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If $n=a^2+b^2, a,b \in \mathbb{Z}$ and $n$ is divisible by a prime number $\ne 2$, then $p$ could be expressed as the sum of two integers or divides both $a$ and $b$.

My approach is the following: let us $c=GCD (a,b)$. Then $a=c \alpha, b=c \beta$ and $GCD(\alpha, \beta)=1$. $n=c^2\alpha^2+c^2\beta^2=c^2(\alpha^2+\beta^2)$. Recall the definition of prime number, we obtain that $p\ | \ c^2$ or $p \ | \ (\alpha^2+\beta^2)$. First case gives us that $p \ | \ a, \ p \ | \ b$.

My glitch is on the 2nd: Since $GCD(\alpha^2,\beta^2)=1$, there exist such numbers $\gamma, \delta \in \mathbb{Z}$, such that $1=\alpha^2 \delta+\beta^2 \gamma$. Furthermore, we have $p \ | \ \alpha^2+\beta^2$ i.e. $\alpha^2+\beta^2 = p\varepsilon$. Multiplying the first equation by $p\varepsilon$, we have: $p\varepsilon \alpha^2 \delta + p\varepsilon \beta^2 \gamma = \alpha^2+\beta^2$. And here I'm stuck.

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From $p \mid a^2+b^2$ implies that $p\equiv 1 \pmod{4}$. And from Fermat's theorem on sum of two squares, we find $p$ can be expressed as sum of two squares.

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