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This question is inspired by Subgroup of Order $n^2-1$ in Symmetric Group $S_n$ when $n=5, 11, 71$.

Find all nonnegative integers $m$ and $n$ such that $m!+1=n^2$.

We know that $(m,n)=(4,5)$, $(m,n)=(5,11)$, and $(m,n)=(7,71)$ are solutions. Are there more solutions? Are there only finitely solutions?

Batominovski
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1 Answers1

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You can see other solutions here:

https://oeis.org/A025494

But for proving it is infinity there isn't any sloutions the problem is open.

See:https://en.wikipedia.org/wiki/Brocard%27s_problem

Taha Akbari
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