I was playing around earlier with the function $f(x)=x^3-x$. I decided that its inverse would be way too messy for me to try to calculate explicitly, but I became interested in the distinct real values $x,y$ for which $f(x)=f(y)$. I noticed that by assuming $y=kx$ I could simplify this a bit to go from the equation $$y^3-y=x^3-x$$ to the equation $$x^2=\frac{1}{k^2+k+1}$$ and so I could parametrize the solutions to $f(x)=f(y)$ as $$x=\pm\frac{1}{\sqrt{k^2+k+1}}, \space\space y=\pm\frac{k}{\sqrt{k^2+k+1}}$$
Now I am interested in similar equations that contain more than just two terms. For instance, consider $f(x)=x^8-x^3-x$. How can I parametrize the solutions to the equation $f(x)=f(y)$ for $x\ne y$? The same trick I used before doesn’t work so well, since there are now too many terms for me to do the same substitution and solve it nicely. Help?
For polynomials of order $n$, I guess the trick you used is this:
$$ f(x)=p(x)=\prod_1^n (x-p_i) $$
and for the requested equation $f(y)=f(x)$, this reduces to:
$$ \prod_{i=1}^n {(y-p_i) \over (x-p_i)}=1 $$
If we introduce $n-1$ auxiliar variables, $k_1,...,k_{n-2},k$:
$$ y=k_i(x-p_i)+p_i, i:1...n-1\\ y={1 \over \prod_{i=1}^{n-2} k_i k}(x-p_n)+p_n $$
we still have $n+1$ variables and $n$ equation, but now we can elliminate $y$, and the $k_i$ with the exception of $x$ and $k$.
And as you can realize, every successive replacement of $k_1...k_{n-2}$ leads to a polynomial equation, this final equation will be a order $n-1$ polynomial in $x$ with coefficients expressed in $k$:
$$ k\prod_{\begin{array}{c}i=1\\i\neq n-1\end{array}}^{n}(k(x-p_{n-1})+p_{n-1}-p_i)=\prod_{\begin{array}{c}i=1\\i\neq n-1\end{array}}^{n}(x-p_i) $$
A hard, but numerically feasible parametrization.