Coprime integer solutions to $ \frac{x^n \pm y^n}{x \pm y}=z^m $ with $n>5 , m>1$ - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T23:11:13Zhttp://mathoverflow.net/feeds/question/69702http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/69702/coprime-integer-solutions-to-fracxn-pm-ynx-pm-yzm-with-n5-m1Coprime integer solutions to $ \frac{x^n \pm y^n}{x \pm y}=z^m $ with $n>5 , m>1$joro2011-07-07T11:25:56Z2011-07-08T09:52:45Z
<p>Are there coprime integer solutions to:
$$ \frac{x^n \pm y^n}{x \pm y}=z^m $$
with $n>5 , m>1$ and excluding $z=0$?</p>
<p>I suppose the abc conjecture implies finitely many solutions.</p>
http://mathoverflow.net/questions/69702/coprime-integer-solutions-to-fracxn-pm-ynx-pm-yzm-with-n5-m1/69782#69782Answer by Mike Bennett for Coprime integer solutions to $ \frac{x^n \pm y^n}{x \pm y}=z^m $ with $n>5 , m>1$Mike Bennett2011-07-08T09:52:45Z2011-07-08T09:52:45Z<p>I suspect this question is very difficult to answer without additional hypotheses.
The case $y=1$ is the classic Nagell-Ljunggren equation, where, unlike the (on the
surface) very similar equation of Catalan, we do not even know whether there exist
finitely many solutions in the variables $(x,z,n,m)$. If we fix $m=2$ and $n > 5$ prime,
say, then it is still a substantial problem to solve the corresponding equation;
Ivorra [Dissertationes Math. 444 (2007)] treats the cases $n \in { 7, 11, 13, 17 }$
via elliptic Chabauty.</p>