A Weaker Version of the ABC Conjecture - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:35:18Z http://mathoverflow.net/feeds/question/112133 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112133/a-weaker-version-of-the-abc-conjecture A Weaker Version of the ABC Conjecture Ryan 2012-11-12T00:39:11Z 2012-11-12T15:13:58Z <p>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:</p> <p>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 $\epsilon > 0$.</p> <p>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)</p> <p>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.</p> http://mathoverflow.net/questions/112133/a-weaker-version-of-the-abc-conjecture/112175#112175 Answer by Tony Huynh for A Weaker Version of the ABC Conjecture Tony Huynh 2012-11-12T13:53:46Z 2012-11-12T13:53:46Z <p>According to <a href="http://math.stackexchange.com/questions/234697/a-weaker-version-of-the-abc-conjecture" rel="nofollow">Rockytheflyingsquirrel</a>, this is still an open problem. I made this answer community wiki so as not to benefit from a squirrel's hard work. </p> http://mathoverflow.net/questions/112133/a-weaker-version-of-the-abc-conjecture/112182#112182 Answer by joro for A Weaker Version of the ABC Conjecture joro 2012-11-12T15:13:58Z 2012-11-12T15:13:58Z <p>Just to point out there are infinitely many coprime solutions to $\frac {\log \left( c \right)}{\log \left( a \: \text{rad} \left( bc \right) \right)} > 1$</p> <p>Take $a=1$ and $b,c$ <a href="https://oeis.org/A060355" rel="nofollow">consecutive powerful numbers</a>.</p> <p>If $n,n+1$ are consecutive powerful numbers so are $4n(n+1),4n(n+1)+1$ so the solutions without epsilon are infinite.</p>