In part D, we concluded that the Pythagorean triple (5,12,13) can’t be generated from the identity (x2 − 1)2 + (2x)2 = (x2 + 1)2, which has only one variable, because the length of the hypotenuse (13 units) and the length of the longer of the other two legs (12 units) are not 2 units apart. This illustrates the triangle from the original triple: Find a two-variable identity by incorporating a second variable, y, into the single-variable identity. Note that x > 1, x > y, and x and y are positive integers.