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Consider the equation: $3^x = 2^y+1$

(with $(x,y) \in \mathbb{N}^2$)

There are two easy to find solutions ((1,1) and (2,3)), but it doesn't look like there are more of them.

Are there more solutions to this equation?

Anne Aunyme
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    Leveque proved in 1950 a more general fact. $a^x-b^y=1$ has at most one solution except in the case $(x,y)=(3,2)$ when it has two solutions. – coffeemath Jul 25 '23 at 18:11
  • Modulo 8 we get $2 \mid x$ for $y \ge 3$. $(3^{\frac{x}{2}} - 1)(3^{\frac{x}{2}} +1) = 2^y$. Then apply unique factorization theorem. – Denis Shatrov Jul 25 '23 at 18:14
  • reference... https://studylib.net/doc/11110117/differences-between-perfect-powers – coffeemath Jul 25 '23 at 18:15
  • @coffeemath perfect! (I was having trouble finding it) – Anne Aunyme Jul 25 '23 at 18:16
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    Catalan's conjecture (the case with difference $1$ is proven in the mean time , only solution $(8/9)$) , whereas the other differences (Pillai's conjecture , the generalization) are unsolved (conjecture : for every difference only finite many solutions) – Peter Jul 25 '23 at 18:34

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Case 1: $y\ge 3$

$3^x = 2^y+1$

$3^x \equiv 1 \pmod 8$

This implies $2|x$ and $\exists\space a$ such that$\space 2a=x$

$3^{2a}=2^y+1$

$3^{2a}-1=2^y$

$(3^a-1)(3^a+1)=2^y$

$\exists \space s,t\in\Bbb{N}$ such that $y=s+t, s>t,\space 3^a+1=2^s,3^a-1=2^t$

$3^a+1=2^s \quad [1]$

$3^a-1=2^t \quad [2]$

subtracting [2] from [1] we have:

$2=2^s-2^t$

$2=2^t(2^{s-t}-1)$

$t=1$ because there is only one factor of $2$ on the left hand side

$2=2(2^{s-1}-1)$

$1=2^{s-1}-1$

$2=2^{s-1}$

$1=s-1$

$2=s$

first solution $y=3,$ $x=2$

case 2: $\quad y=0,$ or$\space y=1,$ or $\space y=2$

by trial and error there is only one solution $y=1,$ $x=1$

quantus14
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