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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.

It is easy to see that this conjecture is implied by the ABC conjecture. However, I was wondering if there is a way to prove it without relying on ABC. Any clue where to start?

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    What is `rad` denoting?2012-11-11
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    rad(x) is the product of the unique prime factors of x. For example, rad(30) is 30, rad(36) is 6, and rad(20) is 10.2012-11-11
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    Can you provide some sources about its importance in music theory?2012-11-11
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    @B.D At the moment, I cannot provide academic sources because this is a relatively new field of study and has not found press in academia. [Gene Smith](http://en.wikipedia.org/wiki/Gene_Ward_Smith) however has shown that this weak version of the ABC conjecture establishes a sort of intuitive complexity metric on musical temperaments. See the "DoReMi" section on [this page](http://xenharmonic.wikispaces.com/ABC%2C+High+Quality+Commas%2C+and+Epimericity) for a more mathematical explanation.2012-11-11
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    I think this would be more approriate on mathoverflow.2012-11-11
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    @Sanchez You think so? My friend told me that mathoverflow might be too dense for my interests, but I'll try over there anyways. Thanks!2012-11-11
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    Posted on MathOverflow: http://mathoverflow.net/questions/112133/a-weaker-version-of-the-abc-conjecture2012-11-12

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Your conjecture is still open. Fix an integer $k$; your conjecture implies that there are finitely many solutions to the equation $$y^m = x^n + k$$ for exponents $n$, $m$ greater than one. For $k = 1$, this is Catalan's conjecture, which is now a theorem, but for $k > 1$ the finiteness of the number of solutions is still unknown; see http://en.wikipedia.org/wiki/Tijdeman%27s_theorem.

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    They are in no sense "nearly equivalent". Solving equations in S-integers is *much* easier; you may as well say that solving $x^n+y^n=z^n$ in $S$-integers is "nearly equivalent" to Fermat's Last Theorem.2012-11-13