<s>[INST] 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.
 [/INST] Hint $\ {\rm mod}\ x\!+\!y\!:\ \color{#c00}{x\equiv -y} \,\Rightarrow\, \color{#c00}x^n+y^n\equiv (\color{#c00}{-y})^n+y^n\equiv 0\,$ by $\,n\,$ odd, and $ $ Congruence Rules.
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