2. Find all prime numbers $p$ such that $17p + 1$ is a perfect square of a natural number.
Solution. Let $17 p+1=n^{2}$. From the fact that 17 and $p$ are prime numbers, we have

$$
17 p=n^{2}-1=(n-1)(n+1)
$$

There are two possible cases

1) $17=n-1, \quad p=n+1$, from which it follows that $n=18$ and thus $p=19$;
2) $p=n-1,17=n+1$, i.e., $n=16$ and $p=15$ which is a contradiction (15 is not a prime number).

Therefore, the only solution is $p=19$.