TITLE: Are there any solutions to the diophantine equation $x^n-2y^n=1$ with $x>1$ and $n>2$?
QUESTION [12 upvotes]: This problem arose when considering storage of cannonballs in n-dimensional pirate ships, as explained in this MSE post. This MO question can also be reduced to the $n=3$ case. If $x,y$ is a solution then $$0<\frac{x}{y}-2^\frac1n<\frac{2^\frac1n}{2ny^n}$$ then by Roth's theorem this has finitely many solutions for fixed $n$. Let $$2^{1/n}=a_0+\frac{1}{a_1+\dots}$$
be the canonical continued fraction of $2^{1/n}$, then $a_0=1$ and $a_1\in\{\lfloor\frac{n}{\ln(2)}\rfloor,\lfloor\frac{n}{\ln(2)}\rfloor-1\}$, and since $\frac{x}{y}$ is a convergent of this continued fraction, $y>\frac{n}{\ln(2)}-1$. There are no solutions with $x^{n}<2^{64}$. It is also sufficient to only consider $n=4$ and odd primes, in FLT fashion.

REPLY [2 votes]: The case $n = 4$ and so $n$ even can be done by hand. Darmon and Merel proved (in 1997) the stronger statement that there aren't even any rational solutions to this equation for $n$ odd besides $(x^n,y^n) = (1,1)$, See their paper "Winding quotients and some variants of Fermat's last theorem," which can be found here:
http://www.math.mcgill.ca/darmon/pub/Articles/Research/18.Merel/paper.pdf