<s>[INST] show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ $n = 4k + 3 $
We start by letting $a \not\equiv 0\pmod n$ $\Rightarrow$ $a \equiv k\pmod n$ .
$\Rightarrow$ $a^{4k+2} \equiv 1\pmod n$
Now, I know that the contradiction will arrive from the fact that if we can show $a^2 \equiv 1 \pmod n $ then we can say $b^2 \equiv -1 \pmod n $ then it is not possible since solution exists only for $n=4k_2+1 $ so $a \equiv b\equiv 0 \pmod n $
So from the fact that $a^{2^{2k+1}} \equiv 1 \pmod n$ I have to conclude something.
 [/INST] Hint: If $p \equiv 3 \pmod{4}$, then $x$ is a quadratic residue $\bmod{p}$ if and only if $-x$ is a quadratic nonresidue $\bmod{p}$.

Edit, to be more direct: 
Lemma: 
Suppose that $p$ is an odd prime, and assume that $x$ is a nonzero quadratic residue $\bmod{p}$.  Then, $-x$ is a quadratic residue $\bmod{p}$ if and only if $p \equiv 1 \pmod{4}$.
Proof: 
First suppose that $p \equiv 1 \pmod{4}$.  By assumption, as $x$ is a nonzero quadratic residue, there exists an integer $a \neq 0$ such that $a^2 = x$.  Since $-1$ is also a quadratic residue, there also exists an integer $b$ such that $b^2 = -1$.  Thus,
$$
-x = -1 x = b^2 a^2 = (ba)^2,
$$
and hence $-x$ is a quadratic residue.
On the other hand, suppose that $p \equiv 3 \pmod{4}$ and $x = a^2$ is a quadratic residue.  Since $x \neq 0$, it follows that $-x \neq 0$ as well.  Then, suppose for a contradiction that $-x$ is also a nonzero quadratic residue, and hence $-x = c^2$ for some $c$.  If this were the case, it would follow that
$$
-1 = -x/x = c^2/a^2 = (c/a)^2
$$
which would imply that $-1$ is a square $\bmod{p}$.  This gives a contradiction, and we conclude that if $p \equiv 3\pmod{4}$ and $x$ is a nonzero quadratic residue, then $-x$ must be a quadratic nonresidue. 
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