TITLE: Can the numbers $2^m 3^n$ have an infinitely long arithmetic sequence?
QUESTION [5 upvotes]: I am asking for proof the set $ \{ 1,2,4,8,\dots \} \times \{1,3,9,27,\dots \} = \{1,2,3,4,6,8,9,12,\dots \}$ does not have infinitely long arithmetic sequences inside.  This is OEIS A036561.
What if we allow for more prime factors $\{2^a 3^b 5^c: a,b,c \in \mathbb{N}\}$ ?

One possibility is to let $X = 2^\mathbb{N} \times 3^\mathbb{N}$ and check that $X \cap [1,n]$ does not have enough elements as $n \to \infty$.

REPLY [5 votes]: No.
Assume contrariwise that such sequences existed. Let $s,s+r,s+2r,s+3r,\ldots$  be the one with minimal $s$. If $\gcd(s,r)>1$ then both $s$ and $r$ are divisible by $p\in\{2,3\}$. Then the sequence $s/p,s/p+r/p,s/p+2r/p,\ldots$ would be another sequence violating the minimality of $s$.
Therefore $\gcd(s,r)=1$. But then Dirichlet's theorem tells us that there are infinitely many prime numbers in that sequence.