Timeline for For $x,y\ge 2$ does $x^4+x^2y^2+y^4$ ever divide $x^4y^4+x^2y^2+1$?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 17, 2014 at 6:05 | comment | added | Aaron Meyerowitz | 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. | |
| Jul 17, 2014 at 6:04 | comment | added | John Bamberg | One idea I've add is to use Zsigmondy's Theorem. There exists a primitive prime divisor $p$ dividing $x^6-y^6=(x^2-y^2)(x^4+x^2y^2+y^4)$ and so $p$ divides $x^6-1$ or $x^6+1$ (as we also know that $p\ne 2$). Still no idea what to do with this yet. | |
| Jul 17, 2014 at 5:30 | comment | added | John Bamberg | @GHfromMO: Ah yes, thanks for that. | |
| Jul 17, 2014 at 4:33 | comment | added | user44191 | @GHfromMO Blah, of course. My apologies; I'm tired. | |
| Jul 17, 2014 at 4:21 | comment | added | GH from MO | @user44191: It is trivial that it is sufficient to show that $p\mid r$ and $q\mid r$ cannot hold simultaneously (because $pq\mid r$ implies $p\mid r$ and $q\mid r$). What you really observed is the fact that the reduced statement is no stronger than the original one (i.e. the impossibility of $pq\mid r$). | |
| Jul 17, 2014 at 4:14 | comment | added | user44191 | 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 $pq|r$. | |
| Jul 17, 2014 at 3:51 | comment | added | GH from MO | @JohnBamberg: I don't think so. Instead, we have reduced the problem to showing that we cannot have that, simultaneously, $(y^7-y^3)x-y^4+1$ is divisible by $x^2+xy+y^2$ and $(y^3-y^7)x-y^4+1$ is divisible by $x^2-xy+y^2$ (for any $x,y\geq 2$). | |
| Jul 17, 2014 at 3:50 | comment | added | Gerry Myerson | I accept the objections. I note (for whatever it might be worth) that $(y^7-y^3)x-y^4+1=(y^4-1)(xy^3-1)$ | |
| Jul 17, 2014 at 3:45 | comment | added | John Bamberg | 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 $x,y\ge 2$)". | |
| Jul 17, 2014 at 3:33 | comment | added | Glasby | @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. | |
| Jul 17, 2014 at 3:27 | comment | added | GH from MO | The $x$-polynomial remainders need not be zero for $p\mid r$ and $q\mid r$ to hold (for a given $x,y\in\mathbb{Z}$). | |
| Jul 17, 2014 at 2:16 | history | answered | Gerry Myerson | CC BY-SA 3.0 |