A Quartic System

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[dqjfrue]

I think the 2nd method could've been even easier if we'd noticed here (6:21) not only (x+y)⁴=... but (x-y)⁴=... also.
We'd come to (x+y)⁴=81 and (x-y)⁴=1, then x+y=±3 and x-y=±1 and get the answer without solving quadratic equations.

[bobkurland]

if you switch to variables u = (x^2 +y^2) and v = xy, it becomes easier to solve. you can get your equations for s and p without solving for x+y.

[HenkVanLeeuwen-io]

The 2nd equation can be solved for y^2 in terms of x using the quadratic formula.

[williamspostoronnim]

Блестяще решенная очень интересная задача!

[VictorPensioner]

(1) xy (x² + y²) = 10
(2) (x² + y²) + 4x²y² = 41
From (1)
(3) (x² + y²) = 10 / xy (x ≠ 0, y≠ 0)
Substitute (3) in (2)
[10/xy]² + 4x²y² = 41
Let
t = x²y²
Then
100/t + 4t = 41
or
4t² - 41t + 100

t₁, ₂ = [41 ± √(41² - 1600)] / 8 = (41 ± √81) / 8 = (41 ± 9)/ 8
Therefore
x²y² = 4
or
x²y² = 50/8 = 25/4

Case 1.
x²y² = 4

From (2)
(x² + y²)² = 41 - 4x²y² = 41 - 16 = 25
For equations
(x² + y²) = 5
x²y² = 4
apply Viete formula to x² and y²:
z² -5z + 4 = 0
z₁, ₂ = [5 ± √(5² - 16)] / 2 = (5 ± √9) / 2 = (5 ± 3)/ 2
Therefore
x² = 4, y² = 1
or
x² = 1, y² = 4
From (1) x and y has the same sign
(x, y) --> (2, 1), (-2, -1), (1, 2), (-1, -2)

Case 2.
x²y² = 25/4

From (2)
(x² + y²)² = 41 - 4x²y² = 41 - 4*(25/4) = 16
For equations
(x² + y²) = 4
x²y² = 25/4
apply Viete formula to x² and y²:
z² -4z + 25/4 = 0
or
4z² -16z + 25 = 0

b² - 4ac= 16² -4*4*25 < 0 no real solutions
All solutions are
(x, y) --> (2, 1), (-2, -1), (1, 2), (-1, -2)

[SidneiMV]

x³y + xy³ = 10
x⁴ + 6x²y² + y⁴ = 41

(x +y)⁴ = x⁴ + y⁴ + 4(x³y + xy³) + 6x²y²
(x + y)⁴ = 41 + 4(10) = 81 = 3⁴

[(x + y)² + 3²][(x + y)² - 3²] = 0

x³y + xy³ = 10
xy(x² + y²) = 10
xy[(x + y)² - 2xy] = 10

(x + y)² = 9
xy(9 - 2xy) = 10
-2(xy)² + 9xy = 10
2(xy)² - 9xy + 10 = 0
xy = (9 ± 1)/4
xy = 5/2 ∨ xy = 2
(x + y)² = 9 => x + y = ±3

x + y = 3
x + y = 2
t² - 3t + 2 = 0
t = (3 ± 1)/2 => t = 2 ∨ t = 1
*(x, y) = {(2, 1), (1, 2)}*

x + y = -3
x + y = 2
t² + 3t + 2 = 0
t = (-3 ± 1)/2 => t = -1 ∨ t = -2
*(x, y) = {(-2, -1), (-1, -2)}*

x + y = 3
x + y = 5/2
t² - 3t + 5/2 = 0
t = (3 ± i)/2
*(x, y) = {[(3 + i)/2, (3 - i)/2], [(3 - i)/2, (3 + i)/2]}*

x + y = -3
x + y = 5/2
t² + 3t + 5/2 = 0
t = (-3 ± i)/2
*(x, y) = {[(-3 + i)/2, (-3 - i)/2], [(-3 - i)/2, (-3 + i)/2]}*

GENERAL SOLUTION

*(x, y) = {(2iⁿ, iⁿ), (iⁿ, 2iⁿ)}*
*(x, y) = {[(3 + i)iⁿ/2, (3 - i)iⁿ/2], [(3 - i)iⁿ/2, (3 + i)iⁿ/2]}*
n ∈ ℤ