MHB How Do You Calculate the Sum of Minimum Values in Polynomials?

[anemone]

Here is this week's POTW:

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Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$, and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$.

Find the sum of the minimum values of $P(x)$ and $Q(x)$.

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Olinguito
2. castor28

Solution from castor28:

Spoiler

We start by observing that the graph of the quadratic function $f(x)=x^2-2ax+b$ is a parabola (concave upwards) with an vertical symmetry axis $x=a$. Therefore, if $f(x_1)=f(x_2)$ with $x_1\ne x_2$, $x_1$ and $x_2$ are symmetrical with respect to the axis, and $x_1+x_2=a$. This is also true in particular if $x_1$ and $x_2$ are the roots of $f(x)$.

Assume that the roots of $P(x)$ are $\alpha$ and $\beta$, and that the roots of $Q(x)$ are $\gamma$ and $\delta$. We may therefore write:
\begin{align*}
P(x) &= x^2 -(\alpha+\beta)x + p\\
Q(x) &= x^2 -(\gamma+\delta)x + q
\end{align*}

The roots of $P(Q(x))$ are the values of $x$ such that $Q(x)$ is equal to a root of $P(x)$; by the remark above, the pairs of values of $x$ giving the same value of $Q(x)$ are symmetrical with respect to the axis of the graph of $Q(x)$. This means that we have $Q(-21)=Q(-17)=\alpha$ and $Q(-23)=Q(-15)=\beta$ (if $\alpha<\beta$). We also have $\gamma+\delta=-38$, and we may now write $Q(x)=x^2+38x+q$.

A similar argument shows that $P(-57)=P(-51)=\gamma$ and $P(-59)=P(-49)=\delta$. We also have $\alpha+\beta=-108$ and $P(x)=x^2+108x+p$.

We now have;
\begin{align*}
P(-51)+P(-49) &= 2p-5798 = \gamma+\delta = -38\\
Q(-17)+Q(-15) &= 2q-702=\alpha+\beta=-108
\end{align*}
giving $p=2880$ and $q=297$. The polynomials are therefore:
\begin{align*}
P(x) &= x^2 + 108x + 2880\\
Q(x) &= x^2 + 28x+297
\end{align*}
The minimum values of $P(x)$ and $Q(x)$ are $P(-54)=-36$ and $Q(-19)=-64$, respectively. The sum of these values is $\mathbf{-100}$.