Question

Use the polynomial identity (x^(2)+y^(2))2=(x^(2)-y^(2))2+(2xy)2 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 use the polynomial identity \((x^2+y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\) to generate a Pythagorean triple, let's start by substituting \(x = 7\) and \(y = 3\).

First, calculate \(x^2\) and \(y^2\):
\[
x^2 = 7^2 = 49
\]
\[
y^2 = 3^2 = 9
\]

Now, we can calculate \(x^2 + y^2\):
\[
x^2 + y^2 = 49 + 9 = 58
\]

Next, we calculate \((x^2 + y^2)^2\):
\[
(x^2 + y^2)^2 = 58^2 = 3364
\]

Now, calculate \(x^2 - y^2\):
\[
x^2 - y^2 = 49 - 9 = 40
\]

Next, calculate \((x^2 - y^2)^2\):
\[
(x^2 - y^2)^2 = 40^2 = 1600
\]

Now, calculate \(2xy\):
\[
2xy = 2 \cdot 7 \cdot 3 = 42
\]

Then calculate \((2xy)^2\):
\[
(2xy)^2 = 42^2 = 1764
\]

To verify the identity, check that \((x^2 - y^2)^2 + (2xy)^2\) equals \((x^2 + y^2)^2\):
\[
(x^2 - y^2)^2 + (2xy)^2 = 1600 + 1764 = 3364
\]

This confirms the identity holds true.

Now we summarize the Pythagorean triple generated by the calculations:
- \(a = x^2 - y^2 = 40\)
- \(b = 2xy = 42\)
- \(c = x^2 + y^2 = 58\)

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

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