Numbers which are a sum of two squares

Question

I know that

I - If $p = 4m + 3$ is a prime that divides a number $n = x^2 + y^2$, then $p$ divides both $x$ and $y$, and thus $p^2$ divides n.

II - If $n=x^2+y^2$, $x,y \in \mathbb{Z}$, and $p = 4m + 3$ divides $n$, then $p^2$ divides $n$ and $\frac{n}{p^2}= \alpha^2 + \beta^2$, $\alpha, \beta \in \mathbb{Z}$.

I must use I and II to conclude that if $n=x^2 +y^2, x,y \in \mathbb{Z}$, then every prime factor of the form $p = 4m + 3$ appears with an even exponent in the prime decomposition of $n$.

I imagine that everything is done in I and II but I can't organize the ideas in order to get a conclusion.

Thanks a lot for any help.

Answer 1

Assume there exists $n\in\Bbb N$ that is the sum of two squares and $p=4m+3$ occurs with an odd exponent $k$ in the prime decomposition of $n$. Then there also exists a minimal such $n$. From either I or II, we have $p^2\mid n$ and $n':=\frac n{p^2}$ is a sum of two squares. As $p$ occurs with the odd exponent $k-2$ in $n'$, this contradicts the minimality of $n$.