A problem by Sahil Shrivastava

For $n \geq 4,$

$n! + 10 \equiv 0 + 10 \equiv 2 \pmod4$ . But squares are $\equiv 0,1 \pmod4$ . Therefore $n! + 10$ cannot be a square for $n \geq 4$

So we only have to consider when $n = 1,2,3,$ .

When $n = 1 , n! + 10 = 11$ which is not a square.

When $n = 2, n! + 10 = 12$ which is not a square.

When $n = 3, n! + 10 = 16$ which is $4^2$

Therefore total number of integral values of $n$ for which $n! + 10$ is a square is $\boxed{1}$ .

GIVE CREDIT TO THE EXAM FROM WHICH IT IS TAKEN AT LEAST

KVPY