Solution:

For $p=3$ we have $p^{2}+8=17$, which is prime.

If $p \neq 3$ then $p$ is not divisible by $3$. The remainder of $p$ when divided by $3$ is either $1$ or $2$. This means that $p=3k+1$ for some integer $k$, or $p=3l+2$ for some integer $l$.

In the first case we get $p^{2}=(3k+1)^{2}=9k^{2}+6k+1$ and in the second, $p^{2}=(3l+2)^{2}=9l^{2}+12l+4$.

In both cases $p^{2}$ gives a remainder $1$ upon division by $3$. Hence $p^{2}+8$ is divisible by $3$ for all prime numbers different than $3$.