Sum of Odd Positive Powers

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

Let $x, y \in \Z_{>0}$ be (strictly) positive integers.

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

Proof
Given that $n \in \N$ be odd, it can be expressed in the form:
 * $n = 2 m + 1$

where $m \in \N$.

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

When $m = 0$ we just have:
 * $x + y = x + y$

which is trivially an identity.

When $m = 1$ we can apply Sum of Two Cubes, which gives:
 * $x^3 + y^3 = \left({x + y}\right) \left({x^2 - x y + y^2}\right)$

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

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

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

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

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

So:

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

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

Hence the result by the Second Principle of Mathematical Induction.