Find all positive integer solutions for $3^x-2^y=1$.
Quickly we can find the solutions $(1,1)$ and $(2,3)$. Now the claim is that for $x \ge 2$ and $y \ge 3$ there are no solutions.
The equation can be expressed as $3^x=2^y+1$ from where we can get to $3^x-3 = 2^y-2 \implies 3(3^{x-1}-1)=2(2^{y-1}-1)$ now from this clearly $2$ is not a multiple of $3$ so it must be that $$2^{y-1}-1 \equiv 0 \pmod 3 \implies 2^{y-1} \equiv 1 \pmod 3$$ and the order of $2$ modulo $3$ is $2$ so $y-1$ is a multiple of two say $y-2=2t$.
Plugging this back to the lhs of the equation we get $$3(3^{x-1}-1)=2(2^{2t}-1)=2(2^2-1)(2^{2(t-1)}+ \dots +1) = 6(2^{2(t-1)}+ \dots +1)$$ now $3$ is not a multiple of $6$ so we must have that $$3^{x-1}-1\equiv 0 \pmod 6 \implies 3^{x-1} \equiv1 \pmod 6$$ but noticing that the only way this can happen is if $x-1=0$ i.e. $x=1$, but this contradicts our assumption that $x \ge2$ thus there are no solutions for $x \ge 2$.
Is this a correct argument? I think there are multiple ways to do the problem, but this seemed most natural to me.
Best Answer
You used two times the following false statement:
The correct statement is
The first time you used it was with $a=2$ and $m=3$ so you got a correct conclusion, but the second time was with $a=3$ and $m=6$ so that conclusion is wrong.
From $$3(3^{x-1}-1) = 6(2^{2(t-1)}+ \dots +1)$$ it follows $$3^{x-1}-1= 2(2^{2(t-1)}+ \dots +1)$$ and you may conclude that $3^{x-1}-1 \equiv 0 \pmod {\color{red}2}$ , not $\pmod 6$ as you got.