3
$\begingroup$

Recently, I have asked a question about the balance of abc triples. Since then I have come up with a different idea of a new criterion that somewhat combines balance and magnitude and has two advantages:

  • it does not imply a (somewhat arbitrarily chosen) threshold, like e.g. the idea of "good abc triples" or the question when to consider a triple "sufficiently balanced".
  • it is symmetric in $a,b,c$, meaning that it can be possibly modelized in terms of (hyper-?)elliptic curves, with things happening in $\mathbb Z$ rather than in $\mathbb N$.

If we denote the usual abc triples by "c-abc triples", my idea would be to introduce a subset called "a-abc triples" or for short, "a-triples" (maintaining the hyphen to avoid grammatical ambiguities), defined as follows:

A triple $(a,b,c)$ with $a<b$ and $a+b=c$ is an a-triple iff $a>\text{rad}(abc)$.

It is natural to define the a-quality of such a triple as $\frac{\log a}{\log\text{rad}(abc)}>1$ .

Since we have automatically $b,c>\text{rad}(abc)$ as well, we could consider equivalently

triples $(a,b,c)\in\mathbb Z^3$ with $a+b+c=0$ and $|a|,|b|,|c|>\text{rad}|abc|$.

It turns out that $95$ of the $241$ known "good" abc triples (i.e. with quality $\geqslant1.4$) are a-triples. The 10 ones with best a-quality are the following:

rk  quality size    merit   a/b     a-quality
66  1.4420  15.51   15.53   0.6363  1.4038
95  1.4316  13.28   12.18   0.8366  1.3948
151 1.4158  23.92   24.63   0.5997  1.3906
173 1.4121  29.38   31.48   0.3006  1.3815
9   1.5270  9.78    11.02   0.1139  1.3723
105 1.4290  10.44   8.74    0.6055  1.3710
240 1.4003  16.79   14.68   0.6427  1.3662
43  1.4526  9.43    8.28    0.3550  1.3629
28  1.4646  21.58   25.80   0.0302  1.3605
72  1.4403  16.98   17.38   0.1058  1.3538

Note that the penultimate one of those is quite imbalanced, but still has a good a-quality. As the size grows, the contribution of the imbalance is mitigated by taking the logs. Or look at the third one in the list (rank 151): "big" in size, very balanced, thus the a-quality is "hardly" smaller than the (c-)quality.

Looking at a-triples might shed some new light on the abc conjecture. My first question:

Are there still infinitely many a-triples?

$\endgroup$
2
$\begingroup$

We believe there are infinitely many a-triples and here are two partial proofs.

  1. The elliptic curve $x^3+y^3=6z^3$ has infinitely many coprime integer solutions. Take $a=x^3,b=y^3,c=6z^3$. We have $\log a \approx \log b \approx \log c$ by the theory of elliptic curves. In addition abc implies $\log a < (1-C) \log c$ can't happen infinitely often for fixed $C > 0$, since this will give infinitely many abc triples violating the abc conjecture.

Since infinitely many prime squares divide $abc$, we have a-triple.

  1. There are infinitely many coprime integer solutions to $y^2=x^3 + 2 z^6$ and take $a=x^3,b=2z^6,c=y^2$. Again we have $a,b,c$ with roughly equal logarithms.

Here is sage session without taking the radical, just log(min(x^3,2z^6,y^2)/log(xyz):

sage: E=EllipticCurve(QQ,[0,2]);P=E.gens()[0]
sage: for k in [ 2 .. 20]:
....:     x1,y1=(k*P).xy()
....:     a=numerator(x1);b=denominator(x1).isqrt();c=a^3+2*b^6;
....:     c=c.isqrt()
....:     A=min(abs(_) for _ in [a^3,2*b^6,-c^2]);ra=prod(u for u,_ in factor(ZZ
....: (2*a*b*c),limit=10^6))
....:     print k,RR(abs(A)).log()/RR(abs(ra)).log(),RR(ra).log(10)

2 0.622930427076373 3.38273726576133
3 0.882982381624213 7.14783363090242
4 0.994840130747121 14.5358637500049
5 0.962868637411342 24.2686482149114
6 1.00233452517062 33.6432311660788
7 0.921720461403240 47.6950749995341
8 0.999678413471366 60.8697877606209
9 0.982053057678340 78.1486267664457
10 0.984848581223119 96.8940057210651
11 0.995456196322965 118.558219981424
12 1.00640878471394 139.820791369602
13 0.983087805421523 164.530570247524
14 0.989966424580658 192.117631841615
15 1.00341504891259 218.904599808444
16 0.986803892310106 249.860860625422
17 0.997502907017965 283.433945971681
18 1.00125521214337 316.773734709346
19 0.997604588937612 354.234286803754
20 0.998545019903716 391.821206879189
| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So do I understand correctly: for your second curve, you have found numerically that for many of the integer solutions, the ratio "without taking the radical" approaches $1$ quite closely, and so you are hoping that more often than not, the radical is small enough that the ratio, if taken "with taking the radical", exceeds $1$? If so, it seems a bit fishy to me... Could you also display the magnitude of each $xyz$, in order to give an idea by which kind of factor the radicals need to be smaller for the respective triple to be indeed an a-triple? $\endgroup$ – Wolfgang May 13 at 15:11
  • $\begingroup$ Also, I don't seem to see a reason why the ratio even "without taking the radical" should not be able to exceed $1$ in theory. But curiously enough, it does not seem to happen. May you try to compute the radical for #12, or are the numbers way too big for that? $\endgroup$ – Wolfgang May 13 at 15:27
  • $\begingroup$ @Wolfgang I take partial radical: the radical of (x^3,2z^6,-y^2) is at most xyz, but in addition rad(xyz) might be smaller and I don't take this into account. You can run the sage session in a browser, check the web for that. $\endgroup$ – joro May 13 at 15:58
  • $\begingroup$ Yes I know. The idea is very clever, maybe there are other elliptic curves with even higher powers and still infinitely many coprime integer solutions, doing an even better job. Hopefully Noam Elkies comes along here... $\endgroup$ – Wolfgang May 13 at 16:06
  • $\begingroup$ @Wolfgang I edited with radical partial factorization up to 10^6 and log_10(xyz). $\endgroup$ – joro May 13 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.