Question

Use the polynomial identity (22 + 42)2 = (x2 - y?) + (2xy)? to generate a Pythagorean
triple when x equals 7 and y equals 3. Which of the following is one of the values of the Pythagorean triple?

Answer 1

To generate a Pythagorean triple using the given polynomial identity \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\), we will substitute \(x = 7\) and \(y = 3\) into the identity.

1. Calculate \(x^2\) and \(y^2\):
- \(x^2 = 7^2 = 49\)
- \(y^2 = 3^2 = 9\)

2. Calculate \(x^2 + y^2\):
\[
x^2 + y^2 = 49 + 9 = 58
\]

3. Calculate \((x^2 + y^2)^2\):
\[
(x^2 + y^2)^2 = 58^2 = 3364
\]

4. Calculate \(x^2 - y^2\):
\[
x^2 - y^2 = 49 - 9 = 40
\]

5. Calculate \(2xy\):
\[
2xy = 2 \times 7 \times 3 = 42
\]

6. Now, we will return to the expression from the polynomial identity:
\[
(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2
\]

7. Evaluate \((x^2 - y^2)^2\) and \((2xy)^2\):
- \((x^2 - y^2)^2 = 40^2 = 1600\)
- \((2xy)^2 = 42^2 = 1764\)

8. Now we can conclude:
\[
(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2
\]
\[
3364 = 1600 + 1764
\]

According to the identity, the Pythagorean triple generated is:
- \(x^2 - y^2 = 40\)
- \(2xy = 42\)
- \(x^2 + y^2 = 58\)

Thus, the Pythagorean triple is \( (40, 42, 58) \).

Among these values, one of the values of the Pythagorean triple is **40**, **42**, or **58**.