Find $(m,n$ )

For $n \ge 1$ let $\displaystyle S_{n} = \sum_{k=1}^{n} n!$ . Then $S_{1} = 1 = 1^{2}, S_{2} = 3, S_{3} = 9 = 3^{2}, S_{4} = 33, S_{5} = 153, ....$ . Thus far we can then identify $(m,n) = (1,1)$ and $(3,3)$ as solutions.

Now for $n \ge 5$ we have that $n! \equiv 0 \pmod{10}$ , so since $S_{4} \equiv 3 \pmod{10}$ we know that $S_{n} \equiv 3 \pmod{10}$ for $n \ge 4$ . But no perfect square is equivalent to $3$ modulo $10$ , so we will not find any valid solutions for $n \gt 3$ , and thus there are just $\boxed{2}$ valid solutions, namely $(1,1)$ and $(3,3)$ .

Hint: $(1,1)\ \text{and} \ (3,3)$

A square must end in the digit $0,1,4,5,6, or 9.$ If $n ≥ 4,$ then $1!+2!+···+n!$ ends in the digit $3$ , so cannot be a square. A simple check for the remaining cases reveals that the only solutions are (1, 1) and (3, 3).