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Solve the system: x⁴ + y⁴ = 82 and x*y = 3 (over the reals).

1. (x, y) = (3, 1), (1, 3), (-3, -1), (-1, -3)
2. (x, y) = (3, 1) only
3. (x, y) = (2, 1.5), (1.5, 2)
4. no real solutions

Correct answer: (x, y) = (3, 1), (1, 3), (-3, -1), (-1, -3)

Solution

x⁴ + y⁴ = (x² + y²)² - 2(xy)² = 82, with (xy)² = 9, so (x² + y²)² = 82 + 18 = 100, giving x² + y² = 10. Now (x + y)² = x² + y² + 2xy = 10 + 6 = 16 -> x + y = +/-4; (x - y)² = x² + y² - 2xy = 10 - 6 = 4 -> x - y = +/-2. Solving combinations with xy = 3: (x, y) = (3, 1), (1, 3), (-3, -1), (-1, -3).

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