Solve equation $$ y^p  (2^p1)^x = 1 $$ where $x,y>0 \in \mathbb{Z}$, $p \in \mathbb{P}$. Is there a elementary method to do it? Thanks. =)

2$\begingroup$ In a way: en.wikipedia.org/wiki/Catalan%27s_conjecture as the only possible solution is 9  8 = 1. $\endgroup$ – Will Jagy Nov 6 '12 at 5:04

$\begingroup$ @Will, and thus there are no solutions. Right? $\endgroup$ – jmc Nov 6 '12 at 5:15

$\begingroup$ Dear Will, but it's toooo hard to show that Catalan's conjecture is right. Can we find a easier proof? Dear Johan, there are only a solution $(x,y)=(1,2)$. $\endgroup$ – Lwins Nov 6 '12 at 5:54

$\begingroup$ That pair doesn't give a solution of the equation, dear Lwins. $\endgroup$ – José Hdz. Stgo. Nov 6 '12 at 6:01

1$\begingroup$ Ok, now I object to the title, because Catalan's conjecture requires $x > 1$. However, if $x = 1$, it is easy to show that $(x,y) = (1,2)$ is the only solution, and is a solution for every $p \in \mathbb{P}$. For $x > 1$, the result follows from Catalan's conjecture (i.e., there are no solutions to your equation). $\endgroup$ – jmc Nov 6 '12 at 6:53
Suppose the equation is $$y^pz^r=1,$$ with $p,r$ odd primes. The classical approach to Catalan's conjecture was to consider two cases (similar to Fermat's last theorem) which go as follows:
First you rearrange the equation as $$(y1)\left(\frac{y^p1}{y1}\right)=z^r$$ and then you consider the $\gcd$ of the factors on the left, it can only take the values $1$ or $p$. The first case $\gcd(y1,\frac{y^p1}{y1})=1$ was shown to have no solutions by Cassels
J.W.S. Cassels, On the equation $a^xb^y=1$, II, Proc. Cambridge Philos. Soc. 56 (1960), 97103
with another proof given by S. Hyyro later. Cassels' proof uses elementary techniques. The punchline is that the second (hard) case is when $r y$ and $pz$.
Coming back to your equation we see that $2^p1\equiv 1 \pmod{p}$, so we are in the first case, and you do not need the full strength of Mihăilescu's proof.

$\begingroup$ This is of course assuming that $x$ has a prime divisor, otherwise you have the trivial solution $(x,y)=(1,2)$ mentioned in the comments. $\endgroup$ – Gjergji Zaimi Nov 6 '12 at 7:13

$\begingroup$ I do not see why $2^{p}  1 \cong 1 \pmod{p}$ forces us into the first case. Maybe I am confused by the fact that your $x$ is not the same $x$ as that of the questioner. $\endgroup$ – jmc Nov 6 '12 at 7:30

$\begingroup$ I'm grateful for your patient. Very nice answer, I think. Thanks. =) $\endgroup$ – Lwins Nov 6 '12 at 7:31

1$\begingroup$ Johan, I renamed the variables. Hope it's clear now. $\endgroup$ – Gjergji Zaimi Nov 6 '12 at 7:36

1$\begingroup$ To Johan, obviously $(y1,\frac{y^p1}{y1})=(y1,p) \mid p$. So show that $p \nmid y1$ is enough. It's trivial because $y \equiv 1 \pmod p \implies (2^p1) \equiv 0 \pmod p$. $\endgroup$ – Lwins Nov 6 '12 at 7:38
I think this gives a nice elementary solution, too. See this: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=505773