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.)

  • 3
    $\begingroup$ 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. $\endgroup$
    – lulu
    May 22 at 22:29
  • 1
    $\begingroup$ A sum of two primes can be twice a prime, but the product of two primes cannot the square of a prime. $\endgroup$
    – dxiv
    May 23 at 7:42

2 Answers 2


(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 $1$.

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    $\begingroup$ 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 $R$. $\endgroup$
    – Rosie F
    May 23 at 8:56
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    $\begingroup$ If this gets any more general, I won't be able to understand it. $\endgroup$ May 23 at 14:49
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    $\begingroup$ I knew there must be something more elegant than my argument, but this is stunning. $\endgroup$
    – A.M.
    May 23 at 16:17
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    $\begingroup$ Thank you very much. "Stunning" - wow. $\endgroup$ 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, $1, a, k$.


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