How to prove $5\mid (6^n -1)$

Question

How can I prove Inductive Step? $5$ divides $6^n -1$.

I've already proved base case and induction Hypothesis but I don't know how to prove Inductive step.

Answer 1

If we look at the induction hypothesis: we suppose $5|6^k-1$ for some $k \in \mathbb{Z}_{\geq 1}$, then we can write $6^k-1 = 5m$ for some $m \in \mathbb{Z}_{\geq 1}$. Now we have $$6^{k+1}-1 =6\cdot6^k-1 = 6\cdot 6^k -6 +5 = 6(6^k-1)+5 = 6(5m)+5 = 5(6m+1).$$ This is a multiple of 5, hence $5|6^{k+1}-1$, and we are done.

Answer 2

Inductive step

We need to show that for any $k\in \mathbb{N},$ if $5\mid6^k-1$ holds, then $5\mid 6^{k+1}-1$ also holds.

Assume that $5\mid 6^{k}-1$ holds, then there exists some $q\in\mathbb{Z}$ such that $6^{k+1}-1=5q.$

It follow that

$$\begin{align*}6^{k+1}-1&=6\cdot6^{k}-1\\&=(5+1)\cdot6^k-1\\&=5\cdot6^k+6^k-1\\&=5\cdot6^k+5q\\&=5(6^k+q).\end{align*}$$

Therefore, if $5\mid 6^{k}-1$ is true, then $5\mid 6^{k+1}-1$ is also true.

So, by the principle of mathematical induction $5\mid6^{n}-1$ for all $n\in\mathbb{N}$.