If $n$ is a natural number, then $4$ divides $5^n-1$. The question is Statement 3.38.

If $n$ is a natural number, then $4$ divides $5^n-1$.

So far all I have is:

*

*$5^k\times5-1$

*$5^k(4+1)-1$

*$4\times5^k+5^k-1$

*$4\times5^k+4m$
I do not get how this equals what we are trying to get to using proof of induction, which is $k+1$. If we factor that out we get $4(5^k+m)$. What do I do with that remaining $m$? And is $4(5^k+m)$ equal to the $k+1$ was this all that I need for my final answer.

Another solution is using the algebraic identity
$$x^n - y^n = (x - y) \left( x^{n - 1} + x^{n - 2} y + x^{n - 3} y^2 + \cdots x y^{n - 2} + y^{n - 1} \right)$$
In our case, $x = 5$ and $y = 1$ so
$$5^n - 1 = (5 - 1) \left( 5^{n - 1} + 5^{n - 2} + \cdots 5 + 1 \right) = 4 \sum_{k = 0}^{n - 1} 5^k$$
which is a multiple of 4.
