Let the given statement be P(n), i.e.,
P (n) : x2n−y2n is divisible by x + y.
For n = 1, P(1) : x2–y2 is divisible by x + y. Hence, P(1) is true.
Assume P(k) is true for some positive integer k, i.e.,
x2k−y2k = (x + y) g , where g ∊ N ... (i)
Now we shall prove that P(k + 1) is true.
For this we have to prove that
x2(k+1)−y2(k+1) is divisible by x + y.
Let us consider,
x2(k+1)−y2(k+1) = x2kx2−y2ky2
= [(x + y)g + y2k] x2–y2ky2 [From (i)]
= (x + y)g x2 + x2y2k – y2ky2 = (x + y)g x2 + y2k(x2–y2)
= (x + y) [gx2+y2k(x – y)] ⇒ x2 (k + 1) – y2 (k + 1) is divisible by x + y.
Hence, P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true ∀ n ∈N.