<s>[INST] For odd $m\ge3$, does it follow: $\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$ Unless I am making a mistake, I am calculating that:
$$\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$$
Here's my reasoning:


*

*$\dfrac{x^m + y^m}{x+y} = x^{m-1} - x^{m-2}y - xy^{m-2} + x^{m-3}y^2 + x^2y^{m-3} + \dots + x^{\frac{m-1}{2}}y^{\frac{m-1}{2}} + y^{m-1}$

*$\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-3} - x^{m-4}y - xy^{m-4} + x^{m-5}y^2 + x^2y^{m-5} + \dots + x^{\frac{m-3}{2}}y^{\frac{m-3}{2}} + y^{m-3}$

*$(xy)\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-2}y - x^{m-3}y^2 - x^2y^{m-3} + x^{m-4}y^3 + x^3y^{m-4} + \dots + x^{\frac{m-1}{2}}y^{\frac{m-1}{2}} + xy^{m-2}$

*So that, $\dfrac{x^m + y^m}{x+y} + (xy)\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$

*I am figuring that $\dfrac{x^{m-2} + y^{m-2}}{x+y}$ has exactly 2 terms less than $\dfrac{x^m + y^m}{x+y}$
Is this reasoning correct?  For each value that I test, it seems correct.
Thanks,
-Larry
 [/INST] $$\frac{x^m+y^m}{x+y}+(xy)\frac{x^{m-2}+y^{m-2}}{x+y}=\frac{x^m+y^m+(xy)(x^{m-2}+y^{m-2})}{x+y}= \\
=\frac{x^m+y^m+x^{m-1}y+xy^{m-1}}{x+y}=\frac{x^m+x^{m-1}y+xy^{m-1}+y^m}{x+y}\\
=\frac{x^{m-1}(x+y)+y^{m-1}(x+y)}{x+y}=x^{m-1}+y^{m-1}$$
</s>