Let $p$ be a prime number of the form $4n+1$, and let $n$ be a natural number. Show that congruence $x^2\cong-1 \mod p$ is solvable.
I tried solving by simplifying $\cong$ sing i.e. $x^2 + 1 = p.m$ or $x^2 = (4n+1).m -1$. But I can't proceed after this
Not sure can you use this, but a quick easy proof will be assuming the existence of a primitive element $a$, with $$(a^{2n})^2\equiv a^{4n}\equiv a^{p-1}\equiv 1$$
Then by factoring: $$(a^{2n}+1)(a^{2n}-1)\equiv 0$$
Since $a$ is primitive, $a^{2n}\not\equiv 1$, so $a^{2n}\equiv-1$.
Then $x=a^n$ gives $x^2\equiv-1$.
