TITLE: $1+n!=m^{2}$ for some n,m$\in\mathbb{N}$
QUESTION [6 upvotes]: I have no idea whether this is known or not and I couldn't find anything related on Google. While I was studying , I come up with this idea $1+n!=m^{2} $ for some $n,m\in\mathbb{N}$
$1+4!=5^{2}$
$1+5!=11^{2}$
$1+?!=?^{2}$ 
and the question is what is the next number? Wolfram Alpha gave me this interesting graph: 
Thanks in advance for your interest.

REPLY [7 votes]: $$4!+1=5^2\\5!+1=11^2\\7!+1=71^2$$
This problem is known as Brocard's Problem, and pairs of integers which satisfy it are known as Brown Numbers. There are no other known solutions up to $10^9!+1$.