For any natural number n,

According to the question,

P(n) = xn – yn is divisible by x – y, x integers with x ≠ y.

So, substituting different values for n, we get,

P(0) = x0 – y0 = 0 Which is divisible by x − y.

P(1) = x − y Which is divisible by x − y.

P(2) = x2 – y2

= (x +y)(x−y) Which is divisible by x−y.

P(3) = x3 – y3

= (x−y)(x2+xy+y2) Which is divisible by x−y.

Let P(k) = xk – yk be divisible by x – y;

So, we get,

⇒ xk – yk = a(x−y).

Now, we also get that,

⇒  P(k+1) = xk+1 – yk+1

= xk(x−y) + y(xk−yk)

= xk(x−y) +y a(x−y) Which is divisible by x − y.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) xn – yn is divisible by x – y, where x integers with x ≠ y which is true for any natural number n.