For odd $m\ge3$, does it follow: $\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$

Question

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

Answer 1

Well, I would do it that way : $$ x^m = x\ x^{m-1} $$ So :

$$\begin{align}\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} &= \frac{x^m + y^m+ (xy)\ x^{m-2} + (xy)\ y^{m-2}}{x+y} \\ &=\frac{x^m + y^m+ y\ x^{m-1} + x\ y^{m-1}}{x+y} \\ &=\frac{ x\ x^{m-1} + y\ y^{m-1}+ y\ x^{m-1} + x\ y^{m-1}}{x+y}\\ &=\frac{ (x+y)\ x^{m-1} + (x+y)\ y^{m-1}}{x+y}\\ &= x^{m-1} + y^{m-1}\end{align}$$

Answer 2

$$\frac{x^m+y^m}{x+y}+(xy)\frac{x^{m-2}+y^{m-2}}{x+y}=\frac{x^m+y^m+(xy)(x^{m-2}+y^{m-2})}{x+y}= \\ =\frac{x^m+y^m+x^{m-1}y+xy^{m-1}}{x+y}=\frac{x^m+x^{m-1}y+xy^{m-1}+y^m}{x+y}\\ =\frac{x^{m-1}(x+y)+y^{m-1}(x+y)}{x+y}=x^{m-1}+y^{m-1}$$

Answer 3

Your argument is fine, possibly a bit more complicated than necessary. And the theorem is true for $m$ even, too, so a better proof wouldn't use the closed form for $\frac{x^m+y^m}{x+y}$, which is only true for $m$ odd.

A more advanced way to prove this uses linear recurrences. Show that $a_n=x^n+y^n$ satisfies:

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

It is easy to show this linear recurrence is true for $b_n=x^n$ and $c_n=y^n$ because $z^2-(x+y)z+xy=0$ has roots $z=x,y$. So it is true for $a_n=b_n+c_n$.

The recurrence is also true if $a_n=\alpha\cdot x^n + \beta \cdot y^n$ for any $\alpha,\beta,x,y$. So:

$$\frac{\alpha x^m + \beta y^m}{x+y} + (xy)\frac{\alpha x^{m-2} + \beta y^{m-2}}{x+y} = \alpha x^{m-1} +\beta y^{m-1}$$

That shows that the $x+y$ in the denominator is not part of the original pattern, $a_1$, but related to the $x,y$ independent of the $\alpha,\beta$.

This can then be generalized to more variables. If $a_n=\alpha x^n+\beta y^n+\gamma z^n$ then:

$$a_n = (x+y+z)a_{n-1} -(xy+yz+xz)a_{n-2} + (xyz)a_{n-3}$$

You can go to more variables, with the coefficients of the linear recurrence being the (alternating) basic symmetric polynomials.