Timeline for Re: Mordell's equation $y^2 = x^3 + k$ and perfect numbers
Current License: CC BY-SA 4.0
51 events
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| Aug 31 at 18:31 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Aug 31 at 15:56 | history | edited | Jose Arnaldo Bebita | CC BY-SA 4.0 |
TeXified post, added minor contextual updates
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 5, 2015 at 19:25 | review | Close votes | |||
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| Apr 5, 2015 at 19:00 | answer | added | Joe Silverman | timeline score: 7 | |
| Apr 5, 2015 at 18:35 | history | edited | Jose Arnaldo Bebita |
added "perfect-numbers" tag
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| Oct 5, 2013 at 23:03 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
minor math edits
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| Apr 13, 2011 at 6:45 | comment | added | Yemon Choi | Arnie: please stop using MO as a place to announce your work | |
| Apr 13, 2011 at 5:17 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
updated post with the current state-of-the-art
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| Mar 3, 2011 at 5:23 | history | edited | Jose Arnaldo Bebita |
added 1
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| Dec 8, 2010 at 20:58 | comment | added | Jose Arnaldo Bebita | Update: I just saw this MathWorld link: mathworld.wolfram.com/MordellCurve.html. My apologies to everyone if I didn't bother to check. But given that the Mordell curve only has a finite number of solutions when $k$ is an integer, (and given Felipe's answer below), I do see where the "weakness" of my argument is: It's in passing from $\mathbb{Q}$ to $\mathbb{Z}$ that we do indeed have to "discover" a specific relationship between "the" k and "the" K. Indeed, I would think that a full characterization for odd numbers of the form $N = {p^k}{m^2}$ to be perfect would be required. | |
| Dec 8, 2010 at 1:43 | comment | added | Jose Arnaldo Bebita | @Kevin, I forgot to say: Thank you also about the CAS [i.e. Computer Algebra System] idea. Although, I prefer doing everything by hand first, then verifying if my calculations are correct using a CAS. WolframAlpha is one (online) CAS that I consider very useful for doing mathematics this way. | |
| Dec 8, 2010 at 1:39 | comment | added | Jose Arnaldo Bebita | @Kevin, my profuse thanks for providing the second counterexample! That being said, I will (of course) still be interested in a theorem/lemma (or any result for that matter) which contains a set of sufficient conditions for finiteness. Lastly, please do try to look at www3.alpha-net.ne.jp/users/fermat/dioph54e.html when you get the time. I got that link just now. I don't really know if there is a problem with my PC (e.g. malware, network issues, etc.), but I have tried searching for that webpage numerous times ever since I saw it. Moving forward, we can now take note of the URL. =) | |
| Dec 7, 2010 at 19:23 | comment | added | Kevin Buzzard | @Arnie: given an explicit elliptic curve, and a computer algebra package, you can easily check for yourself whether it has finitely or infinitely many points on it. You suggest that you're interested in 350 elliptic curves. About half of them will have finitely many solutions and the other half will have infinitely many (to a very rough first order approximation). I hope this information helps somehow! If you actually care about which curves have infinitely many points then I urge you to learn how to use a standard package like mwrank (within SAGE) to compute these things yourself. | |
| Dec 7, 2010 at 19:19 | comment | added | Kevin Buzzard | @Arnie: your revised criterion is also not right; $K=2$ is a counterexample. I'm afraid that I can no longer take this criterion seriously: you firstly stated it as an iff criterion but neither implication is true, and furthermore you have given no hint of either a proof or a reference :-( | |
| Dec 7, 2010 at 11:34 | comment | added | Jose Arnaldo Bebita | @KConrad, in (2), I actually meant to say "Check for finiteness" when I said "Check each resulting value of $K$..." but I am guessing the context is clear enough... | |
| Dec 7, 2010 at 11:27 | comment | added | Jose Arnaldo Bebita | That being said, thank you for your comment @KConrad. Although, I will have to say, that I did state that you need to do two things in order to prove what I "conjectured": (1) Come up with a natural and/or suitable "relationship" between $k$ and $K$ (in the context of the theory of perfect numbers, crossing over to elliptic curve theory). That is, give an interpretation for $K$ that naturally "relates" to the interpretation for $k$. (2) Check each resulting value of $K$ in the range implied by the "interpretation" you have given in (1). If you can do both, the conjecture is proved. | |
| Dec 7, 2010 at 11:23 | comment | added | Jose Arnaldo Bebita | ... of generating new solutions from old. Indeed, I have not checked for finiteness of solutions for any $K$ in the range I've given. Nonetheless, I am convinced that the theory of perfect numbers can serve as a "baseline" for further developments in number theory (e.g. the Euclid-Euler model allows one to distinguish between the even and odd cases by looking at the "exponent" of the Mersenne and Euler primes, respectively.) For more on this, you can have a look at arnienumbers.blogspot.com/2010/12/…. | |
| Dec 7, 2010 at 11:13 | comment | added | Jose Arnaldo Bebita | Or to put it in another way: The fact that it is conjectured that there are infinitely many even perfect numbers contrasted to the "conjecture" that there are no OPNs is an instance of the "law of excluded middle" -- provided, that there is NO relationship (direct or indirect, explicit or implicit) between "the" $k$ and "the" $K$. I will have to categorically state that I do not know how to check when a particular elliptic curve (corresponding to a particular value of $K$ for Mordell's equation) has finitely many solutions [although I am familiar with chord-tangent method... | |
| Dec 7, 2010 at 11:02 | comment | added | KConrad | The posted question presents an idea in the direction of showing there are finitely many perfect numbers. However, the even perfect numbers correspond to Mersenne primes and it is a standard conjecture that there should be infinitely many such primes. More precisely, there are heuristic predictions for the frequency of Mersenne primes and I think the data so far fit the heuristic at least up to an order of magnitude. Everyone expects there to be infinitely many even perfect numbers, so a strategy for proving there are just finitely many perfect numbers is suspicious. | |
| Dec 7, 2010 at 10:10 | history | edited | Jose Arnaldo Bebita | CC BY-SA 2.5 |
corrected an error with my arithmetic @darn@
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| Dec 7, 2010 at 10:03 | history | edited | Jose Arnaldo Bebita | CC BY-SA 2.5 |
contextualized my post
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| Dec 7, 2010 at 9:11 | comment | added | Jose Arnaldo Bebita | Apologies for the delay in responding. @Kevin, I see your point now. I didn't realize that the "iff" mattered here. (Indeed, I was under the naive assumption that BSD would be proven ahead of any such "iff" criterion). Notwithstanding, I still have not gotten hold of the Japanese paper I alluded to the other day. I do believe however, given your counterexample, that the conditions given by (1), (2) and (3) are sufficient for showing that Mordell's equation has finitely many solutions. Editing my post now to clarify the context for the number-theory problem that I am considering. | |
| Dec 5, 2010 at 13:13 | comment | added | Kevin Buzzard | @Arnie: you're barking up the wrong tree. I don't care what the class number is. $7$ isn't $1$ or $2$ mod 4, so (2) fails, so "(1) and (2) and (3)" fails regardless of the truth status of (3). So according to the "criterion" you're claiming, the curve should have infinitely many points. And my computer says it doesn't. Let me again stress that I am a bit skeptical about a genuine provable "iff" criterion of this form. | |
| Dec 5, 2010 at 11:40 | comment | added | Jose Arnaldo Bebita | @Kevin, I did verify that the class number of $Q(\sqrt{7})$ is $1$, per WolframAlpha. However, it is still a problem for me to get hold of the paper I alluded to (e.g. I do not even recall the author's name nor the title of the paper). At any rate, I will get back to you with an update ASAP, earliest would be tomorrow. | |
| Dec 5, 2010 at 11:33 | comment | added | Jose Arnaldo Bebita | @Kevin, for k = 7, does 3 not divide the class number of the (real) quadratic field $Q(\sqrt{k})$? | |
| Dec 4, 2010 at 12:28 | comment | added | Kevin Buzzard | @Arnie: I should clarify: there might be wonderful simple criteria---but they will depend on BSD being true. | |
| Dec 4, 2010 at 12:27 | comment | added | Kevin Buzzard | I would worry more about the fact that I have given you a counterexample to the criterion you're claiming! | |
| Dec 4, 2010 at 12:26 | comment | added | Kevin Buzzard | Arnie: you are living in a dream world. You cannot expect to find a wonderful simple criterion for this equation to have finitely many solutions. There are plenty of methods for computing class numbers of real quadratic fields. There are also plenty of methods for computing ranks of elliptic curves such as the one in your question. You are asking these questions but it's hard to believe that the answers can help you. Say for example that I just tell you some explicit formula whose answer is the class group---such formulas exist. What will you do now? | |
| Dec 4, 2010 at 11:54 | comment | added | Jose Arnaldo Bebita | @Kevin: I am still trying to get hold of the paper containing the result I alluded to. Will let you know ASAP. @Franz: Thanks for pointing that out. In particular, may I ask if there is a known closed-form for the class number of the real quadratic field $Q(\sqrt{k})$? | |
| Dec 4, 2010 at 11:12 | comment | added | Franz Lemmermeyer | Conditions such as the one on the class number usually are made to make the proof work. If the class number is divisibly by 3, on the other hand, then this does not mean that there are solutions, it just means that you have to work harder for solving the problem. | |
| Dec 4, 2010 at 10:55 | comment | added | Kevin Buzzard | [$k=7$: only finitely many points, but $k$ isn't 1 or 2 mod 4] | |
| Dec 4, 2010 at 10:54 | comment | added | Kevin Buzzard | Arnie: the criterion you write is not correct. $k=7$ is a counterexample, at least according to my computer. | |
| Dec 4, 2010 at 10:48 | comment | added | Jose Arnaldo Bebita | @Kevin, I will try to get hold of that Japanese(?) paper and will get back to you with an update ASAP. (It is a bit hard to try searching for it using Google, etc. because of the differences in alphabetic characters [i.e. English vs. Nihongo].) But that being said, my profuse thanks for taking the time out to read my post. Your effort is appreciated! =) | |
| Dec 4, 2010 at 10:42 | history | edited | Jose Arnaldo Bebita | CC BY-SA 2.5 |
final edit
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| Dec 4, 2010 at 10:35 | comment | added | Jose Arnaldo Bebita | @Kevin, editing my post now. | |
| Dec 4, 2010 at 9:38 | comment | added | Kevin Buzzard | Why don't you write the correct criterion in the statement of the question? That way people will understand much better what you are asking. As I have already said, currently a logically correct answer to the question is "number of solutions is finite iff number of solutions is finite". | |
| Dec 4, 2010 at 9:36 | comment | added | Kevin Buzzard | So, for example, the way you wrote your criterion, it seems to me that you're saying that if $k$ is 3 mod 4 then there will always be infinitely many solutions. I am not sure that this sort of statement is accessible. This is exactly the sort of statement that is super-hard to prove. It might be true, but I am a bit skeptical. In fact it's not even true is it---consider $k=7$. | |
| Dec 4, 2010 at 9:33 | comment | added | Kevin Buzzard | Let me also make some other comments. If one wants to prove a statement of the form "number of solutions is finite if and only if [some condition $C$ is true for $k$]" then the hard part of the proof will be proving that if $C$ fails then there are infinitely many solutions. This is exactly the part of BSD that we can't get to, because solving Diophantine equations is hard. In particular I'm saying any iff statement for $k$ should either be not too deep (i.e. basically equivalent), or should assume BSD, or should be amazing. | |
| Dec 4, 2010 at 9:07 | comment | added | Kevin Buzzard | @Arnie: I'm not sure you've stated the criterion correctly. In fact I'm quite confused. You implied in the question that the criterion in the paper only applied for $k=1$ or 2 mod 4. What you have written in the comment above seems to apply for all $k$ (in the sense that it seems to give information for all $k$). Furthermore condition (3) seems to imply that $k>0$ and condition (1) seems to imply that $k<0$ is the case of interest. | |
| Dec 4, 2010 at 8:46 | comment | added | Jose Arnaldo Bebita | @Kevin, my profuse thanks for your detailed response. As I am not an expert in elliptic curve theory, I would have to disclose what exact approach I am taking, and from where I am taking off. The (Japanese[?]) paper that I allude to in my previous comment states that Mordell's equation will have finitely many rational solutions iff (1) $−k$ is not of the form $(3t^2)+1$ or $(3t^2)−1$, (2) $k≡1(mod4)$ (xor) $k≡2(mod4)$, and (3) 3 does not divide the class number of the [real quadratic field] $Q(\sqrt{k})$. I was trying to see if there was a (more) general result... | |
| Dec 4, 2010 at 8:29 | comment | added | Kevin Buzzard | @Arnie: having said all that, there could be some analogue of Tunnell's theorem: perhaps there are some weight 3/2 modular forms kicking around such that the special value of the $L$-function is a coefficient of the form. It's not as simple as Tunnell's case becuase we're not dealing with quadratic twists so I don't think there would be a result that's as clean as his theorem, but perhaps for a given $k_0$ you could get a relatively easy criterion for all numbers of the form $k_0n^3$ if BSD is true. | |
| Dec 4, 2010 at 8:26 | comment | added | Kevin Buzzard | ...curve for any given explicit small k and you'll get an answer very quickly. You ask "when does the equation have finitely many solutions"---but this isn't really a mathematics question because you don't give a criterion which makes any given answer acceptable to you: e.g. clearly "it has finitely many solutions iff it has finitely many solutions" is not an acceptable answer, even though it's logically correct. How about "it has finitely many solns if the L-function doesn't vanish at 1". Is that any good? It sounds useless in practice! Why not just compute the solutions! | |
| Dec 4, 2010 at 8:21 | comment | added | Kevin Buzzard | @Arnie: Diophantine equations are "hard". There will be other criteria, some unconditional, some conditional on the truth of some parts of the Birch--Swinnerton-Dyer conjecture. There may well be some easy criteria when $k$ is prime; one can do the descent and computer Selmer ranks sometimes. But the bottom line is that all these equivalent, or conditionally equivalent, criteria, for general $k$, will involve mathematics that you cannot explain to a high-school kid. You can get as fancy as you like but the bottom line is that you can ask a computer algebra package for the rank of the... | |
| Dec 4, 2010 at 4:10 | comment | added | Jose Arnaldo Bebita | @Kevin, I do think there is such a criterion (although, the paper I allude to is in Japanese). =) @Kimball - your assumptions are correct. =) | |
| Dec 4, 2010 at 1:24 | comment | added | Kimball | Well, it seems the OP is in IT, and I think it's not surprising a non number theorist might expect a simple answer. | |
| Dec 3, 2010 at 20:37 | comment | added | Kevin Buzzard | Are you really expecting a criterion simpler than the statement itself? I'm not sure you're going to get one! | |
| Dec 3, 2010 at 20:11 | vote | accept | Jose Arnaldo Bebita | ||
| Dec 3, 2010 at 20:07 | answer | added | Igor Rivin | timeline score: -3 | |
| Dec 3, 2010 at 20:04 | answer | added | Felipe Voloch | timeline score: 16 | |
| Dec 3, 2010 at 19:49 | history | asked | Jose Arnaldo Bebita | CC BY-SA 2.5 |