Let $n=4k+3$ is a prime number and $a^2+b^2\equiv 0\pmod{n}$, prove that $a\equiv b \equiv 0\pmod{n}$.
Proof: If $a\equiv 0\pmod{n}$, because $a^2+b^2\equiv 0\pmod{n}$, $b\equiv 0\pmod{n}$.
If $a\not\equiv 0\pmod{n}$, because $n$ is a prime number, according to Fermat’s Little Theorem, we have $$a^{n-1}\equiv 1\pmod{n}\tag{1}$$
Because $a^2+b^2\equiv 0\pmod{n}$, we have $b\not\equiv 0\pmod{n}$. Similarily, we $$b^{n-1}\equiv 1\pmod{n}\tag{2}$$ Therefore, we have the following: $$a^{n-1}+b^{n-1}\equiv 2\pmod{n}\tag{3}$$
Because $n=4k+3$, and $a^2+b^2\equiv 0\pmod{n}$, we have $$a^{n-1}+b^{n-1}\equiv a^{4k+2}+b^{4k+2}\equiv (a^2)^{2k+1}+(b^2)^{2k+1}$$ $$\equiv (a^2+b^2)\sum_{i=0}^{2k}(-1)^{i}(a^2)^{2k-i}(b^2)^i\equiv 0\pmod{n}\tag{4}$$
Therefore the assumption of $a\not\equiv 0\pmod{n}$ leads conflicting equation $(3)$ and $(4)$. So, we can only have $\boxed{a\equiv b\equiv 0\pmod{n}}$.