Sum of Odd Positive Powers

Theorem
Let $$n \in \N$$ be an odd positive integer.

Let $$x, y \in \Z_+$$ be integers.

Then $$x + y$$ is a divisor of $$x^n + y^n$$.

Proof
Let $$n \in \N$$ be odd, so that we can express it in the form $$n = 2 m + 1$$ where $$m \in \N$$.

We need to show that $$x^{2m + 1} + y^{2m + 1} = \left({x + y}\right) \left({x^{2 m} + \cdots + y^{2 m}}\right)$$.

When $$m = 0$$ we just have $$x + y = x + y$$ which is trivially an identity.

When $$m = 1$$ we have the identity:
 * $$x^3 + y^3 = \left({x + y}\right) \left({x^2 - xy + y^2}\right)$$

whose truth is apparent after some simple algebra.

Now we assume that:
 * $$\exists k \in \N: \forall j: 1 \le j \le k: x^{2j + 1} + y^{2j + 1} = \left({x + y}\right) P_{2 j} \left({x, y}\right)$$

where $$P_{2 j} \left({x, y}\right)$$ is a polynomial of order $2 j$ in $$x$$ and $$y$$.

We need to show that:
 * $$x^{2k + 3} + y^{2k + 3} = \left({x + y}\right) P_{2k + 2} \left({x, y}\right)$$

where $$P' \left({x,y}\right)$$ is another polynomial in $$x$$ and $$y$$.

Now:
 * $$\left({x^{2 k + 1} + y^{2 k + 1}}\right) \left({x^2 + y^2}\right) = x^{2k + 3} + y^{2k + 3} + x^2 y^{2 k + 1} + x^{2 k + 1} y^2$$

So:

$$ $$

But $$\left({x^{2 k - 1} + y^{2 k - 1}}\right)$$ itself is of the form $$\left({x + y}\right) P_{2k - 2} \left({x, y}\right)$$.

So:
 * $$x^{2k + 3} + y^{2k + 3} = \left({x + y}\right) \left({\left({x^2 + y^2}\right) P \left({x,y}\right) - x^2 y^2 P_{2k - 2} \left({x, y}\right)}\right)$$

Hence the result by the Second Principle of Mathematical Induction.