There are arbitrarily long arithmetic progressions of primes e.g. $5, 11, 17, 23, 29$ for a $5$-length progression, but no (infinite) arithmetic sequence of primes with common difference $d\neq 0$, as $d\in\mathbb{Z}$ is an obvious constraint and $(a+nd)_{n\in\mathbb{N}}$ contains $a+ad=a(1+d)$.
A natural question is then: what is the longest geometric progression of primes? If $r>1$ is an integer then you can't get a progression longer than $1$, as $ar$ has at least three distinct factors: $1, ar, a, r$ (possibly $a=r$). But what about arbitrary $r\in\mathbb{R}$? You can get a sequence of $2$ e.g. $2, 3$ by taking first term $a=2$ and common ratio $r=1.5$. But it doesn't seem to be possible to get more.
So my question is:
Prove that if $a,ar,ar^2,\dots,ar^n$ is a list of prime numbers then either $r=1$ or $n\le 1$.
(Self-answering because I'm surprised not to find this question asked before; it seems elementary but interesting.)
Best Answer
(Edited for more generality)
If $p, q, r$ are ANY three primes in geometric progression, then $q^2=pr$ so, by prime factorization, $p=q=r$.
Therefore the ratio of the progression is $1$.