Solution:
If $p > 5$ then the units digit of $p$ must be 1, 3, 7 or 9.
- If the units digit of $p$ is 1 or 9 then the units digit of $p^{2}$ is 1. Therefore the units digit of $p^{2} - 6$ is 5. Since $p^{2} - 6 > 5$ this means that $p^{2} - 6$ is not prime.
- If the units digit of $p$ is 3 or 7 then the units digit of $p^{2}$ is 9. Therefore the units digit of $p^{2} + 6$ is 5. Since $p^{2} + 6 > 5$ this means that $p^{2} + 6$ is not prime.
Therefore we must have $p \leq 5$ . Hence $p$ must be 2, 3 or 5.
- If $p = 2$ then $p^{2} + 6 = 10$ is not prime.
- If $p = 3$ then $p^{2} + 6 = 15$ is not prime.
- If $p = 5$ then $p^{2} \pm 6$ are 19 and 31 which are both prime.
Therefore the only answer is $p = 5$ .

Solution:
Consider the following product modulo 5.
$$p(p^{2} - 6)(p^{2} + 6) = p^{5} - 36p \equiv p^{5} - p \pmod {5}$$
By Fermat's Little Theorem, this product is 0 (mod 5). So if $p$ , $p^{2} - 6$ and $p^{2} + 6$ are all prime numbers then at least one of them must be equal to 5.
- If $p = 5$ then $6p^{2} - 1 = 29$ and $6p^{2} + 1 = 31$ . This is one solution.
- If $6p^{2} - 1 = 5$ then $p = \pm 1$ . Neither 1 nor $-1$ is prime, so this case leads to no solutions.
- If $6p^{2} + 1 = 5$ then $p$ is not an integer. No solutions in this case.
Therefore the only solution is $p = 5$ .