Longest geometric progression of primes

Question

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.)

Answer 1

(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 $1$.

Answer 2

Assume for a contradiction that $r\neq 1$ and $n\ge 2$. We have that $a,ar,ar^2$ are primes.

Thus, $a$ is prime, and an integer. $ar$ is not an integer multiple of $a$ (as it is prime), so $r$ is not an integer.

Since $a,ar$ are integers, $r$ is rational, and in simplest form must have denominator $a$, since $a$ is prime and $r\neq 1$. That is $r=\frac{k}{a}$ for some $k\in\mathbb{Z}$. And $k=ar$ is prime.

Hence, $ar^2=\frac{ak^2}{a^2}=\frac{k^2}{a}$ is an integer. This means $k^2$ is a multiple of $a$ and hence ($a$ is prime) so is $k$. But $k$ is prime and $k\neq a$ (as $r\neq 1$). So we have a contradiction: $k$ has at least $3$ distinct factors, $1, a, k$.