Chapter 12.5, Problem 27E

To determine

To Prove: $x-y$ is a factor of $x^n-y^n$ for all natural numbers $n$ .

Explanation of Solution

Given information:

Use Mathematical induction to prove the above mentioned formula.

Concept used:

1.By using Mathematical induction technique we need to prove that this formula holds True for all natural numbers. The first step is to prove True for $n=1$ .

2.Second step by using induction hypothesis of mathematical induction we assume for $n=k$ given formula is True.

3.Third step is we need to prove that above formula is True for $n=k+1$ by using formula is True for $n=k$ .

Proof:

First, we need to prove the above formula is True for $n=1$ .

Given formula is

  $x-y$ is a factor of $x^n-y^n$ for all natural numbers $n$

Put $n=1$

We get

  $\begin{array}{l}{x}^n-y^n=x^1-y^1\\ \Rightarrow x^n-y^n=x-y\end{array}$

  $x-y$ is factor of $x-y$ .

It True for $n=1$ .

Let us assume it is True for $n=k$

So

  $x-y$ is a factor of $x^k-y^k$ is True

Therefore

We get

  $x^k-y^k=c\left(x-y\right)$

Where $c$ is an integer

We need to prove it is True for $n=k+1$

  $\begin{array}{l}{x}^n-y^n=x^{k+1}-y^{k+1}\\ \Rightarrow x^n-y^n=x^k\left(x-y\right)+\left(x^k-y^k\right)y\end{array}$

We know

  $x^k-y^k=c\left(x-y\right)$

Using that

We get

  $\Rightarrow x^n-y^n=x^k\left(x-y\right)+c\left(x-y\right)y$

Taking $\left(x-y\right)$ common from both terms

We get

  $\Rightarrow x^n-y^n=\left(x-y\right)\left(x^k+cy\right)$

Therefore

  $x-y$ is a factor of $x^n-y^n$ is True for $n=k+1$ .

And hence the proof by Mathematical induction.