Is there a solution for the equation x^m-y^n=k in which k > 1? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T05:41:30Z http://mathoverflow.net/feeds/question/29677 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29677/is-there-a-solution-for-the-equation-xm-ynk-in-which-k-1 Is there a solution for the equation x^m-y^n=k in which k > 1? Hashem sazegar 2010-06-27T05:50:24Z 2010-07-23T20:44:19Z <p>The Catalan conjecture state that $x^m-y^n=1$ has only the solution $x=3, m=2, y=2, n=3$. This conjecture was proved by Preda Mihailescu in 2004, but I want to know about the equation mentioned above. Is there a asolution of this?</p> http://mathoverflow.net/questions/29677/is-there-a-solution-for-the-equation-xm-ynk-in-which-k-1/29678#29678 Answer by Gerry Myerson for Is there a solution for the equation x^m-y^n=k in which k > 1? Gerry Myerson 2010-06-27T05:59:49Z 2010-06-27T06:15:12Z <p>That would depend on $k$. I'm assuming you want all variables positive integers with $m>1$ and $n>1$. It may have been proved that for any given $k$ there are only finitely many solutions. </p> <p>Edit: in view of the other answers, it appears my "may have been proved" was overly optimistic. </p> http://mathoverflow.net/questions/29677/is-there-a-solution-for-the-equation-xm-ynk-in-which-k-1/29679#29679 Answer by Péter Komjáth for Is there a solution for the equation x^m-y^n=k in which k > 1? Péter Komjáth 2010-06-27T06:07:07Z 2010-06-27T06:07:07Z <p>There is a conjecture that for every positive natural number <i>k</i> there are just finitely many solutions of the above equation. As far as I know, this is open for <i>k</i>>1. In fact, Erdős conjectured that the difference between a full power <i>x</i> and the next full power is at least <i>x<sup>c</sup></i> for some positive constant <i>c</i>. </p> http://mathoverflow.net/questions/29677/is-there-a-solution-for-the-equation-xm-ynk-in-which-k-1/29680#29680 Answer by Kevin O'Bryant for Is there a solution for the equation x^m-y^n=k in which k > 1? Kevin O'Bryant 2010-06-27T06:08:54Z 2010-07-23T20:44:19Z <p>Pillai's conjecture is that for each $k$, there are only finitely many solutions.</p> <p>The ABC conjecture implies Pillai's conjecture as follows. First, I state a form of the ABC conjecture. Given three relatively prime positive integers $A+B=C$, the quality of the triple $(A,B,C)$ is $\log(C)/\log(R)$, where $R$ is the product of the primes that divide $ABC$. For example, the quality of $(5,27,32)$ is $\log(32)/\log(30)$. One strong form of the ABC conjecture is that there are only finitely many triples (of relatively prime positive integers) with quality greater than $1.001$.</p> <p>Now a solution to $x^m-y^n=k$ has $(A,B,C)=(k,y^n,x^m)$, and so $R\leq k y x$. Thus, the quality of the triple is at least $$\frac{m\log(x)}{\log(k)+\log(x)+\log(y)}\approx \frac{m\log(x)}{\log(k)+(\frac mn +1)\log(x)}.$$ As $x\to\infty$, this gives a quality approaching $\frac{mn}{m+n}$, and for $m n>4$ this is $\geq 1.2$, which can only happen finitely many times by ABC. Also, $\frac{mn}{m+n}=1$ if $m=n=2$, so that case has to be handled separately.</p>