Find all primes of the form $a^{2}-1$, with $a \geqslant 2$ natural.
3 is the only prime of this form. Indeed, if a prime $p$ is of the form $a^{2}-1=(a-1)(a+1)$, it is divisible by $a-1$ and $a+1$. Therefore, these must be 1 and $p$. Since they are distinct and $a-1$ is the smallest of them, $a-1=1$ and $p=a^{2}-1=2^{2}-1=3$. Conversely, $2^{2}-1=3$ is indeed prime.