# Proving that $5^n - 1$ is divisible by $4$ by mathematical induction.

I have done it, but I am not sure that the inductive step is right. Can anybody please clear me about it?

Basic steps as:

Taking $$n=1$$: $$p(1)=5-1=4$$.

Inductive hypothesis: Assume the statement is true for $$p(k)$$. $$5^k - 1$$ is divisible by $$4$$.

Inductive steps: We must show $$p(k+1)$$ is true when $$p(k)$$ is true. \begin{align*} & 5^k -1 + 5^{k+1} -1\\ & 5^k -1 + 5.5^{k} -1\\ & (5^k -1) + 4 \end{align*}

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The idea is $$5^{k+1} - 1 = 5\cdot 5^{k} - 1 = 4\cdot(5^k) + (5^k - 1)$$ We have $$4\cdot(5^k)$$ is obviously divisible by $$4$$ and $$5^k - 1$$ is divisible by $$4$$ by inductive hypothesis

1. For n=1 the statement is true.
2. Assume that $$5^n-1$$ is divisible by 4.
3. Let us consider $$5^{(n+1)}-1$$. We have to prove that it is divisible by4 also.

We have that $$\begin{gathered} 5^{n + 1} - 1 = 5 \cdot 5^n - 1 = 5 \cdot 5^n - 5 + 5 - 1 = \hfill \\ \hfill \\ = 5\left( {5^n - 1} \right) + 4 \hfill \\ \end{gathered}$$ Since by inductive step 2 you have that $$5^n-1=4k$$ whit k integer you have that $$5\left( {5^n - 1} \right) + 4 = 5 \cdot 4k + 4 = 4\left( {5k + 1} \right)$$ and this prove step 3

$$4|5^0-1=0$$

and, for integer $$\ n>0\$$ we have

$$5^n-1\ =\ (5^n-5^{n-1}) + (5^{n-1}-1) \ =\ 4\cdot 5^{n-1} + (5^{n-1}-1)$$

where $$\ 5^{n-1}-1\$$ is divisible by $$\ 4\$$ by induction.