Conditions for solutions to $p^a = q^b$

Question

When (for what conditions on $p,q$) can we solve this equation for integers. I know that $p = q^b$ can only be solved when $p$ is an integer to the power of $b$. By analysing the prime factorisation multiplicities. can we generalise this further?

Answer 1

If we write $p = \prod_i r_i^{k_i}$ and $q = \prod_i r_i^{l_i}$ where the $r_i$ are the primes, then we can solve $p^a = q^b$ if and only if the vectors $\vec{k}$ and $\vec{l}$ are linearly dependent. Equivalently, either there is some $\lambda\in\mathbb{Q}$ such that $\lambda k_i = l_i$ for all $i$, or $p=1$.

Indeed, the first condition is equivalent to $p^\lambda = q$, so we can write $\lambda=\frac{a}{b}$ ($b$ is nonzero because $p$ is not $1$). The second case covers $p=1$, in which case $b=0$ always works.