Are the only solutions to $n! = x^{2} - 1$ when $n = 4, 5, 7$ ?

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

Your equation [ n!=x^2-1 ] is equivalent to [ n!+1=x^2, ] i.e. “when is $n!+1$ a perfect square?” This is the **Brocard–Ramanujan equation** (often called **Brocard’s problem**). ([Wikipedia][1])

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_398 : answer(sorry) ↔ {n | ∃ m, n ! + 1 = m ^ 2} = {4, 5, 7}

formal-conjectures/398.lean ↗

oeis

A141399 — Positive integers k such that the distinct primes that divide k or k+1 form a set of consecutive primes. In other words, k is included if and only if k*(k+1) is contained in sequence A073491.1,2,3,5,8,9,14,15,20,24,35,80,125,224,384,440,539,714,1715,2079,2400,3024,4374,9800,12375,123200,194480,633555 A146968 — Brocard's problem: positive integers n such that n!+1 = m^2.4,5,7

links

ABC conjecture · reference

#936Are

2^{n} \pm 1

and

n! \pm 1

powerful (i.e. if

p ∣ m

then

p^{2} ∣ m

) for only finitely many

n

?A146968

status