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.

This is solution from Prasolov V.V. Zadachi po algebre, arifmetike i analizu (В. В. Прасолов. Задачи по алгебре, арифметике и анализу.) This book (in Russian) is freely available at http://www.mccme.ru/free-books/ 
The problem appears there as Problem 31.2.
We want to show that if $p=4k+3$ is a prime number and $p\mid a^2+b^2$, then $p\mid a$ and $p\mid b$.
My translation of the solution given there:

Suppose that one of the numbers $a$, $b$ is not divisible by $p$. Then the other one is not divisible by $p$, either. Thus from Fermat's little theorem we get $a^{p-1} \equiv 1 \pmod p$ and $b^{p-1} \equiv 1 \pmod p$. This implies $a^{p-1}+b^{p-1} \equiv 2 \pmod p$.  
On the other hand, the number $a^{p-1}+b^{p-1}=a^{4k+2}+b^{4k+2}=(a^2)^{2k+1}+(b^2)^{2k+1}$ is divisible by $a^2+b^2$, and thus it is divisible by $p$.

