I want to show that we can find a real polynomial $p(y)$ such that $p\left(\frac{x-1}{x}\right) = \frac{x^n-1}{x^n}$ if and only if $n$ is an odd positive integer.
I imagine starting with our assumption on $n$ would be the easier of the two paths for this bidirectional proof, so I would like to start by assuming $n$ is odd and a positive integer. I started to play around with $p$ under this assumption about $n$: $$p\left(\frac{x-1}{x}\right) = \frac{x^n - 1}{x^n} = \frac{(x-1)(\sum_{i=0}^{n-1} x^i)}{x^n}$$ $$= \frac{x-1}{x}\frac{(\sum_{i=0}^{n-1} x^i)}{x^{n-1}}.$$ It seems like the $\frac{x-1}{x}$ seems like a convenient representation, but the second term seems difficult to morph into a function of $\frac{x-1}{x}$. Any recommendations, as well as any suggestions on proving the other direction?
only ifpart. $p(x) = -x^2 + 2 x$ works for $n = 2$. – dxiv Sep 22 '16 at 22:55