Prove that $x^{2} \equiv -1 \pmod p$ has no solutions if prime $p \equiv 3 \pmod 4$. Assume: $p$ is a prime that satisfies $p \equiv 3 \pmod 4$
Show: $x^{2} \equiv -1 \pmod p$ has no solutions $\forall x \in \mathbb{Z}$.
I know this problem has something to do with Fermat's Little Theorem, that $a^{p-1} \equiv 1\pmod p$. I tried to do a proof by contradiction, assuming the conclusion and showing some contradiction but just ran into a wall. Any help would be greatly appreciated.
 A: Suppose $x^2\equiv -1\pmod{p}$. Then $x^4\equiv 1\pmod{p}$. Since $p = 4k+3$, we have
$$x^{p-1} = x^{4k+2} = x^2x^{4k} \equiv -1(x^4)^k\equiv -1\pmod{p},$$
which contradicts Fermat's Little Theorem.
A: It is equivalent to $\nexists n: \frac{n^2+1}{p}\in \mathbb{N}$, which is a case of the sum of squares theorem.
A: Hint $\bmod P\! = 4K\!+\!3\!:\;$ $\ \color{#c00}{X^{\large 2} \equiv -1} \;\Rightarrow\; 1\equiv X^{\large P-1} \equiv (\color{#c00}{X^{\large 2}})^{\large 2K+1}\equiv (\color{#c00}{-1})^{\large 2K+1} \equiv -1$
Alternatively $\ X^{\large 4}\equiv 1\equiv X^{\large 4K+2}\Rightarrow\, 1 \equiv X^{\large \,\!(4,\,4K+2)}\equiv X^{\large 2}\equiv -1\,\Rightarrow\, P\mid2\, \Rightarrow\!\Leftarrow$
Remark $ $ The proof is a special case of Euler's Criterion.
For a converse, and a group-theoretic viewpoint, see here.
A: Isn't this easier than using FLT.
Consider a solution x then x is even or odd. Substituting this into x^2=-1 (mod p) leads to a contradiction in both cases.
A: A solution to the given congruence would imply there exists an element of multiplicative order 4 in the finite field of size p which is false because (p-1)/2 is odd and therefore p-1 is not divisible by 4.
