A quadrilateral has vertices $P(a,b)$, $Q(b,a)$, $R(-a, -b)$, and $S(-b, -a)$, where $a$ and $b$ are integers with $a>b>0$. The area of $PQRS$ is $16$. What is $a+b$?
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12  \qquad\textbf{(E)}\ 13$
Note that the slope of $PQ$ is $\frac{a-b}{b-a}=-1$ and the slope of $PS$ is $\frac{b+a}{a+b}=1$.  Hence, $PQ\perp PS$ and we can similarly prove that the other angles are right angles.  This means that $PQRS$ is a rectangle. By distance formula we have $(a-b)^2+(b-a)^2*2*(a+b)^2 = 256$. Simplifying we get $(a-b)(a+b) = 8$. Thus $a+b$ and $a-b$ have to be a factor of 8. The only way for them to be factors of $8$ and remain integers is if $a+b = 4$ and $a-b = 2$. So the answer is $\boxed{\textbf{(A)}\ 4}$
Solution by I_Dont_Do_Math