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
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 23, 2014 at 0:48 | comment | added | John Bamberg | @GeoffRobinson: no proof at all yet. A few times I thought I had it but there's always been a gaping hole. What I claimed above is false. Take $k=81$ and $z=6$. However, it yields an irrational when plugged into the formula for $x$. The best lead we have at the moment is Geoff's equation with the product of the four gcd's. | |
| Jul 23, 2014 at 0:00 | comment | added | Geoff Robinson | If you have a complete proof ((by computer or not) that the divisibility is impossible when neither of $x,y$ is $1,$ , then you should give it as an answer. | |
| Jul 22, 2014 at 23:56 | comment | added | John Bamberg | @ZackWolske: nothing clever, just brute force. We know that $k$ needs to be odd, and substituting $k=2n+1$ gives us $-3n^2z^4+(2n+1)(z^4-1)^2$. | |
| Jul 22, 2014 at 15:30 | comment | added | Zack Wolske | @John: does that computer search use Baker's theorem to bound $k$? Or something clever with numbers of the form $a^2+ab+b^2$? | |
| Jul 21, 2014 at 6:30 | comment | added | John Bamberg | 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 $k=1$". | |
| Jul 21, 2014 at 5:17 | comment | added | Geoff Robinson | @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. | |
| Jul 21, 2014 at 4:25 | comment | added | Aaron Meyerowitz | 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. | |
| Jul 20, 2014 at 21:23 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
added 836 characters in body
|
| Jul 20, 2014 at 10:53 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
deleted some text
|
| Jul 20, 2014 at 10:21 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
typos
|
| Jul 20, 2014 at 10:04 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Gave example
|
| Jul 20, 2014 at 9:58 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Gave example
|
| Jul 19, 2014 at 18:09 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Slight expansion
|
| Jul 19, 2014 at 12:34 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
minor clarification
|
| Jul 19, 2014 at 11:30 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
minor correction
|
| Jul 19, 2014 at 1:27 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
minor explanation
|
| Jul 19, 2014 at 0:44 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Expanded answer-also minor layout change
|
| Jul 18, 2014 at 7:39 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
typo
|
| Jul 17, 2014 at 15:02 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
typo
|
| Jul 17, 2014 at 13:31 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
minor clarification
|
| Jul 17, 2014 at 9:30 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Expanded original answer
|
| Jul 17, 2014 at 7:40 | history | answered | Geoff Robinson | CC BY-SA 3.0 |