Find all prime numbers $p$ such that $p^{2}+8$ is prime number, as well.

Remark. A number $p$ is prime if it has exactly 2 divisors: 1 and $p$. Numbers $2,3,5,7,11,13, \ldots$ are prime, while 4 and 2007 are not.

Hint. Write down first several prime numbers (hint - you can copy them from the paragraph above), calculate $p^{2}+8$ for them, and look at those that happen to be composite. Notice further that they all have a common divisor.
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 $p=3 k+1$ for some integer $k$, or $p=3 l+1$ for some integer $l$. In the first case we get $p^{2}=9 k^{2}+6 k+1$ and in the second, $p^{2}=9 l^{2}-6 l+1$. 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 .