Suppose the little red square is 1 by 1, and the adjoining chartreuse is x by x. Then the green is x+1 by x+1. And the aquamarine is x+2 by x+2 and the cyan is x+3 by x+3.

But the yellow? It's cyan+aquamarine–green–chartreuse, or (x+3) + (x+2) – (x+1) – x = 4, regardless of x!

The rest are easy:

Now we have a rectangle whose width is (2x+1)+(x+11)=3x +12 according to the top two squares, and (x+2)+(x+3)+( x+7)=3x+12, according to the bottom three. But the left three squares give the height as (2x+1)+(x+1)+(x+2)=4 x+4, while the right two give (x+11)+(x+7)=2x +18.

Is this bad? No, it's just what we need to determine x. If we have a rectangle, then the two sides are equal:

4x+4 = 2x+18

Get rid of the 4 on the left by subtracting it from both sides:

4x = 2x+14

And get rid of the 2x on the right by subtracting it from both sides:

2x = 14

Finally, dividing both sides by 2 gives the answer:

x = 7

And we can fill it all in:

Very nearly square. (Caution. It is so difficult to find a true "squared square" that it was believed impossible for centuries.)

Now you can go back and solve the original puzzle. Bill Gosper