21
$\begingroup$

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 $(x,y)=(6,5)$?

$\endgroup$
4
  • 6
    $\begingroup$ It may or may not be useful to notice that $$x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2)$$ and $$x^4y^4+x^2y^2+1=(x^2y^2+xy+1)(x^2y^2-xy+1)$$ $\endgroup$ Jul 17, 2014 at 1:59
  • 8
    $\begingroup$ You wonder if we ever have that (1) $u^2+uv+v^2$ divides $u^2v^2+uv+1$ with the added requirement (2) that $u,v$ are perfect squares. Without this final requirement it happens rarely, but sometimes ($13$ times for $v \lt u \le 10000$). Perhaps figuring out when (1) happens would be a good start. $\endgroup$ Jul 17, 2014 at 6:47
  • $\begingroup$ Suppose one considers specific instances for x, e.g. 2401 + 49y^2 + y^4 divides (or not) 2401y^4 + 49y^2 +1. Can one come up with nice conditions on y to predict when this happens? Can such conditions be "uniformized" over families of x? Gerhard "Likes Solving By Plugging In" Paseman, 2014.07.19 $\endgroup$ Jul 19, 2014 at 19:22
  • $\begingroup$ Taking that particular example, one gets that the desired divisibility implies -2400(49y^2+2402) is divisible by the "smaller" quartic. Since the factor is odd, this reduces to 3*5^2(49y^2+2402), which quickly limits the size of feasible y. Perhaps one can show quickly this way that x cannot be small. Gerhard "For All But Infinitely Many" Paseman, 2014.07.19 $\endgroup$ Jul 19, 2014 at 19:25

2 Answers 2

7
$\begingroup$

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 $y=\pm1$.

$\endgroup$
11
  • 1
    $\begingroup$ The $x$-polynomial remainders need not be zero for $p\mid r$ and $q\mid r$ to hold (for a given $x,y\in\mathbb{Z}$). $\endgroup$
    – GH from MO
    Jul 17, 2014 at 3:27
  • 1
    $\begingroup$ @Myerson It is important to distinguish integers and indeterminants. Let $x$ and $y$ be integers greater than 1, and let $X$ and $Y$ be indeterminants. Consider $f(X,Y), g(X,Y)\in\mathbb{Q}(Y)[X]$. Certainly $f(X,Y)\equiv 0\pmod {g(X,Y)}$ implies $f(x,y)\equiv 0\pmod {g(x,y)}$ for all integers $x,y>1$, but the converse is false. For example, $(X^2+2)\pmod {XY}$ is nonzero, but $(2^2+2)\pmod {2\cdot3}$ is zero. For this reason, I do not see that Gerry has solved the problem. $\endgroup$
    – Glasby
    Jul 17, 2014 at 3:33
  • $\begingroup$ You have reduced the problem to showing that "the integers $(y^7-y^3)x-y^4+1$ and $(y^3-y^7)x-y^4+1$ do not divide $x^4y^4+x^2y^2+1$ (for all $x,y\ge 2$)". $\endgroup$ Jul 17, 2014 at 3:45
  • 1
    $\begingroup$ Note that $x, y$ must be relatively prime in order for $x^4 + x^2 y^2 + y^4$ to divide $x^4 y^4 + x^2 y^2 + 1$; I claim that $p = x^2 + xy + y^2$ and $q = x^2 - xy + y^2$ are also relatively prime. Assume not; then the common divisor also divides $2p - q = (x + y)^2$ and $2q - p = (x - y)^2$; we would therefore also get a common divisor of $x + y, x - y$. As the relevant common divisor can't be 2 (by checking parity), we would have a common divisor of $x, y$, which is impossible. Therefore $p, q$ are relatively prime. It's therefore sufficient to show that $p|r, q|r$ to have that $pq|r$. $\endgroup$
    – user44191
    Jul 17, 2014 at 4:14
  • 3
    $\begingroup$ Of course $pq \mid r$ implies $pq \mid r-pq=(x^4-1)(y^4-1)$ which factors quite a bit further. Not that it is clear if that helps. $\endgroup$ Jul 17, 2014 at 6:05
6
$\begingroup$

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 $3$.

$\endgroup$
7
  • 2
    $\begingroup$ And then (for whatever it is worth) the first thing is $\gcd(y^2-1,x^4+x^2+1)=\gcd((y-1)(y+1),(x^2+x+1)(x^2-x+1))$ which is the product of four things $\gcd(y \pm 1,x^2 \pm x +1)$ (since the two things on the right are odd and coprime while the two things on the right can share at most a factor of $2$.) Similarly for the third factor. Over $\mathbb{Z}[i]$ one could do something similar for the second and fourth. $\endgroup$ Jul 21, 2014 at 4:25
  • $\begingroup$ @AaronMyerowitz: Yes, that is correct, there are several ways to refine things further, either over $\mathbb{Z},$ or $\mathbb{Z}[\omega]$, or $\mathbb{Z}[i],$ though I had not tried the last one. $\endgroup$ Jul 21, 2014 at 5:17
  • $\begingroup$ A similar thing to observe is that if we ask what the solution to the quartic $$ x^4 y^4 + x^2 y^2 + 1 - k (x^4 + x^2 y^2 + y^4) = 0 $$ is in terms of $x$ (over the reals), we find that the four solutions (except when $k=y^4$) are $$ x=\pm \sqrt{ \frac{(k-1)y^2 \pm \sqrt{4k-(3+2k+3k^2)y^4+4ky^8}}{2(k-y^4)}}. $$ So then we ask when $k-\frac{3+2k+3k^2}{4}z^4+k z^8$ is a perfect square (where $k,z\in \mathbb{Z}^+$), and it seems (by computer) the answer is "if and only if $k=1$". $\endgroup$ Jul 21, 2014 at 6:30
  • $\begingroup$ @John: does that computer search use Baker's theorem to bound $k$? Or something clever with numbers of the form $a^2+ab+b^2$? $\endgroup$ Jul 22, 2014 at 15:30
  • $\begingroup$ @ZackWolske: nothing clever, just brute force. We know that $k$ needs to be odd, and substituting $k=2n+1$ gives us $-3n^2z^4+(2n+1)(z^4-1)^2$. $\endgroup$ Jul 22, 2014 at 23:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.