Proof

Let \(P(n)\) be the statement, “\(6^n - 1\) is a multiple of 5.” We will prove that \(P(n)\) is true for all \(n \in \N\text{.}\)

Base case: \(P(0)\) is true: \(6^0 -1 = 0\) which is a multiple of 5.

Inductive case: Let \(k\) be an arbitrary natural number. Assume, for induction, that \(P(k)\) is true. That is, \(6^k - 1\) is a multiple of \(5\text{.}\) Then \(6^k - 1 = 5j\) for some integer \(j\text{.}\) This means that \(6^k = 5j + 1\text{.}\) Multiply both sides by \(6\text{:}\)

\begin{equation*} 6^{k+1} = 6(5j+1) = 30j + 6\text{.} \end{equation*}

But we want to know about \(6^{k+1} - 1\text{,}\) so subtract 1 from both sides:

\begin{equation*} 6^{k+1} - 1 = 30j + 5\text{.} \end{equation*}

Of course \(30j+5 = 5(6j+1)\text{,}\) so is a multiple of 5.

Therefore \(6^{k+1} - 1\) is a multiple of 5, or in other words, \(P(k+1)\) is true. Thus, by the principle of mathematical induction \(P(n)\) is true for all \(n \in \N\text{.}\)