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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 $p\equiv1 \mod p$.

Jyrki Lahtonen
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Nom
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  • Yes, sorry, $p\neq 2,$ I will edit that, thank you! – Nom May 05 '20 at 14:18
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    That polynomial is equal to $$(x^2+y^2)^4+4x^2y^2(x^2-y^2)^2.$$ A sum of two squares. Show that this implies that $-1$ is a quadratic residue modulo $p$. You need to cater for the possibility that both those squares are themselves multiples of $p$. I think some kind of a descent follows. – Jyrki Lahtonen May 05 '20 at 14:19
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    @Aqua I appreciate the vote of confidence. There is the problem with the descent part. Also, I will search f0r duplicates before checking it out further. – Jyrki Lahtonen May 05 '20 at 14:22
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    @JyrkiLahtonen if $p|a^2+b^2=(a+bi)(a-bi),$ then wlog $p|a+bi$ which means that $a+bi\equiv 0 \pmod{p},$ so $i=\sqrt{-1}$ is defined modulo $p.$ That means that -1 is a quadratic residue mod $p,$ so $p\equiv 1 \pmod{p}.$ Does that sound right to you? – Nom May 05 '20 at 14:29
  • @JyrkiLahtonen, how did you find that expression as a sum of two squares? – lhf May 05 '20 at 14:35
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    A fair question @lhf :-) Something about it suggested that it might be. Only even powers, general appearance? May be that was an intuition thing from having seen my share of questions like this? I thought it would be a field norm of some kind, when similar conclusions about the prime factors can often be drawn? – Jyrki Lahtonen May 05 '20 at 14:43
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    @JyrkiLahtonen, all good reasons. Thanks. Yet not trivial. – lhf May 05 '20 at 14:44
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    Correct, Nom. That is more or less the reasoning. – Jyrki Lahtonen May 05 '20 at 14:46
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    Nom, if you don't mind, please add the contents of that comment of yours into the question body. That will show the passers-by that you worked on the hint. And therefore it will pre-empt further votes to close the question for lacking context. – Jyrki Lahtonen May 05 '20 at 14:48
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    As you are a new user our expectations may be somewhat unclear to you. I recommend that you study the guide for new askers. – Jyrki Lahtonen May 05 '20 at 14:50
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    Like that :-) ${}$ – Jyrki Lahtonen May 05 '20 at 14:56

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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 $p$.

Jyrki Lahtonen
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