If $9 \mid 2^b-2^a$, then $7\mid2^b-2^a$ 
Prove that if $9 \mid 2^b-2^a$, then $7\mid2^b-2^a$.

I am not sure how to prove this statement, but it seems that from $9 \mid 2^b-2^a$ we have $b-a = 6n$. Then what should I do from here to prove the statement?
 A: $9 \mid 2^b-2^a$
Let $a \leq b$, otherwise just exchange $a$ and $b$ and note that if $ x \mid -y \Rightarrow x \mid y$
$2^b -2^a=2^a(2^{b-a}-1)$
So, $9 \mid (2^{b-a}-1)$, as $9 \nmid 2^a$, because $9=3^2$ and so they don't have any common prime factors.
$2^{b-a} \equiv 1 \pmod 9$ and $\phi(9)=6$
$2$ is a primitive root modulo $9$, as all the reminders obtained on raising $2$ to a power are co-prime to $9$.
$$\begin{array}\\
 2^1=2 \equiv 2 \pmod 9\\
 2^2=4 \equiv 4 \pmod 9\\
 2^3=8 \equiv 8 \pmod 9\\
 2^4=16 \equiv 7 \pmod 9\\
 2^5=32 \equiv 5 \pmod 9\\
 2^6=63 \equiv 1 \pmod 9\\
\end{array}$$
$\Rightarrow b-a=6n$, where $n \in \mathbb{Z}$
Since $7 \nmid 2^a$, $7 \mid 2^b-2^a \Rightarrow 7 \mid (2^{b-a}-1)$, or $2^{b-a} \equiv 1 \pmod 7$
Now, since $\phi(7)=6$ and $b-a=6n$
$\Rightarrow 2^{6n} \equiv (2^6)^n \equiv (1)^n \equiv 1 \pmod 7$
So $7 \mid 2^b-2^a$, iff $9 \mid 2^b-2^a$
A: If $9\mid(2^b-2^a),$ then, taking $2$ out, we may suppose that $a=0.$ So $9\mid 2^b-1.$
But we see the powers of $2$ are $2,4,-1,-2,-4,1.$ Thus the order of $2$ modulo $9$ is $6.$ Hence $6\mid b.$
Also $2^6\equiv1\pmod7$ by Fermat's little theorem. Therefore $2^b-1\equiv1-1\equiv0\pmod7.$
Hope this helps.
A: $32-2\cdot9
=32-18
=14
=2\cdot 7
$.
Also
$32-3\cdot9
=32-27
=5
$,
so this works for
$5$ also.
