Argue by contradiction : $n\in \mathbb N \to \; 4|(5^{n}-1)$ I was working for various method to solve this :$n\in \mathbb N \to \; 4|(5^{n}-1)$ now I want to solve it only by "Argue by contradiction"

As a reverse induction using contradiction:
Note that for $n=0$ you have $4|(1-1)$ and for $n=1$ you have $4|(5-1)$.
So choose $m$ as the smallest positive integer for which $4\not | (5^m-1)$ 
in which case $4\not | (5^m-1-4\times 5^{m-1})$ and so $4\not | (5^{m-1}-1)$ 
but by the definition of $m$ we know  $4 | (5^{m-1}-1)$, leading to a contradiction, 
implying there is no such $m$
so  $4 | (5^n-1)$ for all positive integers $n$.
