TITLE: How prove this arithmetic progression cannot have terms of the form $2^m+3^n$ only?
QUESTION [9 upvotes]: Let $a_{i}\in \mathbb{N}$, $a_{i}\ge 1$, $i=1,2,\ldots,7$ be arbitrary, and such that
$a_{n+1}-a_{n}=d\neq 0,n=1,2,\cdots,6 $.
Then there exists $a_{\ell}$, $\ell=1,2,\ldots,7$, such that
$$a_{\ell}\notin \{x:x=2^m+3^n,\;m,n\in \mathbb{Z},\;m,n\ge 0\}$$

REPLY [2 votes]: A number of the form $2^m + 3^n$ can only take on $163$ possible values mod $4095$, and in fact there are no arithmetic progressions of length $7$ mod $4095$ in this set.  So $d$ is divisible by $4095$.  But as long as $n \ge 2$, there are no numbers of the form $d=(2^w-2^x)+(3^y-3^z)$ with $w \ne x$ and $y \ne z$ which are divisible by $4095$.  The cases $n=0$ and $n=1$ are seemingly more straightforward to handle.