gpt pro 5.2 attempt on Math-6126

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.