The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$
$\textbf{(A) } 20 \qquad  \textbf{(B) } 21 \qquad  \textbf{(C) } 22 \qquad  \textbf{(D) } 23 \qquad  \textbf{(E) } 24$
We solve each equation separately:


$z^2=4+4\sqrt{15}i$
Let $z=a+bi$ for some real numbers $a$ and $b.$ 
Substituting and expanding, we get 
\begin{align*} (a+bi)^2&=4+4\sqrt{15}i \\ \left(a^2-b^2\right)+2abi&=4+4\sqrt{15}i. \end{align*}
Equating the real parts and the imaginary parts, respectively, we get
\begin{align*} a^2-b^2&=4, &&(1) \\ ab&=2\sqrt{15}. &&(2) \end{align*}
We rearrange $(1)$ and square $(2):$
\begin{align*} a^2&=b^2+4, \hspace{4mm} &&(1\star) \\ a^2b^2&=60. &&(2\star) \end{align*}
Substituting $(1\star)$ into $(2\star),$ we obtain $\left(b^2+4\right)b^2=60.$ Since $b^2\geq0$ for all real numbers $b,$ either inspection or factoring gives $b^2=6.$ Substituting this into either $(1\star)$ or $(2\star)$ produces $a^2=10.$ Since $ab>0$ from $(2),$ we have $(a,b)=\left(\sqrt{10},\sqrt{6}\right),\left(-\sqrt{10},-\sqrt{6}\right).$ 
The solutions to $z^2=4+4\sqrt{15}i$ are $\boldsymbol{z=\sqrt{10}+\sqrt{6}i,-\sqrt{10}-\sqrt{6}i}.$
$z^2=2+2\sqrt{3}i$
By the same process, we have $(a,b)=\left(\sqrt3,1\right),\left(-\sqrt3,-1\right).$ 
The solutions to $z^2=2+2\sqrt{3}i$ are $\boldsymbol{z=\sqrt3+i,-\sqrt3-i}.$

Note that the problem is equivalent to finding the area of a parallelogram with consecutive vertices $(x_1,y_1)=\left(\sqrt{10}, \sqrt{6}\right),(x_2,y_2)=\left(\sqrt{3},1\right),(x_3,y_3)=\left(-\sqrt{10},-\sqrt{6}\right),$ and $(x_4,y_4)=\left(-\sqrt{3}, -1\right)$ in the coordinate plane. By the Shoelace Theorem, the area we seek is \[\frac{1}{2} \left|(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)\right| = 6\sqrt2-2\sqrt{10},\] so the answer is $6+2+2+10=\boxed{\textbf{(A) } 20}.$
~Rejas (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)