Square root of minus one modulo a prime ($x^2\equiv -1(\mathrm{mod}p)$)

The congruence is soluble exactly for $p=2$ or $p\equiv 1(\mathrm{mod}4)$. For odd $p$, Fermat's little theorem applied to a prospective root forces $(p-1)/2$ even. Conversely, when $p\equiv 1(\mathrm{mod}4)$, pair opposite factors in $(p-1)!$ and use the Wilson theorem to obtain $(((p-1)/2)!{)}^2\equiv -1(\mathrm{mod}p)$.