## Task $31 / 73$

We are looking for all prime numbers $p$ for which $p^{4}-1$ is not divisible by 15.
If $p^{4}-1$ is not to be divisible by 15, then it must not be divisible by 3 or not by 5. Now, for every natural number $n \neq 0 \bmod 3$, it holds that $n \equiv \pm 1 \bmod 3$ and thus $n^{4} \equiv 1 \bmod 3, n^{4}-1 \equiv 0$ $\bmod 3$.

Similarly, for every natural number $n \neq 0 \bmod 5$, it holds that $n \equiv \pm 1 \bmod 5$ or $n \equiv \pm 2 \bmod 5$. In each case, therefore, $n^{4} \equiv 1 \bmod 5$ and thus $n^{4}-1 \equiv 0 \bmod 5$. It follows immediately that only the prime numbers $p_{1}=3$ and $p_{2}=5$ can be solutions. The test proves that these two numbers indeed satisfy the requirement.