For $x,y\ge 2$ does $x^4+x^2y^2+y^4$ ever divide $x^4y^4+x^2y^2+1$?

Question

For a problem in combinatorics, it comes down to knowing whether there exist integers $x,y\ge 2$ such that $$ x^4+x^2y^2+y^4\mid x^4y^4+x^2y^2+1. $$ Note that $x^6(x^2-y^2)(x^4+x^2y^2+y^4)+(x^2y^2-1)(x^4y^4+x^2y^2+1)=x^{12}-1$ and so we can look at a perhaps simpler problem: Are the only solutions to $$ x^4+x^2y^2+y^4\mid x^{12}-1 $$ (where $x,y\ge 2$) given by $(x,y)=(5,6)$ and $(x,y)=(6,5)$?

Answer 1

It seems to me that you need both $$p=x^2+xy+y^2$$ and $$q=x^2-xy+y^2$$ to divide $$r=x^4y^4+x^2y^2+1$$ Considering all expressions as polynomials in $x$, the remainder when you divide $r$ by $p$ is $$(y^7-y^3)x-y^4+1$$ and the remainder when you divide $r$ by $q$ is $$(y^3-y^7)x-y^4+1$$ if Maple and I are on the same page. These remainders are both zero if and only if $y=\pm1$.

Answer 2

I do not know whether there is any advantage to considering this problem in the ring of Eisenstein integers $\mathbb{Z}[\omega],$ where $\omega = e ^{\frac{2 \pi i}{3}},$ which is a PID. Then we have to ask when we can have $(x - \omega y)(x +\omega y)(x - \omega^{2}y )(x+\omega^{2}y)$ dividing $(xy- \omega)(xy +\omega)(xy- \omega^{2})(xy +\omega^{2})$ in $\mathbb{Z}[\omega],$ where $x,y$ are rational integers. I have not been able to pursue this to provide further insight myself, but someone else might.

Later remark: It is easy to check that the power of $1-\omega$ dividing both expressions is the same: it is $0$ if $3$ divides $xy,$ and $2$ if $3$ does not divided $xy.$ Hence we can omit the prime $1-\omega$ from our considerations, and we only need to worry about primes in $\mathbb{Z}[\omega]$ such that $N(\pi)$ is a rational prime congruent to $1$ (mod $3$). If $\pi$ is such a prime dividing the leftmost product, we note that $\pi$ divides exactly one of the terms in the rightmost product (and, in fact, $\pi$ also divides exactly one term in the leftmost product). This leads (if the required divisibility holds ) relatively easily to the observation (already made by the OP) that the leftmost expression divides $y^{12}-1$ (and/or $x^{12}-1,$ there is symmetry in $x$ and $y$), but it is unclear to me at present whether this viewpoint provides any more useful information.

Later edit: I noticed that Aaron Meyerowitz's observation (in a comment after Gerry Myerson's answer) that if the required divisibility holds, then $x^{4} + x^{2}y^{2} +y^{4}$ divides $(x^{4}-1)(y^{4}-1)$ can be derived this way. That is not particularly surprising, and the direct derivation is easier. However, perhaps less obvious is that we also have that $x^{4} + x^{2}y^{2} +y^{4}$ divides $(y^{8}+y^{4}+1)(x^{8}+x^{4}+1)$. While $(x^{4} + x^{2}y^{2} +y^{4})^{2}$divides $(x^{12}-1)(y^{12}-1),$ it is not immediately obvious to me that this last claimed divisibility is a consequence of that- for example, there might a priori be a prime $\pi$ such that $x^{4}-1$ is divisible by some higher than expected power of $\pi$- so I outline a proof:

Note that if $\pi$ is a prime in $\mathbb{Z}[\omega]$ with $N(\pi) \equiv 1$ (mod $3$), then if $\pi^{m}$ divides both $x^{4}-1$ and $x^{2}- \omega y^{2},$ we have $\pi^{m}$ divides $\omega^{2}y^{4}-1,$ so that $\pi^{m}$ divides $y^{4}-\omega^{4}.$ It follows that $N(\pi)^{m}$ divides $y^{8}+y^{4} + 1.$ Hence it follows that (in $\mathbb{Z}$), ${\rm gcd}(x^{4}-1,x^{4} + x^{2}y^{2} + y^{4})$ divides $y^{8}+y^{4}+1$ (as before, the power of $3$ is taken care of). Similarly ${\rm gcd}(y^{4}-1,x^{4} + x^{2}y^{2} + y^{4})$ divides $x^{8}+x^{4}+1$. Since $x^{4}+ x^{2}y^{2} + y^{4}$ divides $(x^{4}-1)(y^{4}-1),$ the claim is established (note that ${\rm gcd}(x^{4}-1,y^{4}-1)$ has the form $2^{a}3^{b}$ if the original divisibility holds (and $(x^{2}-1)(y^{2}-1) \neq 0$).

We can continue this analysis: we see (if the original divisibilty holds) that $x^{4}+x^{2}y^{2}+y^{4}$ divides ${\rm gcd}(y^{2}-1,\frac{x^{6}-1}{x^{2}-1}){\rm gcd}(y^{2}+1,\frac{x^{6}+1}{x^{2}+1}) {\rm gcd}(x^{2}-1,\frac{y^{6}-1}{y^{2}-1}){\rm gcd}(x^{2}+1,\frac{y^{6}+1}{y^{2}+1}).$

Additional edit: Conversely, it is easy to check that the rightmost product divides $3(x^{4}+y^{4}+ x^{2}y^{2})$ given that $x$ and $y$ are coprime. Also, the righmost product divides $3(x^{4}y^{4}+x^{2}y^{2}+1).$

It follows that $x^{4} + y^{4} + y^{2}x^{2}$ divides $x^{4}y^{4}+y^{2}x^{2}+1$ if and only if ${\rm gcd}(x,y) = 1$ and $x^{4}+x^{2}y^{2}+y^{4}$ is equal to
${\rm gcd}(y^{2}-1,\frac{x^{6}-1}{x^{2}-1}){\rm gcd}(y^{2}+1,\frac{x^{6}+1}{x^{2}+1}) {\rm gcd}(x^{2}-1,\frac{y^{6}-1}{y^{2}-1}){\rm gcd}(x^{2}+1,\frac{y^{6}+1}{y^{2}+1})$ when $xy$ is divisible by $3$ or $3(x^{4}+x^{2}y^{2}+y^{4})$ is equal to ${\rm gcd}(y^{2}-1,\frac{x^{6}-1}{x^{2}-1}){\rm gcd}(y^{2}+1,\frac{x^{6}+1}{x^{2}+1}) {\rm gcd}(x^{2}-1,\frac{y^{6}-1}{y^{2}-1}){\rm gcd}(x^{2}+1,\frac{y^{6}+1}{y^{2}+1})$ when $xy$ is not divisible by $3$.