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One observes that \begin{equation*} 4!+1 =25=5^{2},~5!+1=121=11^{2} \end{equation*} is a perfect square. Similarly for $n=7$ also we see that $n!+1$ is a perfect square. So one can ask the truth of this question:

  • Is $n!+1$ a perfect square for infinitely many $n$? If yes, then how to prove.
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    The pair $(m, n)$ of the form $n! + 1 = m^2$ are called *Brown Numbers*. It has been conjectured that there only exists *three* pairs. You can do a little bit of research.2017-10-02

3 Answers 3

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This is Brocard's problem, and it is still open.

http://en.wikipedia.org/wiki/Brocard%27s_problem

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    According to Wikipedia's [ABC Conjecture entry](http://en.wikipedia.org/wiki/Abc_conjecture), the original proof was in error, but as of March 2013, a proposed correction has been posted.2013-06-23
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The sequence of factorials $n!+1$ which are also perfect squares is here in Sloane. It contains three terms, and notes that there are no more terms below $(10^9)!+1$, but as far as I know there's no proof.

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My intuition would be that there are very few. There are just not many squares and even fewer factorials. OEIS A025494 lists the squares which are a sum of distinct factorials, which is less restrictive than what you ask and says the list is probably finite. In particular, there are no more below 31!