There is no such square whose area is equal to the difference between the areas of its circumscribed and inscribed circles. Am I right?
Imagine a square with length $L$.
This square has an area of $L^2$.
Its inscribed circle has radius $\frac{L}{2}$.
Hence, it has an area of $π \frac{L^2}{4}$.
By Pythagoras' theorem, the circumscribed circle has a radius of $\sqrt 2L$.
So it has an area of $2πL^2$.
A.T.Q. :
\begin{align}
\text{Area of circumscribed circle - Area of inscribed circle} &= \text{Area of a square} \\
2πL^2 - π\frac{L^2}{4} & \neq L^2
\end{align}
Does this complete my proof?