Primes dividing an integer polynomial $x^8 + 8x^6y^2 -2x^4y^4 + 8x^2y^6 + y^8$ are congruent to 1 mod $4$ I need to prove that if $p\neq 2$ divides the integer polynomial
$$x^8 + 8x^6y^2 -2x^4y^4 + 8x^2y^6 + y^8\in\mathbb{Z}[x,y],$$ $gcd(x,y)=1,$ then $p\equiv 1\pmod{4}.$ Any ideas?
Edit (the OPs reaction to the hint that this is a sum of two squares):

If $p|a^2+b^2=(a+bi)(a−bi)$, then wlog $p|a+bi$ which means that $a+bi≡0\pmod p$, so $i=\sqrt{−1}$ is defined modulo $p$. That means that $-1$ is a quadratic residue mod $p$, so $p\equiv1 \mod p$. 


A proof can be based on the following
Fact. If an odd prime $p$ is a factor of the sum $x^2+y^2$ such that $x$ and $y$ are not multiples of $p$, then $p\equiv1\pmod4$.
Proof. Without loss of generality $p\nmid y$. Therefore we can divide the congruence
$$
x^2\equiv -y^2\pmod p
$$
by $y^2$, and conclude that there exists an integer $z\equiv xy^{-1}$ such that $z^2\equiv-1\pmod p$. This is well known
 to be possible only when $p\equiv1\pmod4$. 

The octic polynomial in the question can be written as a sum of two squares:
$$
x^8 + 8x^6y^2 -2x^4y^4 + 8x^2y^6 + y^8=(x^2+y^2)^4+4x^2y^2(x^2-y^2)^2.
$$
If neither $a=(x^2+y^2)^2$ nor $b=2xy(x^2-y^2)$ is divisible by $p$, then the other cannot be either. Hence the Fact bites, and allows us to conclude that $p\equiv1\pmod4$.
But if $p\mid a$, then $p\mid x^2+y^2$, and we can repeat the argument. Observe that the assumption $\gcd(x,y)=1$ rules out the possibility that both $x$ and $y$ would be divisible by $p$.
