I posted this question over at stackexchange, where a user informed me that it was probably more appropriate for mathoverflow. Here's to hoping that the answer is out there:

The ABC conjecture states that there are a finite number of integer triples (a,b,c) such that $\frac {\log \left( c \right)}{\log \left( \text{rad} \left( abc \right) \right)}>1+\epsilon $, where $a+b=c$ and $\epsilon > 0$.

I am however more interested in a weaker version of the ABC conjecture where the following inequality holds true: $\frac {\log \left( c \right)}{\log \left( a \: \text{rad} \left( bc \right) \right)}>1+\epsilon $. This weaker conjecture has a number of applications in music theory - specifically concerning temperament theory. For instance, it establishes a type of intuitive complexity metric on various temperaments, and then lets us bound a finite number of these temperaments underneath a given complexity. (if you are not familiar with temperament theory, you can think of these "temperaments" as z-module homomorphisms from one free abelian group to another of lower rank)

It is easy to see that this conjecture is implied by the ABC conjecture. However, I was wondering if this weaker version is already proven? And if not, what is the best approach to a proof that does not rely on ABC? I'm not very familiar with number theory so I don't know where to start.

  • 2
    $\begingroup$ Can you post a link to the stackexchange question and vice versa too? $\endgroup$ – Tony Huynh Nov 12 '12 at 1:07
  • 7
    $\begingroup$ For future reference, this is a music theory question with a good fit for MO. Thanks for pointing out the musical reason for asking about this weaker conjecture, it makes a potentially borderline question very interesting. $\endgroup$ – David Roberts Nov 12 '12 at 1:30
  • 9
    $\begingroup$ @David, what would make this question "potentially borderline" without the music-theory motivation? $\endgroup$ – Joël Nov 12 '12 at 3:03
  • 2
    $\begingroup$ If anyone is interested in the music theory aspect of this, that is discussed here: xenharmonic.wikispaces.com/… $\endgroup$ – Gene Ward Smith Nov 12 '12 at 4:19
  • 4
    $\begingroup$ Given the state of interest in ABC among non-specialists, if someone asked 'what if I do x to the statement of the ABC conjecture?', then it would be (for me) merely a curiosity. I suppose 'borderline' was a bit strong, but perhaps I'm just on the watch out for repeats of what happened with recent ABC-related questions. $\endgroup$ – David Roberts Nov 12 '12 at 4:21

According to Rockytheflyingsquirrel, this is still an open problem. I made this answer community wiki so as not to benefit from a squirrel's hard work.


Just to point out there are infinitely many coprime solutions to $\frac {\log \left( c \right)}{\log \left( a \: \text{rad} \left( bc \right) \right)} > 1$

Take $a=1$ and $b,c$ consecutive powerful numbers.

If $n,n+1$ are consecutive powerful numbers so are $4n(n+1),4n(n+1)+1$ so the solutions without epsilon are infinite.

  • $\begingroup$ The fact that you keep multiplying by 4 seems promising. Are you sure that sequence tends to 1? Gerhard "Ask Me About System Design" Paseman, 2012.11.12 $\endgroup$ – Gerhard Paseman Nov 12 '12 at 22:21
  • 1
    $\begingroup$ In fact this sequence deserves more attention. Can one prove that from the sequence (n,n+1) one has infinitely many prime members of (2n+1)? We may have the opportunity of putting two conjectures head-to-head. Gerhard "Ask Me About System Design" Paseman, 2012.11.12 $\endgroup$ – Gerhard Paseman Nov 12 '12 at 22:42
  • 1
    $\begingroup$ @Gerhard it is interesting how relatively big the 3-full part of n(n+1) can be. If it is sufficiently big it will disprove both this and the abc conjecture. $\endgroup$ – joro Nov 13 '12 at 6:16
  • 1
    $\begingroup$ Re: "tends to 1". I suppose in general it tends to 1. Repeatedly multiplying by 4 is exponential in 2. Unfortunately squaring $2n+1$ gives doubly exponential growth. Doubt the product of the smaller $n$ can compensate the doubly exponential growth. Well might be wrong. $\endgroup$ – joro Nov 13 '12 at 8:16

protected by Todd Trimble Aug 2 '14 at 23:04

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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