy (any edge); the issue is only: finitely vs infinitely many edges, i.e., whether ∪d_Qℤ eventually periodic. Right, so the real target isn’t “cheap transversal”, it’s bounding d_Q for ALL edges: Claim FINAL: every edge Q has d_Q ≤ some bound M₀ ⇒ finitely many edges (d_Q distinct… d_Q ≤ M₀ finitely many values, and edge ↔︎ d_Q injective ✓). To bound d_Q, use (v) with X := T(Q) from T-trick: X ∩ Q = ∅, X hits every other edge ⇒ only edge disjoint from X is Q ⇒ (v): d_Q ≤ N·∏X = N·∏T(Q). Need ∏T(Q) bounded independent of Q. T(Q) = ∪{x∈Q}(R_x∖{x}): private edges R_x: d{R_x} can be bounded?? Apply (v) to X = (Q∖{x}) ∪ hmm: private edge existence with SIZE BOUND: X := Q∖{x}: is some edge disjoint from X? Yes: the private R_x (exists by minimality (ii)+(iii)) ⇒ (v*) applies: ∃ edge E, E∩(Q∖{x}) = ∅, d_E ≤ N·∏(Q∖{x}) ≤ N·d_Q. E∩Q = {x} or ∅: ∅ impossible (i) ⇒ E is a private edge for x with d_E ≤ N·d_Q =: cheap private edges!! So R_x choose with d_{R_x} ≤ N d_Q ⇒ ∏T(Q) ≤ ∏x d{R_x} ≤ (N d_Q)^{|Q|} ⇒ d_Q ≤ N·(N d_Q)^{|Q|}: useless (RHS bigger). Need sharper. Hmm.

Improve: instead of covering all other edges via private-edge-union, cover via: X := (E₀ ∖ {x}) ∪ (R_x ∖ {x’})…: Think about what configurations make ALL edges huge except structure… Let’s think about d₁ = min edge d, edge E₁. All edges hit E₁. For x ∈ E₁ private edges cheap: d ≤ N d₁. Build finite family ℱ = {E₁} ∪ {cheap private edges R_x, x ∈ E₁}: all d ≤ N d₁. Now consider X := ∪ℱ-based traps: any edge Q: hits E₁ and each R_x. Suppose 𝒬 infinite ⇒ ∃ edges with d huge. Take Q huge. Q∩E₁ =: S ∋ some x. Q hits R_x for all x ∈ E₁… Consider X = (E₁ ∖ S) ∪ ∪_{x ∈ S}(R_x ∖ {x}) =: trap(S): X ∩ Q: Q∩E₁ = S disjoint from E₁∖S ✓; Q∩(R_x∖x): could be nonempty ✗ hmm Q might intersect R_x elsewhere. So trap misses Q only if Q∩R_x ⊆ {x} for x ∈ S… not guaranteed. Since infinitely many edges and finitely many “intersection patterns” with the FINITE set V := ∪ℱ?? Edges’ intersections with V: finitely many possible traces! Infinitely many edges share the same trace τ ⊆ V. Take two edges Q, Q’ with same trace, both huge… For the trap: want finite X with exactly-one-disjoint-edge… but with same trace τ, X built ⊆ V can’t separate Q from Q’: any X ⊆ V disjoint from Q is disjoint from Q’ too: then (v) gives: some edge disjoint from X with d ≤ N∏X: that edge could be ANY of the trace-τ class: gives min over class ≤ N∏X: fine, no contradiction. To separate, X must include primes outside V distinguishing Q, Q’. Use Q’ itself: X := (stuff) ∪ (Q’ ∖ Q)? Let’s design: want X: X∩Q = ∅, every other edge meets X, then d_Q ≤ N∏X ⇒ need ∏X small: X must hit the infinite family cheaply while avoiding Q. Candidate X: E₂ ∖ Q ∪ …where E₂ cheap edge: E₂∩Q =: S₂ ≠ ∅. X ⊇ E₂∖S₂ hits all edges that meet E₂ outside S₂. Remaining: edges E with E∩E₂ ⊆ S₂: these contain points of S₂ ⊆ Q. For each s ∈ S₂: family ℰ_s = edges ∋ s with E∩E₂ ⊆ S₂… cover them avoiding Q: they all contain s: but s ∈ Q can’t use. Use cheap private edges at s: R_s (d ≤ Nd₁, R_s∩E₂ = {s}): every E ∈ ℰ_s hits R_s: if E∩R_s ⊆ {s}… recursion again — Sunflower feeling: the obstruction to cheap-avoidance of Q is a sequence of nested “private” families all pinned at points of Q. Since Q is finite, and each pin-point spawns bounded-cost expansions… Let me formalize as: define reachable cheap family ℛ: E₁ ∈ ℛ; if F ∈ ℛ cheap… all edges obtained as “cheap private” recursively, all with d ≤ bound? Each application multiplies bound by N and product… R_x from (v) applied to X = F∖{x}: d ≤ N·∏(F∖x) < N·d_F. Depth k: d ≤ N^k d₁^{2k}-ish: any FIXED depth = bounded ⇒ finite family. Plan: show some fixed depth D(|Q|?) suffices to trap Q, with |Q| bounded?? |Q| unbounded ⇒ bound not uniform ⇒ fail. Hmm but maybe |Q| ≤ log₂ d_Q and cost like d₁^{2{|Q|}} vs d_Q: could win if d_Q ≥ tower… not clean.

Time to reconsider: maybe there’s a fundamentally better idea I’m missing. Let me search memory harder: I’m now fairly convinced this is ISL 2024 N7 (appeared as “IMO 2025 shortlist”?) — the user says IMO 2026 P6 — plausibly this is a genuinely recent very hard problem (P6). Known solution idea (from AoPS threads I might recall): uses the notion of “eventually the set of a_n equals union of arithmetic progressions”, proof via extremal/minimality: consider the smallest B such that infinitely many terms divisible by B?? Let me try: Let b = smallest positive integer ≥ 2 such that A contains infinitely many multiples… every a ∈ A… hmm. Alternatively: consider s := liminf-style: define the set R of integers r ≥ 2 with rℤ ⊆ … i.e., such that all sufficiently large multiples of r are in A: A ⊇ d_Qℤ: edges give such r = d_Q. Conversely if all large multiples of r ∈ A… then P(r-multiples) ⊇ hmm r with rad r ⊇ edge. So R = {r: rad(r) ⊇ some edge}. min R = d₁…

Completely different, “dynamic” approach to periodicity: Show directly ∃T, L with a_{n+T} = a_n + L for all n, via: consider L = some fixed highly divisible number; claim m ∈ A ⇒ m + L ∈ A for m ≥ N: need gcd(m + L, a) > 1 ∀a: use edge inside P(m): E ⊆ P(m), if all primes of E divide L then E ⊆ P(m+L) ✓!! So need L divisible by ∏(all primes in all edges) — again finiteness of ∪edges. And converse m+L ∈ A ⇒ m ∈ A same. So SUFFICES: the union of all edges is finite, equivalently: finitely many primes q that “matter”: q ∈ edge means: q necessary somewhere. And actually we need exactly: ∃ finite prime set 𝒫 with: ∀m ≥ N: m ∈ A depends only on P(m) ∩ 𝒫… equivalent to edges ⊆ 𝒫.

Now, primes in edges: q ∈ Q edge ⇒ private edge R with R∩Q = {q} ⇒ ∃ elements a ∈ A with q | a and (P(a) = R-based) hmm. Density viewpoint: A = ∪ d_Qℤ, lower density ≥ 1/N ≥ 1/a₁… and: consider density contributed: if edges with huge d: they contribute tiny density but must EXIST because of… why would greedy ever “create” them? Use (v) NOW with clever single X to bound ALL edges: Let 𝒬_bad = edges containing a prime > Y (Y := N·∏(primes ≤ some Z)…). Hmm let me try assuming: only need bound PRIMES in edges, then separately bound d? If all edges ⊆ 𝒫 = primes ≤ Y: edges are subsets of finite set ⇒ FINITE ✓✓!! So just bound primes in edges!! Claim: q ∈ ∪edges ⇒ q ≤ some bound. Take edge Q ∋ q, q huge. Private edge R_q: R_q∩Q = {q} — CHEAP version: (v) with X = Q∖{q}: ∃edge E_q, E_q∩Q = {q}, d_{E_q} ≤ N·∏(Q∖{q}) = N d_Q/q. But q ∈ E_q ⇒ d_{E_q} ≥ q ⇒ q ≤ N d_Q / q ⇒ q² ≤ N d_Q. Interesting: every prime q ∈ Q: q ≤ sqrt(N d_Q). Iterating on E_q: q ∈ E_q, d_{E_q} ≤ N d_Q/q ⇒ apply same: q ≤ sqrt(N d_{E_q}) ≤ sqrt(N·N d_Q/q) ⇒ q ≤ (N²d_Q/q)^{1/2} ⇒ q³ ≤ N² d_Q. Iterate k times: q^{k+1} ≤ N^k d_Q ⇒ q ≤ N^{k/(k+1)} d_Q^{1/(k+1)} → N as k → ∞?!! Let me verify the iteration: Claim: if edge F ∋ q then ∃ edge F’ ∋ q with d_{F’} ≤ N·∏(F ∖ {q}) = N d_F / q. Proof: X := F∖{q}: is some edge disjoint from X? Need private edge for q in F: exists iff F minimal transversal: ✓ all edges are minimal transversals ((ii),(iii) ⇒ shown). So ∃edge disjoint from F∖{q} ⇒ (v) ⇒ ∃edge E: E∩(F∖q) = ∅, d_E ≤ N·∏(F∖q). E∩F ≠ ∅ (i) ⇒ E∩F = {q} ⇒ q ∈ E ✓. Set F’ := E: d_{F’} ≤ N d_F/q, q ∈ F’. ✓!! Iterate: d_{F_{k+1}} ≤ N d_{F_k}/q with F₀ = Q. Then d_{F_k} ≤ N^k d_Q / q^k. But d_{F_k} ≥ q (q ∈ F_k) ⇒ q ≤ N^k d_Q/q^k ⇒ q^{k+1} ≤ N^k d_Q for ALL k ⇒ if q > N: LHS/RHS → ∞ ⇒ contradiction ⇒ q ≤ N!!! WOW. So every prime in every edge is ≤ N = a₁. Hence ∪edges ⊆ {primes ≤ a₁} finite ⇒ 𝒬 finite (antichain of subsets of finite set) ⇒ A = [N,∞) ∩ ∪{Q}d_Qℤ, periodic mod M := ∏{p ≤ a₁} p. DONE!!! Beautiful — and (v) was the key greedy fingerprint. Let me double check (v*) derivation and the whole chain rigorously.

Check base facts: 1. For all i < j: gcd(a_j, a_i) > 1 (definition). So all terms pairwise non-coprime. ✓ 2. C := {m ∈ ℤ_{≥2}: gcd(m, a_n) > 1 ∀n}. Every term ∈ C ✓. 3. a_{n+1} = min(C ∩ (a_n, ∞)): “≤”: every element of C∩(a_n,∞) satisfies the step condition ⇒ a_{n+1} ≤ min. “≥”: a_{n+1} ∈ C∩(a_n,∞). ✓ Also need C∩(a_n,∞) nonempty: contains multiples of a₁ ✓ hmm multiples of a₁: gcd(ka₁, a_i) ≥ gcd(a₁,a_i) > 1 ✓ for i ≠ 1… i = 1: gcd = a₁ > 1 ✓. So C ⊇ a₁ℤ ∩ [2,∞): also guarantees sequence well-defined (min exists: also need at each finite stage m exists ✓ same argument). ✓ 4. A := {a_n: n ≥ 1} = C ∩ [a₁, ∞): ⊆ ✓ (a₁… every term ≥ a₁ ✓, in C ✓). ⊇: c ∈ C, c ≥ a₁: a_n → ∞ so c ∈ [a_n, a_{n+1}) for some n or c = … if c = a₁ ✓; if a_n < c: and c ≤ … take n max with a_n < c: c ≤ a_{n+1} but a_{n+1} = min C∩(a_n,∞) ≤ c ⇒ c = a_{n+1} ✓.

Definition: for m ≥ 2, m ∈ C ⟺ gcd(m, a) > 1 ∀a ∈ A (constraints from all terms = from A ✓).
Hypergraph: P(m) = set of primes of m. m ∈ C ⟺ P(m)∩P(a) ≠ ∅ ∀a∈A.
Transversal := finite prime set X with X∩P(a) ≠ ∅ ∀a ∈ A. So m ∈ C ⟺ P(m) transversal.
𝒬 := minimal transversals (minimal under ⊆ among transversals; every transversal contains a minimal one since finite). Antichain ✓ by def.
Every edge Q ∈ 𝒬: m := (∏Q)^k with k s.t. m ≥ a₁: m ∈ C, m ≥ a₁ ⇒ m ∈ A. ✓ (P(m) = Q transversal.)
Pairwise intersecting: Q, Q’ ∈ 𝒬: m = (∏Q)^k ∈ A, m’ = (∏Q’)^j ∈ A ⇒ gcd… m’ ∈ A ⊆ C? m’ ∈ A and m ∈ A: gcd(m, m’) > 1?? m ∈ C ⇒ gcd(m, a) > 1 ∀a ∈ A ∋ m’ ✓ ⇒ Q∩Q’ ≠ ∅ ✓. Also: every edge Q hits P(a) ∀ a ∈ A (transversal def) ✓.
m ∈ C ⟺ P(m) ⊇ some edge? (⇐): P(m) ⊇ Q: X∩P(a): Q∩P(a) ≠ ∅? Q is transversal ⇒ ✓ so m ∈ C ✓. (⇒): P(m) transversal ⇒ contains minimal transversal = edge ✓. So C = {m ≥ 2: ∃Q ∈ 𝒬, ∏Q | rad… d_Q | m-multiples}: m ∈ C ⟺ ∃Q: every prime of Q divides m ⟺ d_Q | m (d_Q squarefree) ✓. So C = ∪_Q d_Qℤ ∩ [2,∞)… (d_Qℤ ∩ [2,∞)).
Minimality/private edges: Q edge, q ∈ Q: Q∖{q} not transversal ⇒ ∃a* ∈ A: P(a*) ∩ (Q∖{q}) = ∅. (Not necessarily an edge disjoint… transversal def is vs P(a), a ∈ A. Let me redo with X-disjointness in terms of elements a ∈ A rather than edges — cleaner!) Redefine everything vs A: X transversal ⟺ hits P(a) ∀a∈A. “X not transversal” ⟺ ∃a ∈ A: gcd-wise P(a)∩X = ∅.
(v*): Let X finite prime set, NOT a transversal, with ∏{q∈X} q =: x̄. Then ∃a ∈ A: P(a)∩X = ∅ and a ≤ a₁·x̄. Proof: let k ≥ 1 minimal with x̄^k ≥ a₁: m := x̄^k. Note m ≤ a₁·x̄ (if k = 1: m = x̄ ≤ a₁x̄ ✓; if k ≥ 2: x̄^{k−1} < a₁ ⇒ m < a₁x̄ ✓). Hmm wait also need… X could be empty: x̄ = 1: m = 1? Handle X ≠ ∅ (fine, we apply with nonempty). Also X nonemptiness: k minimal with x̄^k ≥ a₁ exists since x̄ ≥ 2. m ≥ a₁, P(m) = X not transversal ⇒ m ∉ C ⇒ m ∉ A. Since m ≥ a₁ and m ∉ A: m ≠ a₁; let n := max{n: a_n < m} (a₁ < m… a₁ ≤ m and m ≠ a₁ ⇒ a₁ < m ✓ n exists). Then a{n+1} > m (def of max… a_{n+1} ≥ m and ≠ m since m ∉ A ⇒ > m ✓). By step-minimality: m > a_n, m < a_{n+1} ⇒ m fails: ∃i ≤ n: gcd(m, a_i) = 1 ⇒ P(a_i) ∩ X = ∅ and a_i ≤ a_n < m ≤ a₁x̄. Set a := a_i ✓✓. [This is the greedy fingerprint — used the ORIGINAL definition, good, since C-min characterization would also do: a_{n+1} = min C∩(a_n,∞) > m ⇒ m ∉ C — no wait that just says m ∉ C, which we know; the fingerprint needs the finite-stage definition: m must be excluded by some i ≤ n: exactly original def ✓.]
Key Lemma: every prime q in every edge satisfies q ≤ a₁. Proof: Let q ∈ Q ∈ 𝒬. Claim: ∀k ≥ 0 ∃ a_k… define sequence of ELEMENTS instead of edges (since (v) yields elements a ∈ A, and P(a) ⊇ some edge but the edge might not contain q!! CAREFUL: step 12/13 give a ∈ A with P(a)∩(F∖{q}) = ∅; then P(a) ⊇ edge E; E∩F ⊆ {q}; E∩F ≠ ∅?? E hits P(m’) for all elements… E is transversal: hits P((∏F)^j) = F (F edge ⇒ (∏F)^j ∈ A) ⇒ E∩F ≠ ∅ ⇒ E∩F = {q} wait E∩F ⊆ P(a)∩F? NO: E ⊆ P(a), so E∩(F∖q) ⊆ P(a)∩(F∖q) = ∅ ⇒ E∩F ⊆ {q}, nonempty ⇒ = {q} ⇒ q ∈ E ✓ and d_E ≤ rad(a) ≤ a ≤ a₁·∏(F∖{q}).] So Lemma iteration: F₀ = Q; given edge F_k ∋ q: X := F_k∖{q}: not a transversal? Need witness: F_k minimal transversal ⇒ ∃a ∈ A with P(a)∩(F_k∖{q}) = ∅ ⇒ X not transversal ✓ (if F_k = {q}: X = ∅: then… handle: if {q} is an edge: every element divisible by q: then A ⊆ qℤ: density ≤ 1/q; gaps ≤ a₁ ⇒ 1/q ≥ 1/a₁ ⇒ q ≤ a₁ ✓ done separately. Hmm gaps: a_{n+1} − a_n ≤ a₁ ✓ proven? a_{n+1} ≤ min multiple of a₁ > a_n ≤ a_n + a₁ ✓.) General case |F_k| ≥ 2: (v) ⇒ ∃edge F_{k+1} ∋ q, d_{F_{k+1}} ≤ a₁·∏(F_k∖{q}) = a₁ d_{F_k}/q. So d_{F_k} ≤ a₁^k d_Q/q^k ∀k (as long as sizes ≥ 2; if some F_k = {q}: handled: q ≤ a₁ ✓ since then… {q} edge case above ✓). Also d_{F_k} ≥ q… need F_k ≥ 2 elements: d ≥ q·(another prime) ≥ 2q, fine ≥ q suffices: q ≤ d_{F_k} ≤ a₁^k d_Q / q^k ⇒ (q/a₁)^k ≤ d_Q/q ∀k. If q > a₁: LHS → ∞: contradiction ⇒ q ≤ a₁ ✓✓✓.
Conclusion: all edges

Continue writing up conclusion & verify details.

All edges ⊆ {primes ≤ a₁} =: 𝒫₀ (finite). Edges = antichain of subsets of 𝒫₀ ⇒ finitely many edges. Let M := ∏_{p ∈ 𝒫₀… or p ≤ a₁} p. Then for m ≥ 2: m ∈ C ⟺ d_Q | m for some edge Q; each d_Q | M ⇒ C is a union of residue classes mod M: m ∈ C ⟺ (m mod M) ∈ S₀ := {r: ∃Q, d_Q | r-ish}: precisely d_Q | m depends on m mod d_Q, and d_Q | M ⇒ depends on m mod M ✓. So: ∀m: m ∈ C ⟺ m + M ∈ C (both ≥ 2). A = C ∩ [a₁, ∞): for m ≥ a₁: m ∈ A ⟺ m + M ∈ A ✓ exact periodicity.
Final: Let T := #(A ∩ (a₁, a₁ + M]) = #{c ∈ C: a₁ < c ≤ a₁ + M}. By periodicity, for ANY m ∈ A: #(A ∩ (m, m + M]) = T? Since A∩(m, m+M] shifts… count of C-elements in (x, x+M] ∩ [a₁,∞): for x ≥ a₁: every residue appears exactly once in (x, x+M] ⇒ count = |S₀-ish| =: T constant, independent of x ≥ a₁… more carefully: #{c ∈ C: x < c ≤ x+M} = #{residues r mod M with r-class ⊆ C} =: T for any x ≥ 2 (since c ranges over one full residue system and membership depends only on residue; need all elements ≥ 2: x ≥ 2 ✓). And all these c ≥ x ≥ a₁ ⇒ in A. So for x = a_n: A∩(a_n, a_n+M] has exactly T elements, i.e., a_{n+1},…,a_{n+T} ≤ a_n + M and a_{n+T+1} > a_n + M… the T elements of A in (a_n, a_n+M] are exactly a_{n+1} < … < a_{n+T}; largest: is a_n + M ∈ A? a_n ∈ A ⇒ a_n + M ∈ A ✓ (periodicity, a_n ≥ a₁). So a_n + M ∈ A∩(a_n, a_n+M] and it’s the largest element there ⇒ a_{n+T} = a_n + M ✓ for ALL n ≥ 1. Set L := M. Also T ≥ 1 ✓ (a₁ + M… a₁ ∈ A ⇒ a₁+M ∈ A ⇒ T ≥ 1). DONE.

Now double-check the delicate points:

Well-definedness of sequence: at each step candidate multiples of a₁ exist ⇒ min exists ✓. a_{n+1} ≤ a_n + a₁ (smallest multiple of a₁ exceeding a_n is ≤ a_n + a₁ and is a valid candidate) — actually gap bound only needed for the {q}-edge case. Alternatively handle {q} edge case via (v*) too? If Q = {q}: means every a ∈ A divisible by q. Want q ≤ a₁. Use gap bound: multiples of a₁ ≥ a₁… hmm A contains… All elements of A divisible by q; a₁ ∈ A ⇒ q | a₁ ⇒ q ≤ a₁ ✓✓✓ trivial! Great, no density needed: a₁ ∈ A.

Actually similarly maybe simplify main lemma?? q ∈ Q: not nec. q | a₁. Fine.

In iteration step 14: F_k edge with q ∈ F_k, |F_k| ≥ 2. X = F_k∖{q} nonempty ✓ (x̄ ≥ 2 needed for (v): ✓). X not transversal: since F_k is a minimal transversal ⇒ ∃a ∈ A, P(a)∩X = ∅ ✓. (v) gives a ∈ A: P(a)∩X = ∅, a ≤ a₁·∏X. Then P(a) contains an edge E (a ∈ A ⊆ C ⇒ P(a) transversal ⇒ ⊇ minimal transversal E). E∩X = ∅. E∩F_k ≠ ∅ (E transversal hits P(m) for m = (∏F_k)^j ∈ A; P(m) = F_k) ⇒ E∩F_k = {q}. d_E = ∏E ≤ ∏P(a) = rad(a) ≤ a ≤ a₁·∏X = a₁·d_{F_k}/q ✓. Set F_{k+1} = E. If |F_{k+1}| = 1: F_{k+1} = {q} ⇒ {q} transversal ⇒ q | every element of A ⇒ q | a₁ ⇒ q ≤ a₁, stop. Else iterate. If never terminates: d_{F_k} ≤ (a₁/q)^k d_Q → 0 if q > a₁: contradiction (d ≥ 2). So q ≤ a₁. Wait also if q ≤ a₁ we’re done anyway; assume q > a₁ derive contradiction: iteration never gives {q}? If it gives {q} ⇒ q ≤ a₁ contradiction ⇒ so |F_k| ≥ 2 always ⇒ d_{F_k} ≥ 2 but → 0 contradiction ✓. Clean.
(v*) proof detail: X finite nonempty set of primes, not a transversal. x̄ = ∏X ≥ 2. k := min{j ≥ 1: x̄^j ≥ a₁}. m := x̄^k. Claim m ≤ a₁·x̄: if k = 1: m = x̄ < a₁·x̄ ✓ (a₁ ≥ 2). If k ≥ 2: x̄^{k−1} < a₁ ⇒ m = x̄^{k−1}·x̄ < a₁x̄ ✓. m ≥ a₁ ✓. P(m) = X not transversal ⇒ ∃a’ ∈ A: gcd(m, a’) = 1 ⇒ m ∉ C ⇒ m ∉ A. Case m = a₁? m ∉ A but a₁ ∈ A ⇒ m ≠ a₁ ⇒ m > a₁. n := max{i: a_i < m} well-defined (a₁ < m, a_i → ∞). All i ≤ n: a_i < m. a_{n+1} ≥ m, and m ∉ A = set of terms ⇒ a_{n+1} > m. Definition of a_{n+1} (smallest integer > a_n with gcd cond ∀i ≤ n): since a_n < m < a_{n+1}, m must violate: ∃i ≤ n: gcd(m, a_i) = 1 ⇒ P(a_i)∩X = ∅, a_i < m ≤ a₁x̄. But hold on: strictly “a_i ≤ a₁·x̄”: a_i < m ≤ a₁x̄ ⇒ a_i < a₁x̄, fine state a < a₁·∏X. ✓

Hmm wait, one more check in (v*) usage in (b): a ≤ a₁·∏X where ∏X = d_{F_k}/q. Then d_E ≤ rad a ≤ a < a₁·d_{F_k}/q ✓.

The claim “every term ≥ a₁”: a_n strictly increasing from a₁ ✓.
Transversal definition requires nonempty X? Edges nonempty since they hit P(a₁) ≠ ∅… minimal transversal E: E∩P(a₁) ≠ ∅ ⇒ nonempty ✓. Also every P(a) is itself finite and a transversal ⇒ minimal transversals exist and 𝒬 ≠ ∅ ✓.
Step 3 needs C∩(a_n,∞) ≠ ∅: multiples of a₁ ✓ (gcd(ka₁, a_i) ≥ gcd(a₁, a_i) > 1 — need gcd(a₁, a_i) > 1 ∀i: i > 1 by pairwise property; i = 1: a₁ ≥ 2 ✓).
Pairwise property gcd(a_i, a_j) > 1 for i ≠ j: from definition (for j > i, a_j chosen with gcd(a_j, a_i) > 1) ✓.
In 14, need m = (∏F_k)^j ∈ A: choose j with (∏F_k)^j ≥ a₁; m ∈ C since P(m) = F_k edge = transversal ✓ ⇒ m ∈ A ✓.
Periodicity conclusion for all n ≥ 1: a_n ≥ a₁ ✓, a_n + M ∈ A. And counting: #{c ∈ C: a_n < c ≤ a_n + M}: membership in C for integers ≥ 2 is periodic mod M: is it? m ∈ C ⟺ ∃Q ∈ 𝒬: d_Q | m. This is a condition on m mod M since d_Q | M ✓ (d_Q squarefree product of primes ≤ a₁, M = ∏{p ≤ a₁} p). Also must double check: C defined only for m ≥ 2; residue class containing 0 mod something… For counting in (a_n, a_n + M]: all integers ≥ 2 ✓. T := #residues r ∈ {0,…,M−1} such that ∃Q: d_Q | r… “d_Q | m ⟺ d_Q | (m mod M)” since d_Q | M ✓. So count in any window of length M of integers ≥ 2 = T, constant ✓. The elements of A in (a_n, a_n+M] are exactly a{n+1} < ⋯ < a_{n+T} (A∩(a_n,∞) = {a_{n+1}, a_{n+2},…} consecutive) ✓ and max is a_n + M since a_n + M ∈ A ∩ (a_n, a_n+M] and it’s the top endpoint… the largest element of A in the window: it IS in the window and any element ≤ a_n + M: so a_{n+T} = largest = must equal a_n + M? The largest element of A∩(a_n, a_n+M] is some c ≤ a_n+M; we know a_n + M ∈ A and ∈ window ⇒ largest = a_n + M ✓ ⇒ a_{n+T} = a_n + M ✓✓.

Also T ≥ 1 and positive integers T, L = M ✓.

One more subtle point: 𝒬 could contain infinitely many edges even within finite prime set? No: subsets of finite set ✓ finite. But actually we don’t even need finiteness of 𝒬 explicitly—just that all d_Q | M ⇒ periodic. Even simpler: m ∈ C depends only on P(m) ∩ {p ≤ a₁}? m ∈ C ⟺ ∃edge ⊆ P(m); edges ⊆ small primes ⇒ condition depends on gcd(m, M): m ∈ C ⟺ gcd(m,M)… ⟺ ∃Q: d_Q | m ✓ depends on m mod M ✓ fine.

Potential issue: the minimal transversal E in (b) obtained inside P(a): E is a minimal transversal of the FULL family {P(a): a ∈ A} — yes 𝒬 defined that way ✓.
Edge case in Key Lemma start: Q edge, q ∈ Q, |Q| = 1: Q = {q}: then q | all elements ⇒ q | a₁ ⇒ q ≤ a₁ ✓ (mentioned).
Also need: in iteration, F_{k+1} ∋ q and F_{k+1} is an EDGE (minimal transversal) so the next iteration’s “X not transversal” witness exists ✓.
The strictly-decreasing chain: d_{F_{k+1}} ≤ a₁ d_{F_k}/q; if q > a₁ then d decreases strictly by factor < 1 ⇒ eventually d_{F_k} < q: but q | d_{F_k} ⇒ d ≥ q: contradiction ✓ (cleaner than →0).

All checks pass. Also should verify the problem statement’s claim “for every positive integer n” — our proof gives exactly that ✓.

Also should I double-check with the 15-example: a₁ = 15, N=15. Edges should be ⊆ primes ≤ 15: we found A = {m ≥ 15… wait a₁ = 15, A = C∩[15,∞): C = multiples of 6, 10, or 15: edges {2,3},{2,5},{3,5}: primes ≤ 15 ✓. M = 2·3·5·7·11·13 = 30030, T = count per period; smaller valid M: 30. Our M = ∏p≤a₁ works (any M with d_Q | M ✓). Sequence: 15,18,20,… a_{n+T} = a_n + 30030 ✓ consistent with computed a_{n+8} = a_n + 30 ✓ (T for 30030: 8·1001).

Sanity check (v) logic once more on example: X = {2}: transversal? Element 15 ∈ A: P∩X = ∅ ⇒ not transversal. (v): ∃a ∈ A, odd, a ≤ 15·2 = 30: a = 15 ✓.

Key lemma sanity: could an edge contain prime 7 in this example? {2,3},{2,5},{3,5} no ✓. Suppose hypothetically edge {2,7}: private for 7: edge E with E∩{2,7} = {7}: d_E ≤ 15·2 = 30, 7 ∈ E, odd: d = 7·odd ≥ 21: E = {7,3}? etc. plausible in general; iteration kills q > a₁ only. Right: lemma bounds q ≤ a₁ = 15, 7 ≤ 15: allowed. Fine—edges within primes ≤ 15 possible in principle; actual 𝒬 determined by sequence.

Now also double check the very first reduction once more: a_{n+1} = min{m > a_n: gcd(m, a_i) > 1 ∀ 1 ≤ i ≤ n}. Prove a_{n+1} ∈ C: for i ≤ n ✓; for i = n+1: gcd(a_{n+1}, a_{n+1}) = a_{n+1} > 1 ✓; for i > n+1: gcd(a_i, a_{n+1}) > 1 since a_i chosen with gcd to all previous incl. a_{n+1} ✓. So every term ∈ C ✓.

And A = C∩[a₁,∞) as terms. Write-up plan:

Notation & preliminaries: - P(m), pairwise coprimality property, C, sequence = enumeration of C∩[a₁,∞). Lemma 1 (structure of C): transversals, minimal transversals 𝒬, C = {m: d_Q | m for some Q}. - every P(a), a∈A is a transversal; minimal transversals are finite nonempty; every transversal contains minimal one. - key facts: each edge hits each P(a); powers of ∏Q are in A; minimality witnesses. Lemma 2 (greedy fingerprint (v*)). Lemma 3 (key): every prime in every edge ≤ a₁. Conclusion: periodicity mod M = ∏_{p≤a₁}p, define T, finish.

Careful definitions: “transversal”: finite nonempty set X of primes with X∩P(a) ≠ ∅ for all a ∈ A. Note P(m) for m ≥ 2 nonempty finite ✓.

Minor point: minimal transversal within P(a): P(a) is transversal, subsets partially ordered, finite ⇒ contains minimal transversal (minimal in the whole poset of transversals? Take X ⊆ P(a) transversal with |X| minimum among sub-transversals… then is X minimal globally? X minimal means no proper subset of X is a transversal ✓ that’s it. Minimal transversal = transversal with no proper sub-transversal. Any finite transversal contains a minimal one by induction ✓.)

Antichain: distinct minimal transversals incomparable ✓ by definition.

Intersecting: for edges Q,Q’: (∏Q)^j ∈ A for suitable j ⇒ Q’ ∩ Q ≠ ∅ ✓.

Now write the final solution cleanly. Also include well-definedness of the infinite sequence (given a₁, all a_n exist) — the problem says “Suppose that…” so sequence given; but min exists anyway; can remark.

Also confirm Lemma 2 statement: X finite nonempty non-transversal set of primes ⇒ ∃ term a ∈ A with gcd(a, ∏X) = 1 and a < a₁·∏X.

Then Lemma 2’: ∃ edge E with E∩X = ∅ and ∏E < a₁∏X. (E ⊆ P(a), rad(a) ≤ a.)

Lemma 3 proof: q prime, q ∈ Q₀ ∈ 𝒬, assume q > a₁. If Q₀ = {q}: q | a₁ ⇒ q ≤ a₁ contra. So build F_k: F₀ = Q₀, given edge F_k ∋ q, F_k ≠ {q}: X := F_k ∖{q} finite nonempty; not transversal (minimality of F_k); Lemma 2’ ⇒ edge F’ with F’∩(F_k∖{q}) = ∅, d_{F’} < a₁ d_{F_k}/q < d_{F_k} (q > a₁). F’ hits F_k (as F’ is a transversal and (∏F_k)^j ∈ A) ⇒ q ∈ F’. If F’ = {q}: contra as before; else set F_{k+1} := F’. Infinite strictly decreasing sequence of positive integers d_{F_k} — contradiction. (Or: d_{F_k} ≥ 2q… just strictly decreasing positive integers bounded below, infinite descent ✓.) Hence q ≤ a₁.

Then edges ⊆ {p ≤ a₁}; d_Q | M := ∏_{p ≤ a₁} p… need at least one prime ≤ a₁ exists: 2 ≤ a₁ ✓ M well-defined. Then C periodic mod M on ℤ≥2: m ∈ C ⟺ ∃Q: d_Q|m; d_Q | M. So m ∈ C ⟺ m+M ∈ C for m ≥ 2 ✓.

T := number of residues r mod M such that d_Q | r for some Q ∈ 𝒬 (equivalently gcd-condition). T ≥ 1 (r = 0). Every interval (x, x+M] with x ≥ a₁ contains exactly T elements of A. a_n + M ∈ A. ⇒ a_{n+T} = a_n + M.

Detail: A∩(a_n, ∞) = {a_{n+1} < a_{n+2} < …} ✓ since A = all terms, increasing. So (a_n, a_n+M] ∩ A = {a_{n+1},…,a_{n+T}} where T elements, and largest = a_n + M ⇒ a_{n+T} = a_n + M. Set L = M, T as above; independent of n ✓.

Also handle n such that… all n ≥ 1 ✓ since a_n ≥ a₁ ∀n.

Also should note T finite ✓.

Ready to write. Length: moderate-long but complete. Include verification example not needed. Also add remark that sequence well-defined.

One more potential subtlety: Lemma 2 requires m ∉ A ⇒ ∃i ≤ n coprime — but what if m ≤ a_1? Handled: m ≥ a₁ and m ≠ a₁ ⇒ m > a₁ ✓ since m ∉ A ∋ a₁. Good.

Another subtlety: In Lemma 2, need X nonempty so x̄ ≥ 2 and k exists ✓; also m = x̄^k could equal some a_i? m ∉ A shown, fine.

Also in Lemma 3, “F_k minimal ⇒ F_k∖{q} not transversal”: if F_k∖{q} were transversal it’d be a proper sub-transversal contradicting minimality ✓; note F_k∖{q} nonempty needed for Lemma 2 (F_k ≠ {q} ✓).

Also E∩F_k ≠ ∅ argument: E is a transversal ⇒ E∩P(a) ≠ ∅ for a = (∏F_k)^j ∈ A, P(a) = F_k ✓.

Double-check “every transversal contains an edge” not needed anywhere else — we used C = ∪d_Qℤ: needs “P(m) transversal ⇒ contains minimal transversal”: P(m) finite ✓ done via finite descent ✓.

Write final answer now.