Problem 1. Determine the prime numbers $p$ for which the number $a=7^{p}-p-16$ is a perfect square.

## Lucian Petrescu
Solution: $p=2$ does not verify. $p=3$ is a solution: $a=7^{3}-3-16=324=18^{2}$. We show that there are no other solutions. Let $p \geq 5$ be a prime number. If $p \equiv 1(\bmod 4)$, then $a \equiv 2(\bmod 4)$, so $a$ is not a perfect square. It is easily observed that $p=7$ is not a solution (by calculating $a$, the last two digits of $a$, or noting that $a \equiv 5(\bmod 7))$. If $p>7$ is a prime of the form $4k+3$, then by Fermat's little theorem, we have $7^{p} \equiv 7(\bmod p)$, so $a \equiv-9(\bmod p)$, which means $p \mid a+9$. If $a$ were a perfect square, then $p$ divides $a+9=b^{2}+3^{2}$ implies $p$ divides $b$ and $p$ divides 3, which is impossible. Therefore, the only solution to the problem is $p=3$.