$n!+1$ 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.

Solution 1:

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

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

Solution 2:

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.

Solution 3:

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!