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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    $\begingroup$ What is rad denoting? $\endgroup$ Nov 11, 2012 at 6:52
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    $\begingroup$ 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. $\endgroup$
    – Ryan
    Nov 11, 2012 at 7:00
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    $\begingroup$ Can you provide some sources about its importance in music theory? $\endgroup$ Nov 11, 2012 at 7:17
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    $\begingroup$ @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 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 for a more mathematical explanation. $\endgroup$
    – Ryan
    Nov 11, 2012 at 7:29
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    $\begingroup$ Posted on MathOverflow: mathoverflow.net/questions/112133/… $\endgroup$ Nov 12, 2012 at 1:25

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Nov 13, 2012 at 4:31

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