# Question 15 Exercise 7.1

Solutions of Question 15 of Exercise 7.1 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 15

Show that $a+b$ is a factor of $a^n-b^n$ for all even positive integer $n$.

### Solution

Since $n$ is even, so we put $n=2 n, \quad m \in \mathbb{Z}^{+}$.

1. For $m=1$, then $$a^{2 n}-b^{2 m}=a^2-b^2=(a+b)(a-b)$$ $\Rightarrow(a+b)$ is a factor of $a^2-b^2$.

Thus it is true for $m=1$ or $n=2$.

2. Let it be true for $m=k$ then $$a^{2 k}-b^{2 k}=Q(a+b)$$ where $Q$ is quotient in the induction hypothesis.

3. For $m=k+1$ then it becomes \begin{align}a^{2(k+1)}-b^{2(k-1)} & =a^{2 k+2}-b^{2 k+2} \\ & =a^{2 k} \cdot a^2-b^{2 k} \cdot b^2 \end{align} Adding and subtracting $a^2 b^{2 k}$ \begin{align} & =a^{2 k} \cdot a^2-a^2 b^{2 k}+a^2 b^{2 k}-b^{2 k} \cdot b^2 \\ & =a^2(a^{2 k}-b^{2 k})+b^{2 k}(a^2-b^2) \\ & =a^2 Q(a+b)+b^{2 k}(a+b)(a-b)\end{align} by induction hypothesis $$=(a+b)[a^2 Q-b^2(a-b)]$$ $\Rightarrow a+b$ is factor of $a^{2 n}-b^{2 m}$ for $m=k+1$.

Thus it is true for $m=k+1$. Hence by mathematical induction it is true for all even positive integers.

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