I've carried out the following exercise taken from the $2024$ Putnam competition but my attempt of solution is different from the one provided in the website, so I'd like to know if it is correct.
A2. For which real polynomial $p$ is there a real polynomial $q$ such that $$ p\bigl(p(x)\bigr) - x = \bigl(p(x) - x\bigr)^2 q(x) $$
I use the fact that a given polynomial with real coefficients of degree $n$ and having complex variable has exactly $n$ roots considering their multiplicity. If this is true, then if $p(z)-z$ has degree $n$ it must have exactly $n$ zeroes considering their multiplicity. Suppose $\tilde{z}$ is such that $p(\tilde z)-\tilde z=0$ (unless $p(z)-z$ is constant it is always possible to find $z$ with this property), then I can consider the following limit, in which I can use de l'Hopital rule: $$\lim_{z\to \tilde z}\frac {p(p(z))-z}{(p(z)-z)^2}=\lim_{z\to \tilde z}\frac{p'(p(z))p'(z)-1}{2(p(z)-z)(p'(z)-1)}$$ then since I want that $q(z)$ is a polynomial, it must be continuous and then the numerator must be zero in the limit that is $p'(\tilde z)=- 1$. $p(z)$ must have the property that for all $z$ such that $p(z)-z=0$ we have $p'(z)+1=0.$ If $p(z)$ has degree $n\geq 2$ then the polynomial $p(z)-z$ has $n$ roots while the polynomial $p'(z)+1$ has $n-1,$ then for the condition to be satisfied at least a root of $p(z)-z$ must be double, then for it $p'(z)-1=0$, but at the same moment $p'(z)+1=0,$ this is impossible. If $n=1,$ $p(z)-z$, not being constant has one zero, while $p'(z)+1$ no zero but equal to 0 if $p(z)=-z+C.$ I notice that if $n(z)$ and $d(z)$ have real coefficents and $\frac{n(z)}{d(z)}$ has infinite limit in some point $z,$ $\frac{n(x)}{d(x)}$ can't be a polynomial.
It remain the case in which $p(x)-x$ is constant but in this is case it is easy to check that is satisfied the property required for $q(x)=\frac1c$ or whatever $q(x)$ if $c=0.$ Is this argument correct or there are some subleties I can't see?
