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.
 A: You have shown that $5^{k+1}-1 = 4(5^k+m)$, and $4$ clearly divides this last quantity, so $4$ divides $5^{k+1}-1$, as desired.  You also need to show the base case $n=0$.
A: The simplest way to do problems of this form is to work modulo $4$ and use the fact that $a\equiv b\pmod k \implies a^n\equiv b^n\pmod k$. In this case, $5\equiv 1\pmod 4 \implies 5^n\equiv 1^n = 1 \pmod 4$ or in other words, $4\mid(5^n-1)$ as required.
A: We can use modular arithemtic.
5 is 1 mod 4, so any positive integer (natural) power of 5 still results in 1 mod 4. 1 is 1 mod 4.
1 mod 4 - 1 mod 4 = 0 mod 4.
0 mod 4 means divisible by 4.
A: 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.
A: Here's the full induction proof:
Let $P(n) = 4 \mid 5^n-1$
Base case: $n=1 \implies 4 \mid 4$, hence $P(n)$ is true. (You can extend the number of base case if you need, but yeah, we'll stop here.)
Inductive step:Assume that $P(n)$ holds for all natural $n : 1 \leq n \leq k$. Thus , $P(k)$ is also true, hence we need to prove that $P(k+1)$ must also be true by the principle of induction.
Notice that $P(k) = 4 \mid 5^k - 1 \implies 5^k - 1 = 4x$. Now on rearranging, you get $5^k = 4x + 1$. Multiply both sides by $5$ and you get $5^{k+1} = 4\times 5\times x + 5 = 4 \times 5 \times x + (4 + 1) = 4m + 1$ where $m = 5x + 1$. Take $1$ from the RHS to the LHS to get  $5^{k+1} - 1 = 4m \implies 4 \mid 5^{k+1}-1 \implies P(k+1)$ is also true. Thus $4\mid 5^n -1 \forall n \in \mathbb{N}$
On examining how far you got, I guess you were stuck on understanding how you can prove $5^{k+1}-1$ is divisible by $4$. The key is that you got $4$ as a factor when factorizing the expression, so basically $4$ should divide $5^{k+1}-1$. If you were so confused as to how to arrive at such a saying, assume $m' = 5^{k+1}-1$, then keep doing as follows:
$m' = 5^{k+1}-1 = 5\times 5^k-1 = 4\times 5^k + 5^k - 1 = 4(5^k + \frac{5^k-1}4) = 4\times (\text{some factor; call it some other name}) \implies 4 \mid 5^{k+1}-1$.
You were basically on the verge of proving it but all of a sudden you lost contact with the solution, it seems.
