Integer solutions to $2^x-1=3^y$.

Question

Does anybody know how to solve it? I know how to solve linear diophantine equations, but equation like this I've never seen before. $$2^x-1=3^y.$$

Answer 1

This is a special case of Mihăilescu's theorem (a.k.a. Catalan's conjecture) that has been illustrated here many times before. There are no nontrivial solutions to $$2^x-3^y=1,$$ nontrivial meaning $x,y\geq2$.

Proof: Clearly $x,y\geq0$. If $x$ is even, say $x=2z$, then $$3^y=2^x-1=(2^z-1)(2^z+1),$$ and so the two factors on the right hand side are two powers of $3$ that differ by $2$, which implies that $x=2$ and $y=1$.

If $x$ is odd then $$3^y\equiv2^x-1\equiv1\pmod{3},$$ which implies $y=0$ and hence $x=1$.