If $n$ is an odd integer, does $x+y$ divide $x^n + y^n$? I believe that the answer is yes.
Here's my thinking:


*

*$x^n + y^n -(x+y)(x^{n-1} + y^{n-1}) = -x^{n-1}y -xy^{n-1}$

*$-x^{n-1}y -xy^{n-1} - (x+y)(-x^{n-2}y -xy^{n-2}) = x^{n-2}y^2 + x^2y^{n-2}$

*So, at each step, we get $(-1)^{c}(x^{n-c}y^{c} + x^{c}y^{n-c})$ with $c$ increasing by $1$ at each step and 

*Since $n$ is odd, eventually, we get:  $(-1)^c(x^{c}y^{c+1} + x^{c}y^{c+1}) = (x+y)(-1)^c(x^{c}y^{c})$


Am I correct?  Am I making any mistakes in my thinking?
If this is correct, is the standard formula:
$$x^n + y^n = (x+y)(x^{n-1} -x^{n-2}y -xy^{n-2} + \dots + (-1)^{\frac{n-1}{2}}x^{\frac{n-1}{2}}y^{\frac{n-1}{2}} + y^{n-1})$$
Thanks,
-Larry 

Edit:  Fixed the equation based on comment.

$\text{ We have } P(x) = x^n + y^n \to P(-y) = (-y)^n + y^n = -y^n +y^n = 0 \to x+y \text{ is a factor of P(x) }$.
