Real Roots of Composite Polynomials: Solving P(Q(x))=0

[anemone]

Let $P(x)=x^2+\dfrac{x}{2}+b$ and $Q(x)=x^2+cx+d$ be two polynomials with real coefficients such that $P(x)Q(x)=Q(P(x))$ for all real $x$. Find all real roots of $P(Q(x))=0$.

 

[anemone]

Spoiler: Solution of other

Since $P(x)$ divides $Q(P(x))=(P(x))^2+cP(x)+d$, it follows that $P(x)$ divides $d$ and hence $d=0$. Thus $P(x)Q(x)=P(x)^2+cP(x)$ and hence $Q(x)=P(x)+c$. This gives $c=Q(x)-P(x)=\left(c-\dfrac{1}{2}\right)x-b$ and thus $c=\dfrac{1}{2}$ and $b=c=-\dfrac{1}{2}$. Consequently, $P(x)=x^2+\dfrac{x}{2}-\dfrac{1}{2}$ and $Q(x)=x^2+\dfrac{x}{2}$ and

$\begin{align*}P(Q(x))&=\left(x^2+\dfrac{x}{2}\right)^2+\dfrac{1}{2}\left(x^2+\dfrac{x}{2}\right)-\dfrac{1}{2}\\&=\dfrac{1}{4}(4x^4+4x^3+3x^2+x-2)\end{align*}$

Clearly $P(Q(-1))=0$ and any other rational root will be $\pm\dfrac{1}{4},\,\pm \dfrac{1}{2}$. Of these, $\dfrac{1}{2}$ is a root. The remaining factor is $h(x)=x^2+\dfrac{x}{2}+1$ and this has no real roots.