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.