Prove that $x^{2} \equiv -1 \pmod p$ has no solutions if prime $p \equiv 3 \pmod 4$.
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.
It is equivalent to $\nexists n: \frac{n^2+1}{p}\in \mathbb{N}$, which is a case of the sum of squares theorem.
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.