If `x!= y`, then for every natural number n, `x^n - y^n` is divisible by

Options

A. `x + y`
B. `x - y`
C. `1`
D. None of these

Answer

B

Explanation

Let `P(n):x^(n)-y^(n)` is divisible by x-y, where x and y are any integers with `x!=y`
Step I We observe that P(1) is true.
`P(1):x^(1)-y^(1)=x-y`
Step II Now, assume that P(n) is true for n=k.
`P(k):x^(k)-y^(k)` is divisible by (x-y).
`:.x^(k)-y^(k)=q(x-y)`
Step III Now, to prove p(k+1) is true
P(k+1) : `x^(k+1)-y^(k+1)`
`=x^(k)*x-y^(k)*y`
`=x^(k)*x-y^(k).y-y^(k)*y`
`=x^(k)(x-y)+y(x^(k)-y^(k))`
`=x^(k)(x-y)+yq(x-y)`
`=(x-y)[x^(k)+yq]`, which is divisible by (x-y). [from stepII]
Hence, P(k+1) is true whenever P(k) is true. So, by the principle of mathematical induction P(n) is true for any natural number n.