# $n!+1$ being a perfect square

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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I remember having this same question while trying to prove that some polynomials were irreducible in an algebra assignment years ago. Namely, $(x-1)(x-2)\cdots(x-n)+1$ is irreducible when $n$ is not $4$. After showing that its reducibility would imply that it is a square, one is led to your question. We didn't know it was an open problem. – Jonas Meyer Oct 27 '10 at 5:18
– barto Mar 18 '15 at 13:08

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

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

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oh ho! great i didnt know about it – anonymous Oct 26 '10 at 16:55
Note that the article says that the number of solutions is finite on the assumption that the ABC conjecture is true. There is now a claimed proof of the ABC conjecture. – Mark Bennet Oct 1 '12 at 12:19
According to Wikipedia's ABC Conjecture entry, the original proof was in error, but as of March 2013, a proposed correction has been posted. – Blue Jun 23 '13 at 18:32

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!

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