To generate a Pythagorean triple using the polynomial identity \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\), we will first calculate the required components using \(x = 11\) and \(y = 4\).
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Calculate \(x^2\) and \(y^2\): \[ x^2 = 11^2 = 121 \] \[ y^2 = 4^2 = 16 \]
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Now compute \(x^2 + y^2\): \[ x^2 + y^2 = 121 + 16 = 137 \]
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Compute \(x^2 - y^2\): \[ x^2 - y^2 = 121 - 16 = 105 \]
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Compute \(2xy\): \[ 2xy = 2 \cdot 11 \cdot 4 = 88 \]
Now we need to find the squares of these values to represent the Pythagorean triple:
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The left-hand side of the identity is: \[ (x^2 + y^2)^2 = 137^2 = 18769 \]
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The right-hand side, using our previously computed values, is: \[ (x^2 - y^2)^2 + (2xy)^2 = 105^2 + 88^2 \]
- Calculate \(105^2\): \[ 105^2 = 11025 \]
- Calculate \(88^2\): \[ 88^2 = 7744 \]
- Thus, the right-hand side is: \[ 11025 + 7744 = 18769 \]
Since both sides match, the values of the Pythagorean triple generated are:
- \( (x^2 - y^2) = 105 \)
- \( 2xy = 88 \)
- \( (x^2 + y^2) = 137 \)
Among the given options, the value of the Pythagorean triple is 137.
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