Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question
What is rad denoting? – Cameron Buie Nov 11 '12 at 6:52
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. – Ryan Nov 11 '12 at 7:00
Can you provide some sources about its importance in music theory? – Benjamin Dickman Nov 11 '12 at 7:17
@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. – Ryan Nov 11 '12 at 7:29
Posted on MathOverflow:… – David Roberts Nov 12 '12 at 1:25

1 Answer 1

up vote 16 down vote accepted

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

share|cite|improve this answer
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. – Rocky the flying squirrel Nov 13 '12 at 4:31

protected by Community Aug 4 '14 at 23:24

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.

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.