Let $P(x)$ and $Q(x)$ be polynomials with positive leading coefficients. If $P(P(x)) = P(x)^5+x^{15}+Q(x)$, minimum value of the degree of $Q(x)$

algebra-precalculuscontest-mathpolynomials

My steps:

Since for two polynomial expressions to be equal, their degrees have to be equal, I decided to compare the degrees of the RHS and LHS. I figured out that the degree of $$P(P(x))$$ is $${deg(P(x))}^{deg(P(x))}$$ And the degree of the LHS can only be 15, $$5 \cdot deg(P(x))$$ or $$deg(Q(x))$$. However, I wasn't able to produce a solution from here. Any help would be appreciated.

Let $$P(x)= ax^n+...$$ and $$Q(x) =bx^m+...$$ where $$a,b>0$$. Then $$a(ax^n+...)^n+... = a^5x^{5n} +...x^{15} +bx^m+...$$ so $$n^2 = \max\{5n,15,m\}\geq 5n\implies n\geq 5$$. If $$n\geq 6$$ then $$n^2 > 5n>15$$ so $$n^2 =\max \{5n,15,m\} =m$$ and thus $$m\geq 36$$.

So let $$n=5$$, then $$P(x)= ax^5+cx^4+dx^3+...$$ and $$m\leq 25$$.

$$Q(x) = a(ax^5+...)^5 -(a^5x^{25}+...)-x^{15}$$

• If $$a\neq 1$$ then $$m= 25$$.

• If $$a=1$$ then $$Q(x) = cP(x)^4+...-x^{15}$$ If $$c\neq 0$$ then degree of RHS is 20 so we put $$c=0$$, then the degree of RHS is $$15$$ or less. So let $$c=0$$ Now we have $$Q(x) = dP(x)^3+...-x^{15}$$ If $$d=0$$ then RHS has degree 15 and negative leading coefficient, so $$d\neq 0$$. If we put $$d=1$$ we get $$Q(x) = P(x)^3+...-x^{15}$$ and now RHS has degree $$<15$$.

So if we set $$P(x) = x^5+x^3$$ we get $$Q(x) =3x^{13} + 3x^{11}+x^9$$, so $$m_{\min} =13$$.