[ Prime numbers and their properties $]$

$p-$ prime number. For which numbers $a$ will the solution to the congruence $a x \equiv 1(\bmod p)$ be the number $a$ itself?
$a^{2} \equiv 1(\bmod p) \Leftrightarrow (a-1)(a+1) \equiv 0(\bmod p) \Leftrightarrow a-1$ is divisible by $p$ or $a+1$ is divisible by $p$.

Answer

$a \equiv \pm 1(\bmod p)$