If $1!+2!+\dots+x!$ is a perfect square, then the number of possible values of $x$ is?

Question

If $1!+2!+\dots+x!$ is a perfect square, then the number of possible values of $x$ is?

I looked for a general way of expanding such a factorial series but I was not able to find one, without that I don't know any other way to approach this problem.

All help is appreciated.

Answer 1

Hint:

Let $$a_n = \sum_{j=1}^nj!$$

Knowing that for $j\ge 5$, the term $j!$ will end with $0$, since the product will contain both $2$ and $5$ it should be straighfroward for you to show that $$a_n = 33 + \sum_{j=5}^n j!$$

This sum therefore ends with digit $3$.

Big Hint

It is easy to check that no perfect square can end with digit $3$ (namely by checking all squares $\mod 10$). This shows that $a_n$ is not perfect square for $n\ge 4$.