China | A Nice Algebra Problem 👇

[@paulortega5317]

How about
let x + y = a, a ≠ 0 ❶
x² - y² = (x - y)(x + y) = (x - y)*a = 64
x - y = 64/a ❷
Using ❶ & ❷: x = (a² + 64)/(2a) and y = (a² - 64)/(2a)
xy = (a⁴ - 64²)/(4a²) = 8
a⁴ - 32a² - 64² = 0
(a² - 16)² = 64² + 16² = 16²*(16 + 1) = 16²*17
a² = 16 + 16*sqrt(17)
a = ± 4*(sqrt(1 + sqrt(17))

[@key_board_x]

x² - y² = 64 → given: xy = 8 → y = 8/x
x² - (8/x)² = 64
x² - (64/x²) = 64
x⁴ - 64 = 64x²
x⁴ - 64x² - 64 = 0 → let: X = x² ← where X ≥ 0
X² - 64X - 64 = 0
Δ = (- 64)² - (4 * - 64) = (64 * 64) + (4 * 64) = 68 * 64 = 17 * 4 * 64 = 17 * 2² * 8² = 17 * 16²
X = (64 ± 16√17)/2
X = 32 ± 8√17 → we keep only the positive value (recall; X ≥ 0)
X = 32 + 8√17
X + 8 = 40 + 8√17
x + y = x + (8/x)
x + y = (x² + 8)/x
(x + y)² = (x² + 8)²/x² → recall: x² = X
(x + y)² = (X + 8)²/X
(x + y)² = (40 + 8√17)² / (32 + 8√17)
(x + y)² = [8 * (5 + √17)]² / [8 * (4 + √17)]
(x + y)² = [8 * 8 * (5 + √17)²] / [8 * (4 + √17)]
(x + y)² = [8 * (5 + √17)²] / (4 + √17)
(x + y)² = [8 * (25 + 10√17 + 17)] / (4 + √17)
(x + y)² = [8 * (42 + 10√17)] / (4 + √17)
(x + y)² = [16 * (21 + 5√17)] / (4 + √17)
(x + y)² = 16 * [(21 + 5√17) * (4 - √17)] / [(4 + √17) * (4 - √17)]
(x + y)² = 16 * [84 - 21√17 + 20√17 - 85] / [16 - 17]
(x + y)² = 16 * [- 1 - √17] / [- 1]
(x + y)² = 16 * (1 + √17)
x + y = ± 4√(1 + √17)

[@superacademy247]

Thanks for detailed explanation