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**dori1123** Let $\displaystyle A_1,A_2$ be two quadratic residues of the $\displaystyle (4k+3)-$prime $\displaystyle p$ that satisfy $\displaystyle 0<A_1<A_2<p$. Prove that $\displaystyle A_1+A_2=0$ (mod $\displaystyle p$) is impossible.

The hint says to use the theorem "$\displaystyle N=a^2+b^2$ if and only if no $\displaystyle (4k+3)-$prime in the factorization of $\displaystyle N$ occurs to an odd power.

Suppose $\displaystyle A_1+A_2=0$ (mod $\displaystyle p$). Since $\displaystyle A_1,A_2$ are two quadratic residues of the $\displaystyle p=4k+3$, $\displaystyle x^2=A_1$ (mod $\displaystyle p$) and $\displaystyle y^2=A_2$ (mod $\displaystyle p$). So $\displaystyle x^2+y^2=0$ (mod $\displaystyle p$). So $\displaystyle x^2+y^2=mp$ for some $\displaystyle m \in \mathbb{N}$.

I also know that $\displaystyle 0<A_1<A_2<p$, so $\displaystyle A_1+A_2<2p \implies A_1+A_2=p$.

I think I should end up contracting the theorem above, but I'm kinda stuck.. can I get some help please?