Timeline for Finding integer points on elliptic curves via divisibility conditions like $(a+b)^2 \mid (2b^3+6ab^2-1)$
Current License: CC BY-SA 3.0
50 events
| when toggle format | what | by | license | comment | |
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| Dec 16, 2014 at 15:14 | comment | added | GH from MO | @KierenMacMillan: Mail sent! | |
| Dec 16, 2014 at 14:04 | comment | added | Kieren MacMillan | @GHfromMO: I'd appreciate it if you could email me your results (see my profile for address). Thanks! | |
| Dec 10, 2014 at 7:50 | comment | added | GH from MO | @KierenMacMillan: Sorry, I got busy with other things and the remote computers I used were rebooted regularly, so I stopped checking. I covered all $\ell<2.1\times 10^{12}$ though, and found no counterexample. In this range, I recorded all triples $(\ell,m,m/\ell)$ with $\ell^2\mid m^3-2$ and $m/\ell<50$. I found 421 such triples, and among them the fractions $m/\ell$ distribute very uniformly: if we order these by size in Excel, their histogram is almost a perfect line. So all the empirical and theoretical evidence suggests that there is a counterexample, but I did not find one. | |
| Dec 10, 2014 at 0:06 | comment | added | Kieren MacMillan | @GHfromMO: Just checking in on progress. Would love to see whatever data you've already got, etc. | |
| Sep 6, 2014 at 4:24 | comment | added | GH from MO | I have checked all $\ell<10^{12}$ by distributing the task between many machines. This is quite a big range. I will continue the search for another week, month, or year (depending when I get tired), and I will post a summary then. At the moment all I can say is that the data is in good agreement with Lucia's heuristic. | |
| Sep 5, 2014 at 23:58 | comment | added | Kieren MacMillan | Are there any updates? Have you reached a technical limit on the search for counterexamples? | |
| Sep 5, 2014 at 12:26 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 27, 2014 at 10:28 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 26, 2014 at 13:19 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 26, 2014 at 13:14 | comment | added | GH from MO | @KierenMacMillan: Yes, this is why I kept the previous example of $0.485$ as well. | |
| Aug 26, 2014 at 13:13 | comment | added | GH from MO | @Joël: Thanks for your comment. I will update the text accordingly. | |
| Aug 26, 2014 at 13:07 | comment | added | Kieren MacMillan | Note that the $\approx 0.350$ example doesn't have $\ell,m$ odd, so it's not a candidate. The hypothetical counterexample record still stands at $\approx 0.485$. | |
| Aug 26, 2014 at 12:59 | comment | added | Joël | 0.350... You're getting close! But you should consider writing explicitly in your answer, near the numerical examples, that an example with $m/l < 0.333$ would be a counter-example to the original OP's conjecture. Right now a new reader has to read carefully Lucia's answer or the comment on yours to find this information. | |
| Aug 26, 2014 at 12:47 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 26, 2014 at 12:36 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 26, 2014 at 3:42 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 26, 2014 at 2:39 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 26, 2014 at 1:14 | comment | added | GH from MO | Consider the odd pairs $(\ell,m)$ satisfying $\ell^2\mid m^3-2$. Then the ratios $m/\ell$ are equidistributed in $(0,\infty)$ in a certain sense (I can be more specific if you wish). In particular, the ratios $m/\ell$ are dense in $(0,\infty)$, hence they are infinitely often smaller than $1/3$. | |
| Aug 26, 2014 at 1:13 | comment | added | GH from MO | @KierenMacMillan: One counterexample does not tell much, beyond disproving your conjecture of course. Infinitely many counterexamples would tell that the abc conjecture cannot be strengthened in a certain way. At any rate, here is a conjecture that I believe is true and would yield infinitely many counterexamples to your conjecture (continued in next comment): | |
| Aug 26, 2014 at 1:05 | comment | added | Kieren MacMillan | @GHfromMO: In your original answer, you suggested that the abc conjecture would imply a slightly weaker conjecture is “probably true”. If you find a counterexample to my stronger conjecture, what will that say?? | |
| Aug 26, 2014 at 1:02 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 26, 2014 at 1:02 | comment | added | Kieren MacMillan | According to your original transformation, isn't $b=(\ell-m)/2$? And why do you think $k \approx b$? | |
| Aug 26, 2014 at 0:25 | comment | added | Lucia | I think $k$ is a about size $b$, which is roughly $\ell-m$. Knowing that this is at least $5$ doesn't help, or hurt, the computations of GH from MO. In the notation of the original question, GH from MO's computations have disproved the conjecture if the condition had been $b<a<8b$. | |
| Aug 26, 2014 at 0:09 | comment | added | GH from MO | @KierenMacMillan: Smaller ratios for $m/\ell$ are harder to find. Please do not change your conjecture to make it harder for us to disprove :-) | |
| Aug 26, 2014 at 0:02 | comment | added | Kieren MacMillan | Well, for example, I think I could invert the current proof, and show that $k \ge 5$. So that would, I think, mean you're looking for a ratio $m/\ell \le 0.2$, yes? So your bound on $m$ for a given $\ell$ would be almost cut in half. | |
| Aug 25, 2014 at 22:31 | comment | added | GH from MO | @KierenMacMillan: I don't know how to make use of any information on the quotient. What Noam's code does is: it cycles over $\ell$, and for each $\ell$ it finds the smallest $m$ such that $\ell^2\mid m^3-2$. | |
| Aug 25, 2014 at 22:23 | comment | added | Kieren MacMillan | If I can push my algebraic proof a little further, I might be able to get a bound on the quotient $k$, which would help reduce [possibly significantly] the number of tests required. | |
| Aug 25, 2014 at 21:26 | comment | added | GH from MO | @KierenMacMillan: Yes. I am sure it exists, but it might be as large as $10^{16}$, for which a supercomputer would be needed. I was already surprised by finding a pair with ratio $0.766$. | |
| Aug 25, 2014 at 21:08 | comment | added | Kieren MacMillan | A counterexample would be $< 0.333...$, am I correct? | |
| Aug 25, 2014 at 20:38 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 25, 2014 at 20:34 | comment | added | GH from MO | Using your code, I found a new example with $\ell^2 \mid m^3 - 2$ and $\ell>m$: $(\ell,m)=(3230984317,2474621651)$. A pleasant feature is that $m$ is odd and $m/\ell\approx 0.766$ is significantly below $1$. | |
| Aug 25, 2014 at 15:48 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 25, 2014 at 15:42 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 25, 2014 at 14:43 | comment | added | GH from MO | @NoamD.Elkies: I am stretching your range with your program, in the hope of hitting a counterexample. See my added section. Perhaps you have access to faster computers. | |
| Aug 25, 2014 at 14:40 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 25, 2014 at 8:48 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 25, 2014 at 8:38 | comment | added | GH from MO | @NoamD.Elkies: Thank you for the code! | |
| Aug 25, 2014 at 2:49 | comment | added | Noam D. Elkies | [The condition if(l%3, (i.e., don't try $\ell$ if $3\mid\ell$) exploits the fact that $x^3-2$ has no roots mod $3^2$. I could have also saved a factor of nearly $3/4$ by requiring that $\ell$ not be divisible by $7$, $13$, or $19$.] | |
| Aug 25, 2014 at 2:46 | comment | added | Noam D. Elkies | OK, gp code follows, though I'm afraid the line breaks and indentations will be lost $-$ or is there a way to retain such formatting in comments? $$ $$ L = 10^8; { forstep(l=3,L,2, if(l%10^6==1,print("<",l-1,">")); if(l%3, F = factor(l); n = #F[,1]; v = vector(n,i,polrootspadic(x^3-2,F[i,1],2*F[i,2])); for(i=1,n, v[i] = lift(v[i]) * Mod(1,F[i,1]^(2*F[i,2]))); forvec(r=vector(n,i,[1,#v[i]]), m = Mod(0,1); for(i=1,n, m=chinese(m,v[i][r[i]])); m = lift(m); if(m<l,print([l,m])) ) ) ) } | |
| Aug 25, 2014 at 1:58 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 25, 2014 at 1:52 | comment | added | GH from MO | @NoamD.Elkies: This is a good idea, thank you! I am using a very simple 3-line code in SAGE, utilizing the function squarefree_part. Can you post your code, e.g. as part of a response here? I could learn from it as I am not at all familiar with gp. | |
| Aug 25, 2014 at 0:59 | comment | added | Noam D. Elkies | It's faster to look for $\ell^2 \mid m^3 - 2$ via the factorization of $\ell$ rather than $m^3-2$. A 30 minute calulation in gp (using polrootspadic) finds that there are no examples of $m<\ell$ with $\ell$ odd and $0 < \ell < 10^8$ besides the known $(\ell,m) = (5,3)$ and $(127,100)$. | |
| Aug 24, 2014 at 21:22 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 24, 2014 at 20:23 | comment | added | GH from MO | @Lucia: Yes. Also, there are refinements of the abc conjecture that can make the $m^{2+\epsilon}$ bound more precise. | |
| Aug 24, 2014 at 20:21 | comment | added | Lucia | Note that the heuristic I gave also suggests that apart from finitely many exceptions, the largest square factor of $m^3-2$ is at most $(m\log m)^2$, say. So this is certainly in a very delicate range! | |
| Aug 24, 2014 at 20:04 | comment | added | GH from MO | $x\ll_\epsilon y$ means that there is a constant $c_\epsilon$ depending only on $\epsilon$ such that $|x|\leq c_\epsilon y$. Another notation for the same relation is $x=O_\epsilon(y)$. | |
| Aug 24, 2014 at 20:02 | comment | added | Kieren MacMillan | Very interesting! I'm familiar with $\ll$ but not $\ll_\epsilon$. What is the meaning? | |
| Aug 24, 2014 at 20:01 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 24, 2014 at 19:39 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Aug 24, 2014 at 19:24 | history | answered | GH from MO | CC BY-SA 3.0 |