Timeline for Does the equation x^10+y^10+z^10=t^2 have positive integer solutions? [closed]
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12 events
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| Nov 21, 2009 at 13:21 | comment | added | JSE | RE coprime: good point! I guess I mean: this question, unlike many others on this site, has an answer. And working out that answer, or even starting to think about what it is, involves interesting mathematics. In fact, I think this is a substantially BETTER question than something more vague like "Can you prove a theorem like Fermat in four variables instead of three?" An equally good question, which also has an answer, would be "What is presently known about equations of the form x^a + y^b + z^c + w^d = 0?" | |
| Nov 21, 2009 at 6:53 | comment | added | Kevin Buzzard | JSE: I think you can only pull off the "not pairwise coprime" trick" if the exponents are coprime. Or am I missing something? You certainly can't produce a solution to x^10+y^10=z^10 in positive integers, coprime or not! I absolutely agree that it seems like a "perfectly good question" in the sense that it's not easy. But it doesn't seem like it in most other senses. | |
| Nov 21, 2009 at 3:40 | comment | added | JSE | This seemed like a perfectly good question to me. By the way, the equation presumably DOES have positive integer solutions, just not one where (x,y,z,t) are coprime. | |
| Nov 21, 2009 at 2:08 | comment | added | David Zureick-Brown | Scratch my paramaterization suggestion. The paramaterization you get is not useful. | |
| Nov 21, 2009 at 1:53 | comment | added | David Zureick-Brown | Also, I think that a more detailed version of this question is indeed appropriate for mathoverflow. | |
| Nov 21, 2009 at 1:51 | comment | added | David Zureick-Brown | The quadratic surface (in P^3) x^2 + y^2 + z^2 = t^2 is rational; geometrically (over $C$) you can form it from blowing up two points in $P^2$ and blowing down the line between them (I'm not sure if this is true over $Q$, but the surface at least contains a few rational lines so it is a reasonable guess). Find an explicit paramaterization away from the blow ups (e.g. x = f(a,b,c)) and then try to solve the equation x^5 = [f(a,b,c)]^2, which may or may not be easier. This is a lot of work and a long shot, but an idea. | |
| Nov 21, 2009 at 1:08 | comment | added | Anton Geraschenko | This question looks totally random. If you provide some motivation for your question, and clarify the meaning of your last question, I'll reopen the question. | |
| Nov 21, 2009 at 1:05 | comment | added | David E Speyer | I suspect this is a variety of general type. The Bombierri-Lang conjecture would then say that the rational points are contained in finitely many curves. Other than that, I have no idea what to say. | |
| Nov 21, 2009 at 1:03 | history | closed | Anton Geraschenko | too localized | |
| Nov 20, 2009 at 23:42 | comment | added | Kevin Buzzard | Diophantine equations are hard. It might be the case that there are no solutions but that this is very hard to prove. I'd be more motivated to think about it if you explained why you were interested. If you've just written down a random equation---well, we can all do that... | |
| Nov 20, 2009 at 23:18 | comment | added | Yemon Choi | I don't understand your last question. Also, why have you picked those particular exponents? | |
| Nov 20, 2009 at 22:36 | history | asked | Konstantin | CC BY-SA 2.5 |