$5 \nmid 2^{n}-1$ when $n$ is odd

Question

I want to prove that $$5 \nmid 2^{n}-1$$ where $n$ is odd.

I used Fermat's little theorem, which says $2^4 \equiv 1 \pmod 5$, because $n$ is odd then $4 \nmid n$ , so it is done.

can you check it and say that my proof is right or wrong.

thanks.

Answer 1

Take $n=2k+1~,~k\geq 0$ since $n$ is odd. Now,

$$2^n=2^{2k+1}=2\cdot 4^k\equiv 2\cdot (-1)^k\equiv \begin{cases}2~,~k\textrm{ is even}\\ 3~,~k\textrm{ is odd}\end{cases}\pmod{5}$$

$$2^n-1=2^{2k+1}-1\equiv\begin{cases}1~,~k\textrm{ is even}\\ 2~,~k\textrm{ is odd}\end{cases}\pmod5\implies 2^n-1\not\equiv 0\pmod{5}~\forall~\textrm{odd }n$$

$$\therefore\quad 5\not\mid 2^n-1~\forall~\textrm{odd }n$$

---

This can also be done by induction, but the approach using modular arithmetic is the easiest.

Answer 2

If you observe the first few odd powers of 2, you can infer that last digit of all odd powers of 2 is either 2 or 8.

$2^1=2,2^3=8, 2^5=32, 2^7=128,...$

Hence $$\quad 5\not\mid 2^n-1~\forall~\textrm{odd }n$$

Answer 3

${\rm mod}\ 5\!:\,\ 2^{1+2j}\equiv 2(4^j)\equiv 2(-1)^j\equiv \pm2 \not\equiv 1$