Prove the following by using the principle of mathematical induction for all n ∈ N: x2n – y2n is divisible by x + y.

Let the given statement be $P(n)$, i.e.,

$P(n): x^{2 n}-y^{2 n}$ is divisible by $x+y .$

It can be observed that $\mathrm{P}(n)$ is true for $n=1$.

This is so because $x^{2} \times 1-y^{2} \times 1=x^{2}-y^{2}=(x+y)(x-y)$ is divisible by $(x+y)$.

Let P(k) be true for some positive integer k, i.e.,

$x^{2 k}-y^{2 k}$ is divisible by $x+y$

$\therefore x^{2 k}-y^{2 k}=m(x+y)$, where $m \in \mathbf{N}$   (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

$x^{2(k+1)}-y^{2(k+1)}$

$=x^{2 k} \cdot x^{2}-y^{2 k} \cdot y^{2}$

$=x^{2}\left(x^{2 k}-y^{2 k}+y^{2 k}\right)-y^{2 k} \cdot y^{2}$

$=x^{2}\left\{m(x+y)+y^{2 k}\right\}-y^{2 k} \cdot y^{2} \quad[U \operatorname{sing}(1)]$

$=m(x+y) x^{2}+y^{2 k}+x^{2}-y^{2 k} \cdot y^{2}$

$=m(x+y) x^{2}+y^{2 k}\left(x^{2}-y^{2}\right)$

$=m(x+y) x^{2}+y^{2 k}(x+y)(x-y)$

$=(x+y)\left\{m x^{2}+y^{2 k}(x-y)\right\}$, which is a factor of $(x+y)$.

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.