2021/017) find integer n such that n!+3 is a perfect square

we know that for a perfect square x $x^2 \equiv\, 0 or \, 4 \pmod 4$

and for $n \ge 4$  $n! + 3 \equiv\, 3 \pmod 4$

so for  $n \ge 4$ there is no solution so only n = 0 to 3 need to be checked

$0!+3 = 4 = 2^2$ is a perfect square

$1!+3 = 4 = 2^2$ is a perfect square

$2!+3 = 5$ is not a perfect square

$3!+3 = 9 = 3^2$ is a perfect square

so n = 0 or 1 or 3