Timeline for The equation $x^m-1=y^n+y^{n-1}+...+1$ in prime powers $x,y$
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 8, 2013 at 21:03 | vote | accept | Michael Zieve | ||
| Nov 6, 2013 at 17:08 | answer | added | Alvin | timeline score: 1 | |
| Nov 6, 2013 at 15:55 | answer | added | Michael Zieve | timeline score: 1 | |
| Nov 6, 2013 at 6:58 | comment | added | Michael Zieve | Thanks! I'll let you know if I use this in a paper. You definitely helped a lot! | |
| Nov 6, 2013 at 2:00 | comment | added | Mike Bennett | If you want $y$ to be an odd prime power, then, assuming $x=4$, you have that $n$ is even and so we can apply Runge's method (assuming you can prove the necessary irreducibility, which I think is OK). One gets a bound like $2^m \ll 12^n$ or something similar, which bounds $y$, unless I've messed up. | |
| Nov 6, 2013 at 0:31 | comment | added | Michael Zieve | Thanks Mike! I wasn't aware of the Nagell-Ljunggren equation. That keyword led me to lots of relevant papers which I'm now studying. I have one further question: the special case $x=4$ is of particular interest for my application. Is it conceivable to prove the desired finiteness in that case? | |
| Nov 5, 2013 at 21:27 | comment | added | Mike Bennett | As another comparison, if you just replace your left-hand-side $x^m-1$ by $x^m$, you would have the Nagell-Ljunggren equation (which also shows up in group theory). This equation is, on the surface, a little easier to attack than the one you are considering, but is still unsolved even in the sense that it is unknown whether the number of solutions (in four variables) is finite. | |
| Nov 5, 2013 at 6:30 | comment | added | Michael Zieve | @Mike: thanks! That's helpful. It seems to me that the Ramanujan-Nagell equation ($2^r-7=s^2$) is equivalent to the case $n=x=2$ of my equation, if we ignore the constraint that $y$ is a prime power. But the Ramanujan-Nagell equation has been solved, so maybe it isn't yet completely crazy to hope for a complete solution to my question? | |
| Nov 5, 2013 at 4:52 | comment | added | Mike Bennett | I suspect this is quite hard (though, as my daughter is quick to point out, I'm frequently mistaken). The arguments that get you off the ground for Catalan are very sensitive and, for instance, don't get you anywhere for the equation $x^n-y^m=2$. The beast you're after is, in the case $n=2$, essentially the Ramanujan-Nagell equation (where the assumption that $x$ is prime does, admittedly, simplify things). | |
| Nov 5, 2013 at 2:30 | comment | added | Michael Zieve | @TheMaskedAvenger: thanks for the pointer. Unfortunately I can't find the book online, and Tsinghua's library doesn't have it. But it must be somewhere in Beijing... I'll keep searching. | |
| Nov 4, 2013 at 23:06 | comment | added | The Masked Avenger | This looks close enough to Catalan's conjecture that you might check out Ribenboim's book. He has a lot of ancillary material; you might find something useful there. | |
| Nov 4, 2013 at 22:52 | comment | added | Michael Zieve | Thanks. I edited accordingly. I definitely don't want to fix $m,n$. On the other hand, I'm happy to use any and all sledgehammers (I've already used the classification of finite simple groups in the group theory argument, so the proof of my application is going to be non-elementary regardless). | |
| Nov 4, 2013 at 22:42 | history | edited | Michael Zieve | CC BY-SA 3.0 |
Clarifying question
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| Nov 4, 2013 at 22:39 | comment | added | Qiaochu Yuan | The wording is a little unclear as to whether $m, n$ are fixed or whether you want "finitely many" to apply to tuples $(x, y, m, n)$. If they're fixed, then finiteness follows for all but finitely many pairs $(m, n)$ by Faltings' theorem, although this is a sledgehammer. | |
| Nov 4, 2013 at 22:27 | history | asked | Michael Zieve | CC BY-SA 3.0 |