For any prime $p$, let $f(p)$ be the least integer such that $f(p)!+1\equiv 0(\mathrm{mod}p)$.Is it true that there are infinitely many $p$ for which $f(p)=p-1$?Is it true that $f(p)/p\to 0$ for almost all $p$?

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

Let $p$ be prime and [ f(p):=\min{n\ge 1:\ n!\equiv -1 \pmod p}. ] By Wilson’s theorem ((p-1)!\equiv -1\pmod p), so $f(p)$ is always defined and (f(p)\le p-1).

llm-hunter · gpt pro 5.2 · unverified

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