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Suppose $N$ is a positive integer, we wish to find if there are finite or infinite many $N$ satisfying: $$ N! + 1 = k^2 $$ where $k$ is a positive integer.

In other words, $N! + 1$ is a perfect square.

We can find that $N = 4, 5, 7$ satisfied this by \begin{aligned} &4! + 1 = 25 = 5^2 \\ &5! + 1 = 121 = 11^2 \\ &7! + 1 = 5041 = 71^2 \end{aligned}

I have tried using computer to test $N$ up to $30000$ and still not find the next one. I would like to know if there are mathematical approach to show that, there does not exist the next $N$ satisfying $N! + 1$ is a perfect square. Thank you.

Yunxuan Zhang
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David Hui
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    Cf. Brocard's problem, OEIS – J. W. Tanner Jan 20 '25 at 15:39
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    Sorry I didn't know this is an unsolved problem. I have seen similar question in facebook, and tried but with no luck (not exactly the same question but similar). Anyway you gave me a big hint, thank you! – David Hui Jan 20 '25 at 15:47
  • Sorry for my poor English. I mean I have seen another question. From the author's intermediate step, the author claimed that there are no further N can make N!+1 be a perfect square after N>5. Many people replied that N=7 also satisfied (So the author made something wrong). I just tried to think any possibility for N>7 for my curiosity only. – David Hui Jan 20 '25 at 16:24
  • You are not the only one who thought about such a possibility. See for example here and many other posts. In fact there is no integer solution of $n!+1=m^2$ for all $7<<109$, as proved by Berndt & Galway (2000). See also wikipedia for more references. – Dietrich Burde Jan 20 '25 at 17:14

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