One Well That Ends Well

If you add all the yellow areas in the original problem together, you will have every orange part of the image above three times, every yellow part two times, and the blue part not at all. So if you take the side of the square as $x$ , you can describe the coverage as

$3x^2-x-x-1=65+93+81+101=340$

This equation has two solutions: $x=11, x=-\frac{31}{3}$

Only the first is applicable, and the green area is $11^2-1=120$

A solution with $x=11$ actually exists and it is pictured below.

If we take the four brown rectangles to overlap within the original square, then the red rectangles will be overlapped twice while the inside deep blue square will be overlapped four times. If the side length is $x$ , the area above will equal to $x^2 + 2x +1$ .

Then we will obtain the equation:

$(x^2-63) + (x^2-93) + (x^2-81)+ (x^2-101) = x^2 + 2x +1$

$3x^2 - 2x - 341 = 0$

$(3x+31)(x-11) = 0$

Thus, $x=11$ , and the green area = $11^2 - 1 = \boxed{120}$ .