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$m!+1=n^2$; $m,n$ positive integers.
I've tried bringing the $1$ to the other side and using the difference of squares, but it seems like an endless analysis, is there a shorter way to solve it?

commie trivial
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Luca Piersante
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    Welcome. Generally, taking residues modulo $k$ for some (several) good choice(s) of $k$ helps in solving these equations. You’re expected to show some effort here, but as a freebie I’ll say that taking residues modulo $n>1$ ($n=1$ clearly has no solutions) yields that, if $m\ge n$, the equation has no solutions, so without loss of generality you should consider the equation when $0<m<n$ and $n>1$. – FShrike Jul 30 '22 at 13:26
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    This is Brocard's problem. See this post as an example and the comments. For the only known solutions see OEIS. – Dietrich Burde Jul 30 '22 at 13:27

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