Question

the pythagorean triple (5,12,13) cant be generated from the identity ((x^2 - 1)^2 + (2x)^2 = (x^2 + 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. 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. select the correct answer. (\bigcirc (x^2 + y^2)^2 + (2xy)^2=(x^2 - y^2)^2) (\bigcirc (x^2 - y^2)^2 + (2xy)^2=(x^2 - y^2)^2) (\bigcirc (x^2 - y^2)^2 + (2xy)^2=(x^2 + y^2)^2) (\bigcirc (x^2 - y^2)^2-(2xy)^2=(x^2 + y^2)^2)

Explanation:

Step1: Expand left-hand side (LHS)

First, expand $(x^2 - y^2)^2 + (2xy)^2$:
$$(x^2 - y^2)^2 = x^4 - 2x^2y^2 + y^4$$
$$(2xy)^2 = 4x^2y^2$$
Adding them together:
$$x^4 - 2x^2y^2 + y^4 + 4x^2y^2 = x^4 + 2x^2y^2 + y^4$$

Step2: Expand right-hand side (RHS)

Expand $(x^2 + y^2)^2$:
$$(x^2 + y^2)^2 = x^4 + 2x^2y^2 + y^4$$

Step3: Verify equality

Compare LHS and RHS:
$$x^4 + 2x^2y^2 + y^4 = x^4 + 2x^2y^2 + y^4$$
LHS = RHS, so the identity holds. Also, test with the given triple $(5,12,13)$: let $x=3$, $y=2$
$$(3^2 - 2^2)^2 + (2*3*2)^2 = (9-4)^2 + (12)^2 = 25 + 144 = 169$$
$$(3^2 + 2^2)^2 = (9+4)^2 = 169$$
The identity generates the triple correctly.

Answer:

$\boldsymbol{(x^2 - y^2)^2 + (2xy)^2 = (x^2 + y^2)^2}$