Infinite sequence $2^{n}-3 (n=2,3,...)$ contains no term divisible by 65

Question

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!

Answer 1

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.

Answer 2

Hint:

Write $12k+4$ as $12n+4$ which needs to be equal $4k+3$ which is odd unlike the former and $k,n$ are arbitrary integers

Answer 3

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.