Question 81
Let $x$, $y$, $n$ be natural numbers with $x \geq 3$, $n \geq 2$ and
$$
x^{2}+5=y^{n}
$$
Show that every prime divisor $p$ of $n$ satisfies the congruence $p \equiv 1 \pmod 4$.
$$
x^{2}+5=y^{n}
$$
Show that every prime divisor $p$ of $n$ satisfies the congruence $p \equiv 1 \pmod 4$.