<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] You're on the right track. By Fermat's little theorem, $2^n\pmod{5}$ and $2^n\pmod{13}$ depend only on $n\pmod{4}$ and $n\pmod{12}$, respectively. You've shown that $2^n\equiv 3 \pmod{5}$ if $n\equiv 3\pmod{5}$, but that's the opposite of what you want to prove; you want to show that $2^n\equiv 3\pmod{5}$ only if $n\equiv 3\pmod{5}$, and similarly for $13$. The result you'll get is that $2^n\equiv 3\pmod{5}$ only if $n\equiv 3\pmod{4}$ and $2^n\equiv 3\pmod{13}$ only if $n\equiv 4\pmod{12}$. Those two conditions can clearly never be simultaneously satisfied.
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