Given that N is a natural number .Then I was trying to find the number of positive integral solutions of the equation $$1!+2!+3!+4!+.....+x!=N^2$$
By hit and trial, $1$ and $3$ are obvious solutions for $x$. But are these the only solutions and how can I find any other solutions if present ?
If $x\ge 4$, then $$1!+2!+3!+4!+5!+\cdots+x!\equiv $$
$$\equiv 1!+2!+3!+4!=33\equiv 3\pmod{5},$$
but $3$ is not a quadratic residue modulo $5$ because, e.g., you can notice that if $k\in\mathbb Z$, then $(5k\pm 2)^2\equiv (\pm 2)^2\equiv 4\pmod{5}$, $(5k\pm 1)^2\equiv (\pm 1)^2\equiv 1\pmod{5}$, $(5k)^2\equiv 0\pmod{5}$, so $0,1,4$ are all the quadratic residues mod $5$. Therefore $x\le 3$ and $x=1$, $x=3$ give the only solutions.