Timeline for Diophantine equation with no integer solutions, but with solutions modulo every integer
Current License: CC BY-SA 2.5
24 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2023 at 17:38 | answer | added | user178594 | timeline score: 5 | |
| Aug 3, 2021 at 8:10 | comment | added | Alon Amit | See this previous MO question for a more specific discussion of proofs of the properties of Selmer’s curve. | |
| Aug 1, 2021 at 15:34 | answer | added | Bogdan Grechuk | timeline score: 4 | |
| Jan 1, 2011 at 23:22 | answer | added | Ravi Boppana | timeline score: 22 | |
| Dec 2, 2010 at 16:05 | answer | added | Christian Elsholtz | timeline score: 5 | |
| Nov 27, 2010 at 19:34 | answer | added | KConrad | timeline score: 22 | |
| Nov 27, 2010 at 15:44 | answer | added | Mikhail Borovoi | timeline score: 3 | |
| Nov 27, 2010 at 11:51 | comment | added | Tim Dokchitser | Another elementary example is $x^2+y^2+z^2+w^2=-1$. Because every positive integer is a sum of 4 squares, this has solutions modulo every $n\ge 2$. | |
| Nov 27, 2010 at 1:38 | comment | added | Emerton | Sorry: "conjugacy of $G$" should read "conjugacy class of $G$". | |
| Nov 27, 2010 at 1:38 | comment | added | Emerton | Dear A. Pacetti, Here is the argument I had in mind (it is not original to me, though); hopefully I am not butchering it: consider the Galois group $G$ of the splitting field of the polynomial. If the poly. $f$ is irred., then $G$ acts transitively on the roots of $f$, and so group theory shows that (if $f$ has degree $> 1$) then there is a conjugacy of $G$ with no fixed point. The $p$ whose Frobenius elements are equal to this conjugacy class then have the property that $f$ has no root modulo $p$. (Such a root would give a fixed point for the Frobenius of $p$. | |
| Nov 26, 2010 at 23:20 | comment | added | A. Pacetti | Is Emerton's assertion easy to prove? (that an irreducible polynomial in one variable cannot have a root modulo every prime). The easy case is the one where the polynomial gives a Galois extension. Then Tchebotarev density thm implies the result, but if the extension is not Galois, taking the Galois closure is not enough... (I guess that the result should say that it cannot have a solution on a set of primes with density 1). | |
| Nov 26, 2010 at 20:47 | comment | added | Kevin Buzzard | PS gah why didn't I follow Qiaochu's duplicate link before posting all that stuff again :-/ | |
| Nov 26, 2010 at 20:37 | comment | added | Kevin Buzzard | Remark: the reason I'm blowing my own trumpet here is that Keith's pdf resorts to the crutch of showing there are solutions mod p by using character sums over finite fields; this is often the approach used, in my experience. The homework I set gives a self-contained approach inspired by the trick used in the quadratic case. | |
| Nov 26, 2010 at 20:30 | comment | added | Kevin Buzzard | A proof that there are no solutions over Q is in Cassels' book on elliptic curves. An elementary proof that there are p-adic solutions for all p, using Hensel's lemma and an argument very much in the spirit of Qiaochu's answer (i.e. using a "cube" analogue of the statement that if $a$ and $b$ aren't squares mod $p$ then $ab$ is) can be found at www2.imperial.ac.uk/~buzzard/maths/teaching/10Aut/M4P32/… : I just set it as homework for my students in fact :-) | |
| Nov 26, 2010 at 20:20 | answer | added | Felipe Voloch | timeline score: 40 | |
| Nov 26, 2010 at 19:43 | answer | added | Will Jagy | timeline score: 8 | |
| Nov 26, 2010 at 19:41 | comment | added | Faisal | Qiaochu & Emerton: Thanks! I'm glad to learn that this curve has a name. | |
| Nov 26, 2010 at 19:41 | vote | accept | Faisal | ||
| Nov 26, 2010 at 18:52 | comment | added | Emerton | The example in your question is a famous example of Selmer. An explanation is given by Keith Conrad here: math.uconn.edu/~kconrad/blurbs/gradnumthy/selmerexample.pdf Note also that the example given below by Qiaochu are reducible. One can show that if $f(x)$ is an integer polynomial in one variable that is irreducible, then it cannot have a solution modulo every prime. The essential point of Selmer's example is that it is irreducible. | |
| Nov 26, 2010 at 18:00 | comment | added | Qiaochu Yuan | @Fedor: "solution" here refers to a point on the corresponding projective curve. | |
| Nov 26, 2010 at 17:51 | comment | added | Fedor Petrov | But this equation has a solution $(0,0,0)$. | |
| Nov 26, 2010 at 17:38 | answer | added | Qiaochu Yuan | timeline score: 28 | |
| Nov 26, 2010 at 17:35 | comment | added | Qiaochu Yuan | Duplicate: mathoverflow.net/questions/2779/… | |
| Nov 26, 2010 at 17:31 | history | asked | Faisal | CC BY-SA 2.5 |