#4 - factorial

For any arbitrary positive integer $a$ , $a^{2} \equiv 0, 1, 4 \mod 5$

And, for $n \geq 5$ , $n! \equiv 0 \mod 5$

let, $p^{2} = 1! + 2! + 3! + ... + n!$

Now, for $n \geq 5$ , $1! + 2! + 3! + 4! + 5! + ... + n! \equiv 1! + 2! + 3! + 4! \equiv 3 \mod 5$ $\Rightarrow p^{2} \equiv 3 \mod 5$

which contradicts the original statement for $n \geq 5$ .

Now we are left with only $n = \left \{ 1, 2, 3, 4 \right \}$ . Its trivial to find that $n = 1$ and $3$ are the solutions.

$\therefore$ the answer is $\boxed{2}$