Explanation:
We can prove that X^n + Y^n is divisible by X + Y for any odd natural number n, where n ≥ 1, using mathematical induction.
Base case: n = 1 When n = 1, we have X^1 + Y^1 = X + Y, which is clearly divisible by X + Y.
Inductive step: Assume that X^k + Y^k is divisible by X + Y for some odd natural number k ≥ 1. We will show that X^(k+2) + Y^(k+2) is divisible by X + Y.
We can expand X^(k+2) + Y^(k+2) using the binomial theorem: X^(k+2) + Y^(k+2) = (X^2)(X^k) + (Y^2)(Y^k)
We can factor out X + Y from the first term and Y + X from the second term, since X + Y = Y + X: X^(k+2) + Y^(k+2) = (X + Y)(X^k + XY^k) + (X + Y)(Y^k + XY^k)
We can then simplify this expression by factoring out XY^k from the first term and X^kY from the second term: X^(k+2) + Y^(k+2) = (X + Y)XY^k + (X + Y)X^kY X^(k+2) + Y^(k+2) = (X + Y)XY^k + (X + Y)X^kY
We can then factor out (X + Y) from this expression to get: X^(k+2) + Y^(k+2) = (X + Y)(XY^k + X^kY)
Since X + Y is a common factor of X^(k+2) + Y^(k+2), we have shown that X^(k+2) + Y^(k+2) is divisible by X + Y.
Therefore, by mathematical induction, we have proven that X^n + Y^n is divisible by X + Y for any odd natural number n, where n ≥ 1.