Let $A (x)$ count the number of composite $u < x$ such that $n! + 1 \equiv 0 (mod u)$ for some $n$ . Is it true that $A (x) \leq x^{o (1)}$ ?

Worked, still open.

number theory · open · formalized (Lean) · 0 attempts

use this record

vela registry pull vfr_37aec80d874a0239

vela reproduce examples/erdos-problems

evidence

unverified AI candidates (2)

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

It is an Erdős question (Erdős Problem #1073): if $A(x)$ counts the **composite** (u<x) for which (u\mid(n!+1)) for some $n$, Erdős asked whether [ A(x)\le x^{o(1)}\qquad(\text{equivalently }A(x)=o(x^\varepsilon)\text{ for every }\varepsilon>0). ] It appears as Problem **F*** in Hardy–Subbarao (2002), and is recorded i…

candidate solution ↗

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.

candidate solution ↗

formal

AMS 11 · open (literature)

theorem erdos_1073 :
    answer(sorry) ↔ ∃ (o : ℕ → ℝ), o =o[atTop] (1 : ℕ → ℝ) ∧ ∀ x, A x ≤ x ^ (o x)

formal-conjectures/1073.lean ↗

oeis

A256519 — Composites c for which an integer 1 < k < c exists such that (c-k)! == -1 (mod c).25,121,169,437,551,667,721,1037,1159,1273,1349,1403,1541,1769,1943,2209,2329,2363,2419,3071,3713,4087,5041,5111,7313,835

links

Wilson's theorem · reference

status