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?

1 answer

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.