My question is the following:
Question: Is it true that for any two integers $a,b \ge 2$, the following five conditions are equivalent?
- There exists an integer $c \ge 2$ that is a common power of $a$ and $b$.
- There exists an integer $d \ge 2$ for which both $a$ and $b$ are powers of $d$.
- $\operatorname{log}_a(b) \in \mathbb{Q}$
- $\operatorname{log}_b(a) \in \mathbb{Q}$
- The integers $a$ and $b$ are divisible by exactly the same prime factors, and the ordered lists of exponents of those prime factors of $a$ and $b$ are proportional.
It is easy to see that Conditions 1, 3, and 4 are equivalent, for the equation $a^m=b^n$ is equivalent to the equation $\operatorname{log}_a(b)=\frac{m}{n}$ and also to $\operatorname{log}_b(a)=\frac{n}{m}$.
Also, the idea in proving that Conditions 1, 2, and 5 are equivalent is to consider the LCMs and GCDs of exponents of corresponding prime factors of $a$ and $b$ when forming $c$ and $d$ respectively.
Write $a = \prod_{k=1}^{e(a)} p_k^{a_k}$, where the $p_k$ are distinct primes and $a_k\geqslant 1$ are their exponents, as in its unique prime decomposition. Moreover, assume without loss of generality that the $p_k$ are increasing.
Similarly, write $b = \prod_{k=1}^{e(b)} P_k^{b_k}$.
From $(3)$, let $\log_a(b) = m/n$, where $m$ and $n$ are integers with $\gcd(m,n) = 1$.
Then $a^{m/n} = b \implies a^m = b^n$. It follows that
$$\prod_{k=1}^{e(a)} p_k^{ma_k} = \prod_{k=1}^{e(b)} P_k^{nb_k}$$
Because the decomposition into prime factors is unique, we conclude that $e(a) = e(b)$ and that for all $k$
$$p_k = P_k \quad\quad \text{and}\quad\quad ma_k = nb_k.$$