Let the statement be $P(n)$ i.e.
$P(n): x^{2 n}-y^{2 n}$ is divisible by $x+y \quad \ldots(i)$
Putting $n=1, x^{2 n}-y^{2 n}=x^2-y^2=(x+y)(x-y)$ which is divisible by $x+y$
$\Rightarrow P(n)$ is true for $n=1$
Let $P(k)$ be true i.e. $x^{2 k}-y^{2 k}$ is divisible by $x+y$
or $x^{2 \mathrm{k}}-y^{2 \mathrm{k}}=m(x+y)$
or $x^{2 k}=m(x+y)+y^{2 k} \quad \ldots(ii)$
We have to prove that $P(k+1)$ is true whenever $P(k)$ is true.
$x^{2(k+1)}-y^{2(k+1)}=x^{2 k+2}-y^{2 k+2}$
$=x^2 \cdot x^{2 k}-y^{2 \mathrm{k}+2}$
Putting the value of $x^{2 k}$ from (ii)
$=x^2\left[m(x+y)+y^{2 k}\right]-y^{2 k+2}$
$=m(x+y) x^2+x^2 y^{2 k}-y^{2 k+2}$
$=m(x+y) x^2+y^{2 k}\left(x^2-y^2\right)$
$=m(x+y) x^2+(x+y)(x-y) y^{2 k}$
$=(x+y)\left[m x^2+(x-y) y^{2 k}\right]$
$\therefore x^{2(k+1)}-y^{2(k+1)}$ is divisible by $(x+y)$
i.e., $P(k+1)$ is true whenever $P(k)$ is true.
Hence $P(n)$ is true for all $n \in \mathrm{N}$