Question

find whether xⁿ+yⁿ is divisible by x-y (y≠0)
or not​

Answer 1

Answer:

The crux of this proof is (as usual) the inductive step.

In this case, if we assume xⁿ - yⁿ is divisible by x - y, how do we show xⁿ⁺¹ - yⁿ⁺¹ is divisible by x - y? It wasn't immediately obvious to me, so I began by expanding:

(x + y)(xⁿ - yⁿ) = xⁿ⁺¹ - yⁿ⁺¹ + yxⁿ - xyⁿ.

This gives us xⁿ⁺¹ - yⁿ⁺¹, as we want, but what about the other two terms? If we take out the common factor of xy, we get:

yxⁿ - xyⁿ = xy(xⁿ⁻¹ - yⁿ⁻¹).

Thus xⁿ⁺¹ - yⁿ⁺¹ = (x + y)(xⁿ - yⁿ) - xy(xⁿ⁻¹ - yⁿ⁻¹).

Now our strategy becomes clear. The inductive step will be to assume the proposition is true for n and n-1, and prove it is true for n+1. The basis step must then be to show the proposition is true for n = 1 and n = 2. This is a little different from the usual form of induction, where we assume the proposition is true for n-1 and show it is true for n, but it is perfectly valid.

Basis step: x¹ - y¹ = x - y is clearly divisible by x - y, if x - y ≠ 0; i.e., if x ≠ y.

Also, x² - y² = (x - y)(x + y) is also divisible by x - y, again if x ≠ y.

For the inductive step, assume the proposition is true for n and n-1. Then:

xⁿ⁺¹ - yⁿ⁺¹ = (x + y)(xⁿ - yⁿ) - xy(xⁿ⁻¹ - yⁿ⁻¹).

By the inductive hypothesis, xⁿ - yⁿ and xⁿ⁻¹ - yⁿ⁻¹ are both divisible by x - y.

Hence xⁿ⁺¹ - yⁿ⁺¹ is the difference of two multiples of x - y, and so is itself a multiple of x - y.

The result now follows by mathematical induction.

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