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Select the correct answer.

In part D, we concluded that the Pythagorean triple [tex]\((5,12,13)\)[/tex] can't be generated from the identity [tex]\(\left(x^2-1\right)^2+(2x)^2=\left(x^2+1\right)^2\)[/tex], which has only one variable, because the length of the hypotenuse (13 units) and the length of the longer of the other two legs (12 units) are not 2 units apart. This illustrates the triangle from the original triple:

Find a two-variable identity by incorporating a second variable, [tex]\(y\)[/tex], into the single-variable identity. Note that [tex]\(x \ \textgreater \ 1\)[/tex], [tex]\(x \ \textgreater \ y\)[/tex], and [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are positive integers.

A. [tex]\(\left(x^2-y^2\right)^2-(2xy)^2=\left(x^2+y^2\right)^2\)[/tex]

B. [tex]\(\left(x^2+y^2\right)^2+(2xy)^2=\left(x^2-y^2\right)^2\)[/tex]



Answer :

GinnyAnswer
To generate and verify a Pythagorean triple using a two-variable identity, we can use the identity:
[tex]\[ (x^2 - y^2)^2 + (2xy)^2 = (x^2 + y^2)^2 \][/tex]
Let's break this down step-by-step and solve it with specific values for [tex]\(x\)[/tex] and [tex]\(y\)[/tex].

### Step-by-Step Solution

1. Choose Values for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
We need to choose [tex]\(x\)[/tex] and [tex]\(y\)[/tex] such that [tex]\(x > y\)[/tex] and both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are positive integers. Let's choose values that are easy to work with and will yield a Pythagorean triple. We select:
[tex]\[ x = 3, \quad y = 2 \][/tex]

2. Calculate the Sides of the Triangle
Using the given values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex], we calculate the three sides of the triangle.
[tex]\[ a = x^2 - y^2 = 3^2 - 2^2 = 9 - 4 = 5 \][/tex]
[tex]\[ b = 2xy = 2 \cdot 3 \cdot 2 = 12 \][/tex]
[tex]\[ c = x^2 + y^2 = 3^2 + 2^2 = 9 + 4 = 13 \][/tex]

3. Calculate the Squares of the Sides
Next, we calculate the squares of these sides to verify the identity.
[tex]\[ a^2 = 5^2 = 25 \][/tex]
[tex]\[ b^2 = 12^2 = 144 \][/tex]
[tex]\[ c^2 = 13^2 = 169 \][/tex]

4. Verify the Pythagorean Identity
Finally, we need to verify that the identity holds true:
[tex]\[ (x^2 - y^2)^2 + (2xy)^2 = (x^2 + y^2)^2 \][/tex]
Plugging in our values, we get:
[tex]\[ 25 + 144 = 169 \][/tex]
Since both sides of the equation are equal, the identity is verified and the given values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] generate a valid Pythagorean triple [tex]\((5, 12, 13)\)[/tex].

### Conclusion
The correct two-variable identity that shows the Pythagorean triple [tex]\((5, 12, 13)\)[/tex] is:
[tex]\[ (x^2 - y^2)^2 + (2xy)^2 = (x^2 + y^2)^2 \][/tex]
With [tex]\(x = 3\)[/tex] and [tex]\(y = 2\)[/tex], we verified that the triple [tex]\((5, 12, 13)\)[/tex] satisfies the equation:
[tex]\[ (3^2 - 2^2)^2 + (2 \cdot 3 \cdot 2)^2 = (3^2 + 2^2)^2 \][/tex]
[tex]\[ (9 - 4)^2 + (12)^2 = (13)^2 \][/tex]
[tex]\[ 5^2 + 12^2 = 13^2 \][/tex]
Thus, the values [tex]\( (5, 12, 13), 25, 144\)[/tex], and [tex]\(169\)[/tex] are correct and the identity is verified.

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