A Divisibility Property of Polynomial Values

Disclaimer: I had some help from free AI, but not by lazily pasting your question onto the prompt. I asked AI which facts from algebraic geometry can help me determine whether $p(x^2+xy+y^2)\,|\,q(x^4+x^2y^2+y^4)$ as complex polynomials, and then a separate question about passing to integers. Along the way, it gave me a big hint via undergraduate ring theory which I should have thought of because I had learnt this stuff.

Edit: It seems someone else has answered, and that answer should be correct. The one I'm giving requires zero computation: aside from verifying one little equation, everything is just applying undergraduate abstract ring theory.

Let $u=x^2+xy+y^2$ and $v=x^2-xy+y^2$. Regard $\mathbb{Q}[u,v]$ as a subring of $\mathbb{Q}[x,y]$.

By the way, the ring homomorphism $\phi:\mathbb{Q}[x,y]\to\mathbb{Q}[u,v]$ sending $x$ to $u$ and $y$ to $v$ is injective. Said another way, if you have a two-variable integer polynomial $f$ such that $f(x^2+xy+y^2,x^2-xy+y^2)$ is the zero element of $\mathbb{Q}[x,y]$, then $f$ is the zero polynomial. That is because $\mathbb{Q}(x,y)$ is an algebraic extension over $\mathbb{Q} (u,v)$ - see the next line - and therefore both have the same transcendence degree over $\mathbb{Q}$.

$\mathbb{Q}[x,y]$ is integral over $\mathbb{Q}[u,v]$ because e.g. $x^4-x^2\,\dfrac{u+v}{2}+\big(\dfrac{u-v}{2}\big)^2=0$.

So, let's suppose that $\dfrac{q(uv)}{p(u)}=\dfrac{q(x^4+x^2y^2+y^4)}{p(x^2+xy+y^2)}\in\mathbb{Q}[x,y]$. We'll assume it until I ask when can we assume it.

This would imply, since this fraction also exists in the field of fractions of $\mathbb{Q}[u,v]$, that $\dfrac{q(uv)}{p(u)}\in\mathbb{Q}[u,v]$.

Therefore, $p(x)|q(xy)$.

This would force $q(cy)$ to be $0$ for any complex root $c$ of $p$ and all complex values $y$. We automatically deduce that $c$ has to be $0$ and that $p(x)$ is just a power of $x$ possibly multiplied by a constant.

Now, let's go back to our supposition $\dfrac{q(uv)}{p(u)}=\dfrac{q(x^4+x^2y^2+y^4)}{p(x^2+xy+y^2)}\in\mathbb{Q}[x,y]$ and ask when can we assume it.

If we have a pair of two-variable integer polynomials $f\neq0$ and $g$ such that for all positive integer pairs $(a,b)$ either $f(a,b)=0$ or $f(a,b)|g(a,b)$, then $f|g$.

The key lemma is to check that for a pair of one-variable integer polynomials $h\neq0$ and $k$ such that $h(d)|k(d)$ for all but finitely many positive integers $d$, we have $h|k$. This is because, leaving aside the trivial case $k=0$, we get $\deg(k)\geq\deg(h)$. And if we denote the remainder of dividing $k$ by $h$ in $\mathbb{Q}[x]$ as $r$, we see that $J(d)h(d)|J(d)r(d)$ for some non-zero integer polynomial $J(d)$ and for all but finitely many positive integers $d$, but $\deg(r)<\deg(h)$, forcing $r$ to be the zero polynomial (because you can compare degrees of a polynomial by how fast they grow as $x$ increases, including if $x$ were restricted to positive integer values).

Now, it cannot be that $f_b$ ( where $f_b(x)=f(x,b)$ ) is the zero polynomial for infinitely many positive integer values of $b$. Think why that is - remember, I assumed $f\neq0$. And so, for all but finitely many positive integer values $b$, we have that $f_b\neq0$ and therefore that $f_b|g_b$.

Then, via a step almost the same as above, we divide $g(x,y)$ by $f(x,y)$ in $\mathbb{Q}(y)[x]$, write it out as $g=Wf+R$ where $\deg(R)<\deg(f)$, multiply both sides by $J(y)$ to make both sides be in $\mathbb{Q}[x,y]$, show that $R=0$, then use Gauss to conclude that $f|g$ in $\mathbb{Q}[x,y]$.

Your factorization $m^4+m^2n^2+n^4=(m^2+mn+n^2)(m^2-mn+n^2)$ is a good start, but I think the gcd of the two factors is not the main point. The key observation is that the divisibility condition for all $m$ forces an actual polynomial divisibility relation.

I claim that the only possibilities are $P(x)=x^d$ for some $d\geq0$.

First use the following lemma. If $F,G\in\mathbb Z[x]$, with $F$ monic and nonconstant, and $F(t)\mid G(t)$ for all sufficiently large integers $t$, then $F(x)\mid G(x)$ in $\mathbb Z[x]$.

Indeed, write $G=FH+R$ with $H,R\in\mathbb Z[x]$ and $\deg R<\deg F$. Then $F(t)\mid R(t)$ for all large $t$. Since eventually $|R(t)|<|F(t)|$, we get $R(t)=0$ for infinitely many $t$, hence $R=0$.

Now fix $n$ and define $F_n(X)=P(X^2+nX+n^2)$ and $G_n(X)=Q(X^4+n^2X^2+n^4)$. The hypothesis gives $F_n(m)\mid G_n(m)$ for every positive integer $m$, so the lemma gives $P(X^2+nX+n^2)\mid Q(X^4+n^2X^2+n^4)$ in $\mathbb Z[X]$.

Let $\alpha$ be a root of $P$. For each $n$, let $u_n,v_n$ be the two roots of $X^2+nX+n^2=\alpha$. Then $u_n+v_n=-n$.

Because of the polynomial divisibility above, we have $Q(\alpha B(u_n,n))=Q(\alpha B(v_n,n))=0$, where $B(X,n)=X^2-nX+n^2$.

On the curve $X^2+nX+n^2=\alpha$, we have $B(X,n)=\alpha-2nX$. Hence $B(u_n,n)+B(v_n,n)=2\alpha-2n(u_n+v_n)=2\alpha+2n^2$.

Therefore two roots of $Q$ have sum $2\alpha(\alpha+n^2)$. But $Q$ has only finitely many roots, so it can only produce finitely many such sums. If $\alpha\neq0$, then $2\alpha(\alpha+n^2)$ gives different values for different positive integers $n$, which is impossible.

Therefore every root of $P$ is $0$. Since $P$ is monic, we get $P(x)=x^d$.

It remains to check that these polynomials really work. Take $P(x)=x^d$ and $Q(x)=x^d$. Let $A=m^2+mn+n^2$ and $B=m^2-mn+n^2$. Since $m^4+m^2n^2+n^4=AB$, we have $Q(m^4+m^2n^2+n^4)=(AB)^d$. Clearly $A^d\mid(AB)^d$, so the condition is satisfied.

Hence the complete family is $P(x)=x^d$ with $d\geq0$.