# A special case of Catalan's conjecture

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

• In a way: en.wikipedia.org/wiki/Catalan%27s_conjecture as the only possible solution is 9 - 8 = 1. – Will Jagy Nov 6 '12 at 5:04
• @Will, and thus there are no solutions. Right? – jmc Nov 6 '12 at 5:15
• 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 $(x,y)=(1,2)$. – Lwins Nov 6 '12 at 5:54
• That pair doesn't give a solution of the equation, dear Lwins. – José Hdz. Stgo. Nov 6 '12 at 6:01
• 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). – jmc Nov 6 '12 at 6:53

Suppose the equation is $$y^p-z^r=1,$$ 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 $$(y-1)\left(\frac{y^p-1}{y-1}\right)=z^r$$ and then you consider the $\gcd$ of the factors on the left, it can only take the values $1$ or $p$. The first case $\gcd(y-1,\frac{y^p-1}{y-1})=1$ was shown to have no solutions by Cassels

J.W.S. Cassels, On the equation $a^x-b^y=1$, II, Proc. Cambridge Philos. Soc. 56 (1960), 97-103

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 $p|z$.

Coming back to your equation we see that $2^p-1\equiv 1 \pmod{p}$, so we are in the first case, and you do not need the full strength of Mihăilescu's proof.

• 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. – Gjergji Zaimi Nov 6 '12 at 7:13
• 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. – jmc Nov 6 '12 at 7:30
• I'm grateful for your patient. Very nice answer, I think. Thanks. =) – Lwins Nov 6 '12 at 7:31
• Johan, I renamed the variables. Hope it's clear now. – Gjergji Zaimi Nov 6 '12 at 7:36
• To Johan, obviously $(y-1,\frac{y^p-1}{y-1})=(y-1,p) \mid p$. So show that $p \nmid y-1$ is enough. It's trivial because $y \equiv 1 \pmod p \implies (2^p-1) \equiv 0 \pmod p$. – 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