smc

$P(x)=(x-a{)}^2-b,Q(x)=(x-c{)}^2-d$. 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 $\boxed{-100}$.