Question

Prove $x^{2n}-y^{2n}$ is divisible by x + y for all natural numbers.

Answer 1

Let the given statement be P(n), i.e.,
$P\left(n\right):x^{2n}-y^{2n}$ is divisible by x + y.
It can be observed that 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.,
$x^{2k}-y^{2k}$ is divisible by x + y ..........(1)
We shall now prove that p(k + 1) is true whenever P(k) is true.
Consider
$x^{2(k+1)}-y^{2(k+1)}$
$=x^{2k}.x^2-y^{2k}.y^2$
$=x^2(x^{2k}-y^{2k}+y^{2k})-y^{2k}.y^2$
$=x^2\left\{m\right(x+y)+y^{2k}\}-y^{2k}.y^2$ [Using (1)]
$=m(x+y)x^2+y^{2k}.x^2-y^{2k}.y^2$
$=m(x+y)x^2+y^{2k}(x^2-y^2)$
$=m(x+y)x^2+y^{2k}(x+y)(x-y)$
$=(x+y)\{m{x}^2+y^{2k}(x-y\left)\right\}$, 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.