Reused writeup on Math-6126 (gpt pro 5.2)

Erdős Problem 1072 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1072

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1072.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1072.tex.

1) FORMAL RESTATEMENT For a prime p, define f(p) := the least integer n >= 1 such that p divides $n! + 1$. Wilson's theorem implies $p-1$! ≡ -1 (mod p), hence p divides $(p-1)!+1$, so f(p) is always defined and f(p) <= p-1.

Question A: Are there infinitely many primes p with f(p) = p-1? Question B: Does f(p)/p -> 0 for almost all primes p? (Here "almost all primes" means: #{p <= x : the property fails} = o(pi(x)) as x -> infinity.)

2) QUICK LITERATURE/CONTEXT CHECK The problem statement itself notes: - f(p) <= p-1 by Wilson. - It is conjectured there are infinitely many primes with f(p)=p-1, but also a heuristic that the number of such primes <= x is o(x/log x). No other external results are assumed here.

3) ATTACK PLAN Two directions are natural: (A) Structural: understand solutions of n! ≡ -1 (mod p) for n < p, and derive constraints on where they can occur. (B) Data: compute f(p) for small primes to sanity-check the range of behaviors. I provide two problem-specific lemmas that give (i) the universal Wilson upper bound and (ii) a sharp characterization of the case when the witnessing factorial is smaller than p.

4) WORK Lemma 1072.1 (Wilson bound). For every prime p, f(p) <= p-1. Proof. Wilson's theorem gives $p-1$! ≡ -1 (mod p). Therefore p divides $(p-1)!+1$, so the minimum n with p | $n!+1$ satisfies n <= p-1. QED.

Lemma 1072.2 (When n! < p, the divisibility forces p = n!+1). Let p be prime and suppose p | $n!+1$ for some integer n >= 1. If n! < p, then p = n!+1 and moreover f(p) = n. Conversely, if n!+1 is itself prime p, then f(p)=n. Proof. If p | $n!+1$ then p <= n!+1. If also n! < p then n!+1 <= p, so n!+1 = p. Now assume p = n!+1. For any m < n we have m!+1 < n!+1 = p, hence p cannot divide m!+1. Therefore f(p)=n. The converse implication is exactly the same argument read in reverse. QED.

Lemma 1072.3 (Complementary factorial identity and a consequence). Let p be an odd prime and let n be an integer with 0 <= n <= p-1. Then n! * $p-1-n$! ≡ $-1$^$n+1$ (mod p). In particular, if 1 <= n <= p-2 and p | $n!+1$, then $p-1-n$! ≡ $-1$^n (mod p), and also the complementary product $n+1$$n+2$...$p-1$ ≡ 1 (mod p). Proof. Start from Wilson: $p-1$! ≡ -1 (mod p). Also $p-1$! = n! * product_{k=n+1}^{p-1} k. Rewrite the complementary product by changing variables: for j=1,...,p-1-n, we have p-j ≡ -j (mod p), hence product_{k=n+1}^{p-1} k = product_{j=1}^{p-1-n} $p-j$ ≡ $-1$^$p-1-n$ * $p-1-n$! (mod p). Therefore -1 ≡ $p-1$! ≡ n! * $-1$^$p-1-n$ * $p-1-n$! (mod p). Multiply both sides by $-1$^$p-1-n$. Since p is odd, $-1$^$p-n$ = $-1$^$n+1$. This gives n! * $p-1-n$! ≡ $-1$^$n+1$ (mod p). Now assume additionally that p | $n!+1$, so n! ≡ -1 (mod p). Substituting into the identity yields $-1$ * $p-1-n$! ≡ $-1$^$n+1$ (mod p), so $p-1-n$! ≡ $-1$^n (mod p). Also, from $p-1$! ≡ n! * product_{k=n+1}^{p-1} k and $p-1$! ≡ -1 and n! ≡ -1, we get $-1$ ≡ $-1$ * product_{k=n+1}^{p-1} k (mod p), hence product_{k=n+1}^{p-1} k ≡ 1 (mod p). QED.

5) VERIFICATION (small cases) I computed f(p) for all primes p < 500 by direct modular multiplication. - Primes p < 500 with f(p) = p-1 are: 2, 3, 5, 13, 17, 31, 37, 41, 53, 73, 89, 97, 101, 107, 113, 151, 157, 167, 173, 181, 197, 211, 223, 229, 241, 281, 283, 313, 331, 337, 349, 353, 373, 409, 421, 433, 439, 457, 487. Count: 39 out of 95 primes < 500. - Examples where f(p) is much smaller than p: f$103$=6, f$269$=9, f$479$=15. These computations are only sanity checks; they do not settle Questions A or B.

6) FINAL UNRESOLVED (i) Strongest proved partial result here: f(p) is always defined and f(p) <= p-1 (Lemma 1072.1). If p divides $n!+1$ with n! < p then necessarily p = n!+1 and f(p)=n (Lemma 1072.2). Also, any solution n < p-1 forces explicit complementary congruences (Lemma 1072.3). (ii) First gap (crisp): Prove or disprove that there are infinitely many primes p for which no n <= p-2 satisfies n! ≡ -1 (mod p) (equivalently f(p)=p-1). Separately, prove or disprove that for almost all primes p there exists some n = o(p) with n! ≡ -1 (mod p). (iii) Top 3 next moves (concrete): 1. Try to show that for most primes p there exists n <= p^alpha (for some fixed alpha<1) with n! ≡ -1 (mod p), by studying distribution of factorial residues modulo p. 2. Search for a mechanism forcing f(p)=p-1 along an infinite family (e.g. show that n!+1 avoids 0 mod p for all n<=p-2 in that family). 3. Extend computations to much larger primes to estimate the frequency of f(p)=p-1 and the empirical distribution of f(p)/p. (iv) Minimal counterexample structure: To refute Question B (f(p)/p -> 0 for almost all primes), one would need a fixed c>0 and a positive-density set of primes p for which n! ≠ -1 (mod p) for every n <= c p.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.