# $n!+1$ 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.
• 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
• – punctured dusk Mar 18 '15 at 13:08
• 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. – Feeds Oct 2 '17 at 3:11

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.