I know the following can be proven with logarithms, but I was hoping it could be proven without them. However, I've tried for several days, and I couldn't think of a proof. So I'd like to see what you guys come up with.
Theorem: Let $a, b, x, y \in \mathbb N_{\ge 1}$, where $a, b \ge 2$.
The diophantine equation:
$\quad a^x = b^y$
has a solution iff $x$ is a multiple of $y$ or $y$ is a multiple of $x$.
edit: Embarassing! As it is false, that explains why I couldn't prove it!
Here's what I have so far:
WLOG let $y \ge x$. By the fundamental theorem of arithmetic, I proved:
- Any prime that divides $a$ must divide $b$ and vice versa
- $\exists k \in \mathbb Q, k \ge 1: y = k x$
My dead end ideas:
WLOG let $\operatorname{gcf}(x,y) = 1$. Assume $x, y$ both have a prime factorization. Contradiction, so $x = 1$?
By the division theorem, $y = xt + r, 0 \le r < x$. Show $r=0$?
Show the $p$-adic order of $x$ is less than or equal to that of $y$ for every prime?
It is false:
$$4^3=8^2$$
but $2$ is not a multiple of $3$ and $3$ is not a multiple of $2$.