Question

The identity (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 can be used to generate Pythagorean triples. What Pythagorean triple could be generated using x = 8 and y = 3?

Answer 1

The triple are: 48, 55 and 73

Explanation:

Given

`(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2`

`x = 8; y=3`

Required

The Pythagorean triple

We have:

`(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2`

Substitute:
`x = 8; y=3`

`(8^2 + 3^2)^2 = (8^2 - 3^2)^2 + (2*8*3)^2`

`(64 + 9)^2 = (64 - 9)^2 + (48)^2`

`(73)^2 = (55)^2 + (48)^2`

`5329 = 5329`

Hence, the triple are: 48, 55 and 73