Question

Prove that: $x^{2n}-y^{2n}$ is divisible by $x+y$.

Answer 1

Let $P\left(n\right):x^{2n}-y^{2n}=(x+y)\times d$ where $d\in N$

For $n=1$

$\begin{array}{c}\mathrm{LHS}={\mathrm{x}}^{2\times 1}-{\mathrm{y}}^{2\times 1}\\ ={\mathrm{x}}^2-{\mathrm{y}}^2\\ (\mathrm{x}+\mathrm{y})(\mathrm{x}-\mathrm{y})\\ =\mathrm{RHS}\end{array}$

$\therefore P\left(n\right)$ is true for $n=1$

Assume $P\left(k\right)$ is true.

$x^{2k}-y^{2k}=(x+y)\times m\text{where}m\in N$

We will prove that $P(k+1)$ is true.

$\begin{array}{c}\text{LHS}=x^{2\times (k+1)}-y^{2\times (k+1)}\\ =x^{2k+2}-y^{2k+2}\\ =x^{2k}{x}^2-y^{2k}{y}^2\\ =(x+y)\left[m{x}^2+y^{2k}(x-y)\right]\\ =(x+y)\times r\end{array}$

Where, $r=\left[m{x}^2+y^{2k}(x-y)\right]$

$\therefore P(k+1)$ is true whenever $P\left(k\right)$ is true.

$\therefore$ By the principle of mathematical induction, $P\left(n\right)$ is true for $n$, where $n$ is a natural number.