Saudi Arabia Mathematical Competitions 2012

We have $5^{n^2}+7=(6-1{)}^{n^2}+7\equiv (-1{)}^{n^2}+1(\mathrm{mod}6)$. It follows that $5^{n^2}+7$ is divisible by $6$ if and only if $(-1{)}^{n^2}+1=0$. This is equivalent to $n^2$ being odd. Therefore, the possible values of $n$ are all odd positive integers.