Are $2^{n} \pm 1$ and $n! \pm 1$ powerful (i.e. if $p ∣ m$ then $p^{2} ∣ m$ ) for only finitely many $n$ ?

Worked, still open.

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

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vela registry pull vfr_37aec80d874a0239

vela reproduce examples/erdos-problems

evidence

unverified AI candidates (2)

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

Let “powerful” mean **squarefull**: a positive integer $m$ such that for every prime (p\mid m) we also have (p^2\mid m).

llm-hunter · gpt pro 5.2 · unverified

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

AMS 11 · open (literature)

theorem erdos_936.variants.two_pow_add_one :
   answer(sorry) ↔ EventuallyNotPowerful (2 ^ · + 1)

n! = x^{2} - 1

n = 4, 5, 7

open