Find divisibility for $n \in \mathbb {Z+}, 2\nmid n, 2^n-1, 2^n +1$ w.r.t. $5$ using congruence arithmetic.

Question

Find divisibility for $n \in \mathbb {Z+}, 2\nmid n, 2^n-1, 2^n +1$ w.r.t. $5$ using congruence arithmetic.

Checking w.r.t. modulo $5$
(a) $ 2^n-1$ :: $2 \equiv 2 \pmod 5\implies 2^n \equiv ........$
(b) $ 2^n+1$ :: $2 \equiv 2 \pmod 5\implies 2^n \equiv ........$

I cannot handle any case w.r.t. modulo $5$, as the only way left for me is to break $n$ into two components such that one is odd and other is even, i.e. $n=n'+1$. But that approach confuses a lot.

Answer 1

• $n$ is odd here. Let $n=2m-1$,

$$2^{2m-1} \equiv (2^2)^m(2^{-1})\equiv (-1)^m(3) \pmod5$$

• Hence, when $n \equiv 1 \pmod{4}$, that is when $m \equiv 1 \pmod{2}$, $2^n \equiv -3 \equiv 2 \pmod{5} $

• when $n \equiv -1 \pmod{4}$, that is when $m \equiv 0 \pmod{2}$, $2^n \equiv 3 \pmod{5} $

• With those , we should be able to evaluate $2^n \pm 1 \pmod{5}$.