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The Pythagorean triple [tex]\((5,12,13)\)[/tex] can't be generated from the identity [tex]\(\left(x^2-1\right)^2+(2 x)^2=\left(x^2+1\right)^2\)[/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.

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.

Select the correct answer:
A. [tex]\(\left(x^2-y^2\right)^2+(2 x y)^2=\left(x^2-y^2\right)^2\)[/tex]

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

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

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



Answer :

GinnyAnswer
To solve the given problem, we need to find a two-variable identity that incorporates a second variable [tex]\( y \)[/tex] into the single-variable identity and satisfies the conditions provided. Given the options, we will carefully evaluate and choose the correct identity step-by-step. Let's consider each of the following identities:

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

Let's analyze the third option in detail to see if it matches our requirements:

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

Let's test this identity using a pair of values for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that are common in Pythagorean triples. Let's use [tex]\( x = 3 \)[/tex] and [tex]\( y = 2 \)[/tex]:

1. Calculate the left side:
[tex]\[ \left(x^2 - y^2\right)^2 + (2xy)^2 \][/tex]
Substituting [tex]\( x = 3 \)[/tex] and [tex]\( y = 2 \)[/tex]:
[tex]\[ \left(3^2 - 2^2\right)^2 + (2 \cdot 3 \cdot 2)^2 = (9 - 4)^2 + (12)^2 \][/tex]
Simplify the expression:
[tex]\[ (5)^2 + (12)^2 = 25 + 144 = 169 \][/tex]

2. Calculate the right side:
[tex]\[ \left(x^2 + y^2\right)^2 \][/tex]
Substituting [tex]\( x = 3 \)[/tex] and [tex]\( y = 2 \)[/tex]:
[tex]\[ \left(3^2 + 2^2\right)^2 = (9 + 4)^2 \][/tex]
Simplify the expression:
[tex]\[ (13)^2 = 169 \][/tex]

Both the left side and the right side equal 169, which confirms that the identity holds true. Thus, the correct two-variable identity that can generate Pythagorean triples is:

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

Therefore, the correct answer is:

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

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