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 $y = 3$?
a. 8, 15, 17
b. 5329, 3025, 2304
c. 48, 55, 73
d. 36, 48, 60

Explanation:

Step1: Calculate \(x^{2}+y^{2}\)

$x = 8,y = 3,x^{2}+y^{2}=8^{2}+3^{2}=64 + 9=73$

Step2: Calculate \(x^{2}-y^{2}\)

$x^{2}-y^{2}=8^{2}-3^{2}=64 - 9 = 55$

Step3: Calculate \(2xy\)

$2xy=2\times8\times3 = 48$

Answer:

48, 55, 73