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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x^{2k + 3} + y^{2k + 3}\)	\(=\)	\(\displaystyle \left({x + y}\right) \left({x^2 + y^2}\right) P_{2k} \left({x, y}\right) - x^2 y^{2 k + 1} - x^{2 k + 1} y^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x + y}\right) \left({x^2 + y^2}\right) P_{2k} \left({x, y}\right) - x^2 y^2 \left({x^{2 k - 1} + y^{2 k - 1} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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.

$\blacksquare$

Sources

• George E. Andrews: Number Theory (1971): $\S 1.1$: Exercise $18$