{x or y = 0}
{x = -y}
The added constraint of x>y limits the above to the negative half of the y-axis, the positive half of the x-axis and the 4th quadrant part of the line y = -x.

Plus there is also a complex solution:
    {y/x = (-1 ± √(3)i)/2}
    equivalent to y/x = {i^(4/3) and i^(8/3)}
I am not sure how the added constraint of x>y affects the complex solutions.  The magnitude (modulus) of y/x = (-1 ± √(3)i)/2 is exactly 1, but the complex number is not precisely equal to 1.  I guess it depends on the definition of "greater than" in reference to complex numbers.

Algebra:
x^5+y^5=(x+y)^5
(x+y)^5 = x^5 + y^5 + 5xy^4 + 10x^2y^3 + 10x^3y^2 + 5x^4y
5xy^4 + 10x^2y^3 + 10x^3y^2 + 5x^4y = 0
5xy((x^3 + y^3) + 2xy(x+y)) = 0
Acknowledge {x=0 or y=0} are solutions,

but if x,y ≠ zero:
(x^3 + y^3) + 2xy(x+y) = 0
note:  (x^3 + y^3) = (x+y)^3 - 3xy(x+y)
(x+y)^3 - 3xy(x+y) + 2xy(x+y) = 0
(x+y)^3 - xy(x+y) = 0
(x+y)((x+y)^2 - xy) = 0
Acknowledge x+y=0 or {x = -y} is a solution,

but if x ≠ -y then:
(x+y)^2 - xy = 0
x^2 + xy + y^2 = 0

y = (-x ± √(-3x^2))/2
or
{y/x = (-1 ± √(3)i)/2}

Edited on March 21, 2023, 11:12 am

Posted by Larry on 2023-03-21 11:09:26