Use the polynomial identity (x2+y2)2=(x2−y2)2+(2xy)2

To generate a Pythagorean triple using the polynomial identity \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\), we start with the values \(x = 7\) and \(y = 4\).

1. Calculate \(x^2\) and \(y^2\):
\[
x^2 = 7^2 = 49
\]
\[
y^2 = 4^2 = 16
\]

2. Now, compute \(x^2 + y^2\):
\[
x^2 + y^2 = 49 + 16 = 65
\]

3. Calculate \((x^2 + y^2)^2\):
\[
(x^2 + y^2)^2 = 65^2 = 4225
\]

4. Now, compute \(x^2 - y^2\) and \(2xy\):
\[
x^2 - y^2 = 49 - 16 = 33
\]
\[
2xy = 2 \times 7 \times 4 = 56
\]

5. Then, compute \((x^2 - y^2)^2\) and \((2xy)^2\):
\[
(x^2 - y^2)^2 = 33^2 = 1089
\]
\[
(2xy)^2 = 56^2 = 3136
\]

6. Now we can verify the identity:
\[
(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2
\]
\[
4225 = 1089 + 3136
\]
\[
4225 = 4225 \quad (\text{True})
\]

7. The Pythagorean triple is given by \((x^2 - y^2, 2xy, x^2 + y^2)\):
\[
(33, 56, 65)
\]

Finally, we write the Pythagorean triple in ascending order:
\[
\{33, 56, 65\} \text{ is a Pythagorean triple.}
\]