To generate a Pythagorean triple using the polynomial identity \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\), we start with the values \(x = 7\) and \(y = 4\).
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Calculate \(x^2\) and \(y^2\): \[ x^2 = 7^2 = 49 \] \[ y^2 = 4^2 = 16 \]
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Now, compute \(x^2 + y^2\): \[ x^2 + y^2 = 49 + 16 = 65 \]
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Calculate \((x^2 + y^2)^2\): \[ (x^2 + y^2)^2 = 65^2 = 4225 \]
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Now, compute \(x^2 - y^2\) and \(2xy\): \[ x^2 - y^2 = 49 - 16 = 33 \] \[ 2xy = 2 \times 7 \times 4 = 56 \]
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Then, compute \((x^2 - y^2)^2\) and \((2xy)^2\): \[ (x^2 - y^2)^2 = 33^2 = 1089 \] \[ (2xy)^2 = 56^2 = 3136 \]
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Now we can verify the identity: \[ (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 \] \[ 4225 = 1089 + 3136 \] \[ 4225 = 4225 \quad (\text{True}) \]
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The Pythagorean triple is given by \((x^2 - y^2, 2xy, x^2 + y^2)\): \[ (33, 56, 65) \]
Finally, we write the Pythagorean triple in ascending order: \[ {33, 56, 65} \text{ is a Pythagorean triple.} \]