# 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 $m \in \N$.

The proof proceeds by strong induction.

For all $m \in \N$, let $\map P m$ be the proposition:

$x^{2 m + 1} + y^{2 m + 1} = \paren {x + y} \paren {x^{2 m} + \cdots + y^{2 m} }$

$\map P 0$ is the case:

$x + y = x + y$

which is trivially an identity.

### Basis for the Induction

$\map P 1$ is the case:

$x^3 + y^3 = \paren {x + y} \paren {x^2 - x y + y^2}$

which follows by Sum of Two Cubes.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

$\exists k \in \N: \forall j: 1 \le j \le k: x^{2 j + 1} + y^{2 j + 1} = \paren {x + y} \map {P_{2 j} } {x, y}$

where $\map {P_{2 j} } {x, y}$ is a polynomial of degree $2 j$ in $x$ and $y$.

from which it is to be shown that:

$x^{2 k + 3} + y^{2 k + 3} = \paren {x + y} \map {P_{2 k + 2} } {x, y}$

where $\map {P_{2 k + 2} } {x, y}$ is a polynomial of degree $2 k + 2$ in $x$ and $y$.

### Induction Step

This is the induction step:

We have that:

$\paren {x^{2 k + 1} + y^{2 k + 1} } \paren {x^2 + y^2} = x^{2 k + 3} + y^{2 k + 3} + x^2 y^{2 k + 1} + x^{2 k + 1} y^2$

So:

 $\displaystyle x^{2 k + 3} + y^{2 k + 3}$ $=$ $\displaystyle \paren {x + y} \paren {x^2 + y^2} \map {P_{2 k} } {x, y} - x^2 y^{2 k + 1} - x^{2 k + 1} y^2$ $\displaystyle$ $=$ $\displaystyle \paren {x + y} \paren {x^2 + y^2} \map {P_{2 k} } {x, y} - x^2 y^2 \paren {x^{2 k - 1} + y^{2 k - 1} }$

But $\paren {x^{2 k - 1} + y^{2 k - 1} }$ itself is of the form:

$\paren {x + y} \map {P_{2 k - 2} } {x, y}$

So:

$x^{2 k + 3} + y^{2 k + 3} = \paren {x + y} \paren {\paren {x^2 + y^2} \map {P_{2 k} } {x, y} - x^2 y^2 \map {P_{2 k - 2} } {x, y} }$

Hence the result by the Second Principle of Mathematical Induction.

$\blacksquare$