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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+\varepsilon $, where $a+b=c$ and $\varepsilon > 0$.

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+\varepsilon $. 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.

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    $\begingroup$ Can you post a link to the stackexchange question and vice versa too? $\endgroup$
    – Tony Huynh
    Commented Nov 12, 2012 at 1:07
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    $\begingroup$ 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 ♦
    Commented Nov 12, 2012 at 1:30
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    $\begingroup$ @David, what would make this question "potentially borderline" without the music-theory motivation? $\endgroup$
    – Joël
    Commented Nov 12, 2012 at 3:03
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    $\begingroup$ If anyone is interested in the music theory aspect of this, that is discussed here: xenharmonic.wikispaces.com/… $\endgroup$
    – Gene Ward Smith
    Commented Nov 12, 2012 at 4:19
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    $\begingroup$ 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 ♦
    Commented Nov 12, 2012 at 4:21

2 Answers 2

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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.

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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.

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  • $\begingroup$ 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
    Commented Nov 12, 2012 at 22:21
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    $\begingroup$ 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
    Commented Nov 12, 2012 at 22:42
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    $\begingroup$ @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
    Commented Nov 13, 2012 at 6:16
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    $\begingroup$ 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
    Commented Nov 13, 2012 at 8:16

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