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 $-49$. What is the sum of the minimum values of $P(x)$ and $Q(x)$? 
$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0$
$P(x) = (x - a)^2 - b, Q(x) = (x - c)^2 - d$. Notice that $P(x)$ has roots $a\pm \sqrt {b}$, so that the roots of $P(Q(x))$ are the roots of $Q(x) = a + \sqrt {b}, a - \sqrt {b}$. For each individual equation, the sum of the roots will be $2c$ (symmetry or Vieta's). Thus, we have $4c = - 23 - 21 - 17 - 15$, or $c = - 19$. Doing something similar for $Q(P(x))$ gives us $a = - 54$.
We now have $P(x) = (x + 54)^2 - b, Q(x) = (x + 19)^2 - d$. Since $Q$ is monic, the roots of $Q(x) = a + \sqrt {b}$ are "farther" from the axis of symmetry than the roots of $Q(x) = a - \sqrt {b}$. Thus, we have $Q( - 23) = - 54 + \sqrt {b}, Q( -21) =- 54 - \sqrt {b}$, or $16 - d = - 54 + \sqrt {b}, 4 - d = - 54 - \sqrt {b}$. Adding these gives us $20 - 2d = - 108$, or $d = 64$. Plugging this into $16 - d = - 54 + \sqrt {b}$, we get $b = 36$.
The minimum value of $P(x)$ is $- b$, and the minimum value of $Q(x)$ is $- d$. Thus, our answer is $- (b + d) = - 100$, or answer $\boxed{\textbf{(A)}}$.