Concept

Difference of two squares

In mathematics, the difference of two squares is a squared (multiplied by itself) number subtracted from another squared number. Every difference of squares may be factored according to the identity in elementary algebra. The proof of the factorization identity is straightforward. Starting from the left-hand side, apply the distributive law to get By the commutative law, the middle two terms cancel: leaving The resulting identity is one of the most commonly used in mathematics. Among many uses, it gives a simple proof of the AM–GM inequality in two variables. The proof holds in any commutative ring. Conversely, if this identity holds in a ring R for all pairs of elements a and b, then R is commutative. To see this, apply the distributive law to the right-hand side of the equation and get For this to be equal to , we must have for all pairs a, b, so R is commutative. The difference of two squares can also be illustrated geometrically as the difference of two square areas in a plane. In the diagram, the shaded part represents the difference between the areas of the two squares, i.e. . The area of the shaded part can be found by adding the areas of the two rectangles; , which can be factorized to . Therefore, . Another geometric proof proceeds as follows: We start with the figure shown in the first diagram below, a large square with a smaller square removed from it. The side of the entire square is a, and the side of the small removed square is b. The area of the shaded region is . A cut is made, splitting the region into two rectangular pieces, as shown in the second diagram. The larger piece, at the top, has width a and height a-b. The smaller piece, at the bottom, has width a-b and height b. Now the smaller piece can be detached, rotated, and placed to the right of the larger piece. In this new arrangement, shown in the last diagram below, the two pieces together form a rectangle, whose width is and whose height is . This rectangle's area is . Since this rectangle came from rearranging the original figure, it must have the same area as the original figure.

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About this result

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