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.

Assume that $a$ and $b$ are coprime. If $a$ and $b$ are odd, replace
$a$ and $ b$ by $(a-b)/2$ and $(a+b)/2$. Then $a^2 + b^2 = pm$, and by reducing $a$ and $b$
modulo $p$ you can make sure that $m < p$. Since $p = 4n+3$ and $a^2 + b^2 = 4k+1$, 
we must have $m = 4j+3$. But then $m$ is divisible by a prime number $q = 4r+3$ strictly
smaller than $p$. Repeat until you find that $3$ is a sum of two squares. 
