Chapter 12.5, Problem 28E

To determine

To Prove: $x+y$ is a factor of $x^{2n-1}+y^{2n-1}$ 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 $x^{2n-1}+y^{2n-1}$ for all natural numbers $n$

Put $n=1$

We get

  $\begin{array}{l}{x}^{2n-1}+y^{2n-1}=x^{2\left(1\right)-1}+y^{2\left(1\right)-1}\\ \Rightarrow x^{2n-1}+y^{2n-1}=x^{2-1}+y^{2-1}\\ \Rightarrow x^{2n-1}+y^{2n-1}=x+y\end{array}$

  $x+y$ 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 $x^{2n-1}+y^{2n-1}$ is True

Therefore

We get

  $x^{2n-1}+y^{2n-1}=c\left(x+y\right)$

Where $c$ is an integer

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

  $\begin{array}{l}{x}^{2n-1}+y^{2n-1}=x^{2\left(k+1\right)-1}+y^{2\left(k+1\right)-1}\\ \Rightarrow x^{2n-1}+y^{2n-1}=x^{2k+2-1}+y^{2k+2-1}\\ \Rightarrow x^{2n-1}+y^{2n-1}=x^{2k+1}+y^{2k+1}\\ \Rightarrow x^{2n-1}+y^{2n-1}=x^2\cdot x^{2k-1}+y^2\cdot y^{2k-1}\\ \Rightarrow x^{2n-1}+y^{2n-1}=x^2\cdot x^{2k-1}+\left(y^2-x^2+x^2\right)\cdot y^{2k-1}\\ \Rightarrow x^{2n-1}+y^{2n-1}=x^2\cdot \left(x^{2k-1}-y^{2k-1}\right)+\left(y^2-x^2\right)\cdot y^{2k-1}\\ \Rightarrow x^{2n-1}+y^{2n-1}=x^2\cdot \left(c\left(x+y\right)\right)+\left(y-x\right)\left(y+x\right)\cdot y^{2k-1}\\ \Rightarrow x^{2n-1}+y^{2n-1}=\left(x+y\right)\left[x^2\cdot \left(c\right)+\left(y-x\right)\cdot y^{2k-1}\right]\end{array}$

Therefore

  $x+y$ is a factor of $x^{2n-1}+y^{2n-1}$ is True for $n=k+1$ .

And hence the proof by Mathematical induction.