Proof of Power of Twos and Threes
Are $(1,2), (2,3), (3,4)$, and $(8,9)$ the only consecutive integers that are a power of two and a power of three? And if they are, how do I prove this?
$\endgroup$ 32 Answers
$\begingroup$Updated to add:
Your question is fully answered in this mathoverflow post, which provides an elementary proof (first published in about 1320).
Orginal answer:
Yes, these are the only solutions. This follows from Catalan's conjecture, which is also called Mihăilescu's theorem since 2002, when it was proved by Preda Mihăilescu. It states that $8$ and $9$ are the only consecutive integers that are both perfect powers.
The proof is difficult. There might well exist more elementary proofs for your special case, but I am not aware of them.
$\endgroup$ 1 $\begingroup$The equation $x^a-y^b=1$ with $x, a, y, b > 1$ has only one solution, namely $x = 3, a = 2, y = 2, b = 3$. This is Catalan's conjecture, solved by Mihăilescu - see here. More generally, there is Pillai's conjecture, see also here.
$\endgroup$
