<s>[INST] 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?
 [/INST] 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.
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