12
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ For $n = 3$ this should follow from the fact that the Mordell-Weil group of the curve $x^3 - 2y^3 = 1$ over $\mathbb{Q}$ is of rank $0$. $\endgroup$ – WhatsUp Dec 13 '16 at 14:48
  • $\begingroup$ For $n = 4$ it is also easy to see that there is no non-trivial solution, by writing the equation as $(x^2 + 1)(x^2-1)=2y^4$. $\endgroup$ – WhatsUp Dec 13 '16 at 14:55
20
$\begingroup$

Delone (1930) and Nagell (1928) showed for any nonzero integer $d$ that the equation $x^3 - dy^3 = 1$ has at most one solution in integers $(x,y)$ besides $(1,0)$, with no constraint on the signs of $x$ and $y$. In particular, since $x^3 - 2y^3 = 1$ has the integral solution $(-1,-1)$, there is no integral solution $(x,y)$ in positive integers.

This theorem was extended to exponent 4 by Ljunggren (1942) and to exponent 5 and higher by Bennett (2001): for $n \geq 3$ and $d \not= 0$, the equation $|x^n - dy^n| = 1$ has at most one solution in positive integers. See Theorem 1.1 of https://www.math.ubc.ca/~bennett/B-Crelle2.pdf (which actually treats a slightly more general equation). In particular, $|x^n - 2y^n| = 1$ has at most one solution $(x,y)$ in positive integers. Since $(x,y) = (1,1)$ fits, it is the only one. Of course $x^n - 2y^n = -1$ when $(x,y) = (1,1)$, so for $n \geq 3$ there is no solution to $x^n - 2y^n = 1$ when $x$ and $y$ are positive integers.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.