A number theory problem by mostafa hubble

Note that 1! and 1!+2!+3! are perfect squares. Then, we will focus on the last digit for $n!$ .

$1! \equiv 1 (mod 10)$

$2! \equiv 2 (mod 10)$

$3! \equiv 6 (mod 10)$

$4! \equiv 4 (mod 10)$

For $n \ge 5$ , we have that $n! \equiv 0 (mod 10)$ . So, $1! + 2! + 3! + 4! + ... + n! \equiv 3 (mod 10)$ for $n \ge 5$ . Since the last digits of a perfect square are always 0, 1, 4, 5, 6, 9, the expression will not be a perfect square if $n\ge 4$ . Thus, answer is 2 ( $n = 1$ , $n=3$ ).