Solve the following

Let P(n) be the given statement.

Now,

$P(n): x^{2 n-1}+y^{2 n-1}$ is divisible by $x+y$

Step 1:

$P(1): x^{2-1}+y^{2-1}=x+y$ is divisible by $x+y$

Step 2 :

Let $P(m)$ be true.

Also,

$x^{2 m-1}+y^{2 m-1}$ is divisible by $x+y .$

Suppose :

$x^{2 m-1}+y^{2 m-1}=\lambda(x+y)$ where $\lambda \in N \quad \ldots(1)$

We shall show that $P(m+1)$ is true whenever $P(m)$ is true.

Now,

$P(m+1)=x^{2 m+1}+y^{2 m+1}$

$=x^{2 m+1}+y^{2 m+1}-x^{2 m-1} \cdot y^{2}+x^{2 m-1} \cdot y^{2}$

$=x^{2 m-1}\left(x^{2}-y^{2}\right)+y^{2}\left(x^{2 m-1}+y^{2 m-1}\right)$$\quad[$ From $(1)]$

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

$=(x+y)\left(x^{2 m-1}(x-y)+\lambda y^{2}\right)$$\quad[$ It is divisible by $(x+y) \cdot]$

Thus, $P(m+1)$ is true.

By the principle of $m$ athematical induction, $P(n)$ is true for all $n \in N$.