<s>[INST] Infinite sequence $2^{n}-3 (n=2,3,...)$ contains no term divisible by 65 Show that the infinite sequence $2^{n}-3 (n=2,3,...)$ contains infinitely many terms which are divisible by $5$ and infinitely many terms which are divisible by $13$, but no terms which are divisible by $65$
My attempt at this:-
By Fermat's theorem,
$$2^{4}\equiv 1\pmod5$$
Raising to the power k we get,
$$2^{4k}\equiv 1\pmod5$$
$$2^{4k+3}\equiv 8\pmod5$$
$$2^{4k+3}\equiv 3\pmod5$$
So, $5\mid 2^{n}-3\quad\forall \quad n=4k+3$ where $k$ is any non-negative integer.
Similarly, by Fermat's theorem
$$2^{12}\equiv 1\pmod{13}$$
$$2^{12k}\equiv 1\pmod{13}$$
$$2^{12k+4}\equiv 16\pmod{13}$$
$$2^{12k+4}\equiv 3\pmod{13}$$
Therefore, $13\mid 2^{n}-3\quad\forall \quad n=12k+4$ 
How do I show that it contains no term which is divisible by 65?
Thank you!
 [/INST] Note that $4k+3$ is of the form $12k+3, 12k+7,\text { or } 12k+11,$ never of the form $12k+4,$ so there are no numbers in common in your sequences.
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