HMMT-Feb 2010 alg, Problem 10

Let $p(x)$ and $q(x)$ be two cubic polynomials such that $p(0)=-24, q(0)=30$, and \[ p(q(x))=q(p(x)) \] for all real numbers $x$. Find the ordered pair $(p(3), q(6))$.

Solution: Note that the polynomials $f(x)=a x^{3}$ and $g(x)=-a x^{3}$ commute under composition. Let $h(x)=x+b$ be a linear polynomial, and note that its inverse $h^{-1}(x)=x-b$ is also a linear polynomial. The composite polynomials $h^{-1} f h$ and $h^{-1} g h$ commute, since function composition is associative, and these polynomials are also cubic. We solve for the $a$ and $b$ such that $\left(h^{-1} f h\right)(0)=-24$ and $\left(h^{-1} g h\right)(0)=30$. We must have: \[ a b^{3}-b=-24,-a b^{3}-b=30 \Rightarrow a=1, b=-3 \] These values of $a$ and $b$ yield the polynomials $p(x)=(x-3)^{3}+3$ and $q(x)=-(x-3)^{3}+3$. The polynomials take on the values $p(3)=3$ and $q(6)=-24$. Remark: The pair of polynomials found in the solution is not unique. There is, in fact, an entire family of commuting cubic polynomials with $p(0)=-24$ and $q(0)=30$. They are of the form \[ p(x)=t x(x-3)(x-6)-24, q(x)=-t x(x-3)(x-6)+30 \] where $t$ is any real number. However, the values of $p(3)$ and $q(6)$ are the same for all polynomials in this family. In fact, if we give the initial conditions $p(0)=k_{1}$ and $q(0)=k_{2}$, then we get a general solution of \[ \begin{gathered} p(x)=t\left(x^{3}-\frac{3}{2}\left(k_{1}+k_{2}\right) x^{2}+\frac{1}{2}\left(k_{1}+k_{2}\right)^{2} x\right)+\frac{k_{2}-k_{1}}{k_{2}+k_{1}} x+k_{1} \\ q(x)=-t\left(x^{3}-\frac{3}{2}\left(k_{1}+k_{2}\right) x^{2}+\frac{1}{2}\left(k_{1}+k_{2}\right)^{2} x\right)-\frac{k_{2}-k_{1}}{k_{2}+k_{1}} x+k_{2} \end{gathered} \] $\fbox{(3,-24)}$.

Answer: (3,-24)