Berkeley Math Circle

Solution:

The answer is $p=3$. This indeed works, since $3^{2}+2=11$.

Consider any other prime number $p \neq 3$. Then it follows that $p^{2} \equiv 1 \pmod{3}$; i.e. that $p$ leaves remainder $1$ or $2$ when divided by $3$. Consequently, $p^{2}+2$ is divisible by $3$. Since $p \geq 2$, we have $p^{2}+2 \geq 7$ as well, thus $p^{2}+2$ cannot be prime in this case.