This puzzle is based on a proof of the Pythagorean theorem, which states that for any right triangle with a hypotenuse of length A  other two sides of lengths B and C then
        A*A = B*B + C*C.

The idea of the proof is to calculate the area of the large square in two ways; as the product of its sides and as the sum of the areas of the pieces.  

In this puzzle we have a square that has been divided into 4 congruent right triangles and a smaller square.  If we call the lengths of the triangle sides A, B and C, such that
   • A is the length of the hypotenuse
   • B is the length of the longer remaining side
   • C is the length of the shorter remaining side
then the area of each right triangle is B*C/2 and the area of the inner square is (B - C)*(B - C).  Thus the area calculated in these two ways must be the same so we have the following equation:
        A*A = 4*(B*C/2) + (B - C)*(B - C)
        A*A = 2*B*C + B*B -2*B*C + C*C
        A*A = B*B + C*C
which is what Pythagoras's theorem states.