Aiscrim wrote: For two quadratic trinomials $P(x)$ and $Q(x)$ there is a linear function $l(x)$ such that $P(x)=Q(l(x))$ for all real $x$. How many such linear functions $l(x)$ can exist? (A. Golovanov) I suppose that "quadratic trinomial" suppose that cofficient of $x^2$ is nonzero. If so : $P(x)=Q(l_1(x))$ and $P(x)=Q(l_2(x))$ implies $Q(l_1(x))=Q(l_2(x))$ and so : either $l_1(x)=l_2(x)$ either $l_1(x)+l_2(x)=c$ constant. Hence at most two such functions, and trivially at least two exist. Hence the answer : $\boxed{2}$

A more complicated, but relevant approach. Clearly a linear function $\ell(x) = ax+b$ (with $a\neq 0$) admit a (also linear) inverse $\ell^{-1}(x) = \dfrac {1} {a} x - \dfrac {b} {a}$. Then, if $P(x)=Q(\ell_1(x)) = Q(\ell_2(x))$, we also have $Q(x) = Q(\lambda(x))$, where $\lambda = \ell_2\circ \ell_1^{-1}$ is linear, thus $\lambda(x) = \alpha x + \beta$ (with $\alpha \neq 0$). By identification of the dominant coefficients follows $|\alpha|=1$. (Then, for $\deg Q = n\geq 2$, and working with complex coefficients, we may have as many as $n$ such functions $\lambda$, for example $\lambda_k(x) = \textrm{e}^{2k\pi\textrm{i}/n} x$ for $Q(x) = x^n$). In the sequel, assume real coefficients. With $\deg Q = n\geq 2$, then for odd $n$ we have $\alpha=1$, leading to $Q(x) = Q(x+\beta)$, thus by simple iteration $Q(x) = Q(x+k\beta)$ for all natural $k$; this clearly forces $\beta = 0$, for which $\lambda = \operatorname{id}$. For even $n$ we may also have $\alpha=-1$, leading to $Q(x) = Q(-x+\beta)$, thus $\lambda \circ \lambda = \operatorname{id}$. But we cannot have $Q(x) = Q(-x+\gamma)$ for $\gamma \neq \beta$, since then $Q(-x+\beta) = Q(-x+\gamma)$, so $Q(x) = Q(x+(\beta -\gamma))$, with the obvious contradiction pointed at in the above. It follows in this case may exist at most two such functions $\lambda$, with an immediate example given by $Q$ made only of monomials of even degree, when $\lambda = \pm \operatorname{id}$ work. In the case at hand, $\deg Q = 2$, which allows the simpler and more direct solution based on the symmetries of the graph of a quadratic trinomial.