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[SGuerra]

Que questão bonita! Eu procurei outro tipo de solução, mas não encontrei! Eu a fiz como você fez! Parabéns pela escolha!!!! Brasil - setembro 2024. What a beautiful question! I looked for another type of solution, but I didn't find it! I made her like you did! Congratulations on your choice!!!! Brazil - September 2024.

[x-ajah]

(1±√3) -> real number
...
(1±√5i) -> complex number

[RyanLewis-Johnson-wqxs]

x^5+y^5=152 x+y=2 (x, y)=(1±Sqrt[3], 1±Sqrt[3]) final answer

[key_board_x]

(x + y)² = x² + y² + 2xy → given: x + y = 2
4 = x² + y² + 2xy

(x + y)³ = (x + y)².(x + y)
(x + y)³ = (x² + 2xy + y²).(x + y)
(x + y)³ = x³ + x²y + 2x²y + 2xy² + xy² + y³
(x + y)³ = x³ + y³ + 3x²y + 3xy²
x³ + y³ = (x + y)³ - 3x²y - 3xy²
x³ + y³ = (x + y)³ - 3xy.(x + y) → given: x + y = 2
x³ + y³ = 8 - 6xy

(x + y)⁵ = (x + y)².(x + y)².(x + y)
(x + y)⁵ = (x² + 2xy + y²).(x² + 2xy + y²).(x + y)
(x + y)⁵ = (x⁴ + 2x³y + x²y² + 2x³y + 4x²y² + 2xy³ + x²y² + 2xy³ + y⁴).(x + y)
(x + y)⁵ = (x⁴ + 6x²y² + 4x³y + 4xy³ + y⁴).(x + y)
(x + y)⁵ = x⁵ + x⁴y + 6x³y² + 6x²y³ + 4x⁴y + 4x³y² + 4x²y³ + 4xy⁴ + xy⁴ + y⁵
(x + y)⁵ = x⁵ + y⁵ + 5x⁴y + 5xy⁴ + 10x³y² + 10x²y³
x⁵ + y⁵ = (x + y)⁵ - 5x⁴y - 5xy⁴ - 10x³y² - 10x²y³
x⁵ + y⁵ = (x + y)⁵ - 5xy.(x³ + y³) - 10x²y².(x + y) → given: x + y = 2
x⁵ + y⁵ = 32 - 5xy.(x³ + y³) - 20x²y² → recall: x³ + y³ = 8 - 6xy
x⁵ + y⁵ = 32 - 5xy.(8 - 6xy) - 20x²y²
x⁵ + y⁵ = 32 - 40xy + 30x²y² - 20x²y²
x⁵ + y⁵ = 32 - 40xy + 10x²y² → given: x⁵ + y⁵ = 152
152 = 32 - 40xy + 10x²y²
10x²y² - 40xy - 120 = 0
x²y² - 4xy - 12 = 0 → let: z = xy
z² - 4z - 12 = 0
Δ = (- 4)² - (4 * - 12) = 16 + 48 = 64 = 8²
z = (4 ± 8)/2
z = 2 ± 4
z = 6 or z = - 2

First case: z = 6
xy = 6 ← this is the product P
x + y = 2 ← this is the sum S
So x & y are the solution of the following equation: a² - Sa + P = 0
a² - 2a + 6 = 0
Δ = (- 2)² - (4 * 6) = 4 - 24 = - 20 = 20i²
a = (2 ± i√20)/2
a = (2 ± 2i√5)/2
a = 1 ± i√5

→ x = 1 + i√5 → y = 1 - i√5

→ x = 1 - i√5 → y = 1 + i√5

Second case: z = - 2
xy = - 2 ← this is the product P
x + y = 2 ← this is the sum S
So x & y are the solution of the following equation: a² - Sa + P = 0
a² - 2a - 2 = 0
Δ = (- 2)² - (4 * - 2) = 4 + 8 = 12
a = (2 ± √12)/2
a = (2 ± 2√3)/2
a = 1 ± √3

→ x = 1 + √3 → y = 1 - √3
→ x = 1 - √3 → y = 1 + √3

[RealQinnMalloryu]

10^10^4^13. 2^5^2^5^4^13^1 1^1^1^1^2^2^1^1 1^2^1^1 2^1 (y➖ 2x+1) (1)+(1)=2 (y ➖ 1x+1).