Does the following result: $$[ \ p(p(\dots p(t))) = t \ ] \Rightarrow [ \ p(t)=t \vee p(p(t))=t \ ]$$
(where $p$ is of course a polynomial) have a name? I need a (preferably elementary) proof of this. The LHS is the polynomial composed with itself $n \in \mathbb{N}$ times.
Hints: What are the degrees of $p(p(t))$ and $p^{\circ n}(t)$ (the $n$-fold self-composition of $p$)? How do they relate to the degree of $p(t)$? Can there be a constant term?
NB. As Andres Caicedo remarks, there can be a constant term; I duly apologise for the oversight that gave rise to my misleading hint.
Try $p(t)=\frac32t^2-\frac{19}2t+17$ and $t=2$.
If $t$ is supposed to be a concrete value rather than the indeterminate in which $p$ is a polynomial (this would be the most natural reading of the question as it is phrase) then the statement is false. For instance take $p(x)=x^2$ and $t$ a fifth root of unity like $\exp(2\pi\mathbf i/5)$, then $p(p(p(t)))=t^{16}=t$ but $p(t)=t^2\neq t$ and $p(p(t))=t^4\neq t$.
If $t$ is a indeterminate and $p(t)\in\mathbf C[t]$, then the statement is still false, taking $p(t)=\alpha t$ with $\alpha=\exp(2\pi\mathbf i/3)$ so that $p(t)\neq t$, $p(p(t))=\alpha^2t\neq t$ but $p(p(p(t)))=\alpha^3t=t$.
