Question

Prove by using the principle of mathematical induction for all $nϵN$ such that $x^{2n}-y^{2n}$ is divisible by x+y.

Answer 1

Let P(n) be the statement given by

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

For n=1, P(1): $x^2-y^2$ is divisible by x+y.

$\therefore$ P(1) is true.

Assume that P(k) is true, for some natural number k, then

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

$\Rightarrow x^{2k}-y^{2k}=m(x+y),$ for some $mϵN$.

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

Now, $x^{2(k+1)}-y^{2(k+1)}=x^{2k+2}-y^{2k+2}=x^{2k}.x^2-y^{2k}.y^2$

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

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

$=(x+y)[m{x}^2+(x-y\left)y^{2k}\right],$ which is divisible by (x+y).

$\Rightarrow P(k+1)$ is true whenever P(k) is true.

Hence, by principle of mathematical induction, P(n) is true $\forall nϵN$