Find integral solutions of $\left|3^{x}-2^{y}\right| = 1$

Question

Find integral solutions of $\left|3^{x}-2^{y}\right| = 1$

In case of $3^{x}-2^{y}=1$, By putting values of x,y to be 0 we get it's not possible and x,y has to be greater than 0. So $3^{x}=1+2^{y} \rightarrow$ $1+2^{y} \equiv 0(\bmod 3)$ which means $y$ is odd. I'm stuck here and don't know what to do next.

See this MO-post, or this duplicate form MSE. – Dietrich Burde Jun 19 '22 at 16:07 — Dietrich Burde, Jun 19 '22 at 16:07

score 0 · Answer 1 · answered Jun 19 '22 at 16:05

0

This is a very famous problem, but surprisingly I cannot find a duplicate.

I give here the sketch of a solution. You may try to fill in the details.

For $3^x - 2^y = 1$:

Try $y = 0, 1$ directly. Now suppose $y \geq 2$. Mod $4$ shows that $x$ is even. Rewrite the equation as $(3^{x/2} + 1)(3^{x/2} - 1) = 2^y$. Both factors on the left hand side are powers of two, while their difference is $2$.

For $2^y - 3^x = 1$:

Try $x = 0$ directly. Now suppose $x \geq 1$. Mod $3$ shows that $y$ is even. Do the same as above.

As a remark, this is a special case of Catalan's conjecture (now a theorem of Mihăilescu).

answered Jun 19 '22 at 16:05

WhatsUp

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One of the duplicates is here. – Dietrich Burde Jun 19 '22 at 16:10
@DietrichBurde Indeed, I expected them to exist but somehow it's hard to find because the formulas are written in slightly different ways. – WhatsUp Jun 19 '22 at 16:13
https://approach0.xyz/search/?q=OR%20content%3A%24%5Clvert3%5E%7Bx%7D-2%5E%7By%7D%5Crvert%3D%201%24&p=1 – Sil Jun 19 '22 at 16:26

Find integral solutions of $\left|3^{x}-2^{y}\right| = 1$

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