# Question #b809d

May 21, 2015

We have :

$\frac{{x}^{2 n + 1} - {y}^{2 n + 1}}{x - y} = \lambda$, where $\lambda \in \mathbb{Z}$

$\implies \left({x}^{2 n + 1} - {y}^{2 n + 1}\right) = \lambda \left(x - y\right)$

Basis :

With $n = 0$, $\left(x - y\right) = 1 \left(x - y\right) = \lambda \left(x - y\right)$

With $n = 1$, $\left({x}^{3} - {y}^{3}\right) = \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right) = \lambda \left(x - y\right)$

Inductive step :

Let's assume $\left({x}^{2 n + 1} - {y}^{2 n + 1}\right) = \lambda \left(x - y\right)$ is true for $n \in \mathbb{N}$.

Let's demonstrate that it is true for $n + 1 \in \mathbb{N}$ :

${x}^{2 \left(n + 1\right) + 1} - {y}^{2 \left(n + 1\right) + 1} = {x}^{2 n + 3} - {y}^{2 n + 3}$

${x}^{2 n + 3} - {y}^{2 n + 3} = {x}^{2 n + 3} + {x}^{2 n + 1} {y}^{2} - {x}^{2 n + 1} {y}^{2} - {y}^{2 n + 3}$

$= {x}^{2 n + 1} \left({x}^{2} - {y}^{2}\right) + {y}^{2} \left({x}^{2 n + 1} - {y}^{2 n + 1}\right)$

$= {x}^{2 n + 1} \left(x + y\right) \left(x - y\right) + {y}^{2} \lambda \left(x - y\right)$

$= \left(x - y\right) \left({x}^{2 n + 1} \left(x + y\right) + \lambda {y}^{2}\right)$

$= {\lambda}_{2} \left(x - y\right)$, where ${\lambda}_{2} = \left({x}^{2 n + 1} \left(x + y\right) + \lambda {y}^{2}\right) \in \mathbb{Z}$.

QED.