Chapter 9.5, Problem 20E

Solution

To prove : using mathematical induction to prove that statement $6$ is a factor of $7^n-1$ holds true for all positive integers.

Given information :

The statement $7^n-1$ .

Formula used :

Follow the steps to prove that the statement is true for any positive integer.

Anchor step: Prove that any statement $P_n$ is true for $n=1$ .

Inductive hypothesis: Assume $P_n$ is true for $n=k$ .

Inductive Step: Prove that $P_n$ is true for $n=k+1$ .

Proof :

Conjecture:

Take $P_n$ to be statement $7^n-1$ .

Anchor step:

Prove that $P_n$ is true for $n=1$ .

For $n=1$ ,

  $\begin{array}{l}{7}^1-1\stackrel{?}{=}6\\ 6=6\end{array}$

Inductive hypothesis:

Assume $P_n$ is true for $n=k$ . Replace $n$ with $k$ in $P_n$ .

  $P_k:7^k-1$ is divisible by $6$ .

Inductive Step:

Prove that $P_n$ is true for $n=k+1$ . This means it is needed to be proved that $P_{k+1}$ must be true.

To prove $P_{k+1}:7^{k+1}-1$ is also divisible by $6$ .

Proceed with the inductive hypothesis.

  $P_k:7^k-1$ is divisible by $6$ .

So,

  $\begin{array}{c}{7}^{k+1}-1=7\cdot 7^k-1\\ =\left(6+1\right)7^k-1\\ =6\cdot 7^k+7^k-1\\ =6\cdot 7^k+\left(7^k-1\right)\end{array}$

As, $7^k-1$ is divisible by $6$ and $6\cdot 7^k$ is also divisible by $6$ .

This implies that $P_{k+1}:7^{k+1}-1$ is also divisible by $6$ .

So, if $P_n$ is true for $n=k$ , then it is also true for $n=k+1$ .

Conclusion:

As $P_n$ is true for $n=1$ and $P_k$ further implies $P_{k+1}$ , this means $P_n$ is true for $n=2$ , $n=3$ and so on.

By mathematical induction, $6$ is a factor of $7^n-1$ is true for all positive integers $n$ .