Solving for $3^x - 1 = 2^y$

Question

Besides $x=2, y=3$, are there any other solutions?

I know that if there is another solution:

• $y$ is odd since $2^y \equiv -1 \pmod 3$
• $x$ is even since $3^x - 1 \equiv 0 \pmod 8$
• $3 | y$ since $-1 \equiv 2^y \pmod 9$

Are there any other solutions? If not, what is the argument for showing that if $3^x > 9$, then $2^y \neq 3^x-1$

Thanks,

-Larry

Answer 1

You don't need the full strength of Catalan's conjecture here.

Two solutions are found easily. $3^1-1=2^1$ gives $x=y=1$. $3^2-1=2^3$ gives $x=2$ and $y=3$.

To prove that these are the only solutions, assume $x > 2$. As you have noted, $x$ must be even: $x = 2z$. Now we have $$ 3^x-1 = 3^{2z}-1 = (3^z-1)(3^z+1) = 2^y. $$ It follows that both $3^z-1$ and $3^z+1$ are powers of $2$. But this is impossible, because both numbers are larger than $2$ (since $z > 1$), and they are only two units apart. QED.

PS: I think I've seen this argument somewhere on math.SE.

Answer 2

Catalan's conjecture (now a theorem, with the alternative name "Mihăilescu's theorem") asserts that there are no other possible solutions to the diophantine equation in integers $a,x,b,y > 1$:

$$a^x - b^y = 1$$

other than the solution $(a,b,x,y) = (3,2,2,3)$. As for the proof, I believe that Mihailescu's original proof is not simple enough to be placed into a single answer. Here is a link to a simpler and more elementary one, though it is still considerably long: Elementary proof of Mihailescu's Theorem.

Answer 3

This link can help you to get an answer to your question.