Can the numbers $2^m 3^n$ have an infinitely long arithmetic sequence? 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 $n \to \infty$.

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.