TITLE: Is $7k-9$ ever a power of $2$?
QUESTION [5 upvotes]: I wonder if $7k-9$ is ever a power of $2$.
My work so far: if $k=2$, then $7k-9 > 4$. Then I checked if $7k-9$ is ever divisible by $4$. For $k=3$ it indeed is. For $k=4, 7k-9 > 16$. So I checked if $7k-9$ is ever divisible by $16$ and for $k= 15$ it indeed is. I could go on and try with $32$ but I don't feel inclined to do so. Is there any simpler method how to determine if $7k-9$ is ever a power of $2$?

REPLY [5 votes]: Hint $\ $ If so, $\rm\ mod\ 7\!:\ \color{#C00}2^n \equiv \color{#0A0}{-2}\ \overset{cubing}\Rightarrow \color{#C00}1 \equiv \color{#0A0}{-1}\ \Rightarrow\Leftarrow$