If I understand this correctly, you are saying that the rectangle with lenght x and width 1 can be partitioned into two unit squares and a "scaled down" rectangle similar to the orginal one. That is, the "scaled down" rectangle has sides of length and for some . The we must have , the width of the original rectangle and [tex]\alpha= x- 2[tex]. Putting that value of into the first equation, . That also gives your ratio, Completing the square by adding 1 to both sides, , so that . Since x, as the lenght of a side of a rectangle, must be positive, , your first solution, is the only solution.
"Taking one square away" we still have but now or which reduces to the same thing.
Your "taking one square away" seems to be starting with a completly different problem in which the original rectangle is made up of only one square together with the "scaled down" rectangle.