# Longest geometric progression of primes

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.)

• Say the distinct primes are $p_1,p_2,p_3$, with common ratio $r$. Clearly $r=\frac {p_2}{p_1}$ . But then you want $p_3=p_2\times \frac {p_2}{p_1}$ which is not an integer.
– lulu
May 22 at 22:29
• A sum of two primes can be twice a prime, but the product of two primes cannot the square of a prime.
– dxiv
May 23 at 7:42

(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 $$1$$.

• I think you can have even more generality. Suppose $R$ is a normed ring with norms in a field $F$, where $||a||$ denotes the norm of $a$, and $F$ is a UFD. Then $||q||^2=||p||\cdot ||r||$. Your result follows because of uniqueness of factorization in $F$, not in $R$. May 23 at 8:56
• If this gets any more general, I won't be able to understand it. May 23 at 14:49
• I knew there must be something more elegant than my argument, but this is stunning.
– A.M.
May 23 at 16:17
• Thank you very much. "Stunning" - wow. May 23 at 17:55

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, $$1, a, k$$.