If $m$ is odd, does $\frac{x^m + y^m}{x+y}$ have an odd number of terms?

Question

Let $m,x,y$ be integers with $m$ positive and odd.

It seems to me that $\dfrac{x^m + y^m}{x+y}$ will always have an odd number of terms.

Here's my reasoning:

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

• All but the last term come in pairs so the total number of terms is odd.

Examples of this are:

• $\dfrac{x^3 + y^3}{x+y} = x^2 + y^2 - xy$

• $\dfrac{x^5 + y^5}{x+y} = (x^4 + y^4) - (x^3y + xy^3) + x^2y^2$

Am I right? If so, what would be a standard way to prove this?

Thanks very much,

-Larry

Answer 1

You can really reduce to $$ \frac{1 + z^{m}}{1+z}, $$ by setting $z = y/x$.

And then, yes, if $m$ is odd there is the identity $$ z^{m} + 1 = (z+1)(z^{m-1} - z^{m-2} + \dots + (-1)^{k} z^{k} + \dots - z + 1), $$ which you can see dividing $z^{m} + 1$ by $z+1$, using induction in case $$ z^{m} + 1 = z^{m} - z^{m-2} + z^{m-2} + 1 = (z^{m-1} - z^{m-2}) (z + 1) + z^{m-2} + 1, $$ for $m > 1$.

Answer 2

For $m$ positive and odd $$\dfrac{x^m + y^m}{x+y} = x^{m-1}y^0-x^{m-2}y^1+x^{m-3}y^2-\cdots+x^{2}y^{m-3}-x^{1}y^{m-2}+x^{0}y^{m-1},$$ which has $m$ terms, and $m$ is odd by supposition.