(In reply to
Second look by Larry)
Lets take another look at your equation (1 - (x+y))^5 = 1 - (x^5+y^5)
Now instead of reducing everything to a polynomial, lets just apply binominal expansion to the left side taking (x+y) as a unit:
1 - 5*(x+y) + 10*(x+y)^2 - 10*(x+y)^3 + 5*(x+y)^4 - (x+y)^5 = 1 - x^5 - y^5
Move the (x+y)^5 to the other side and do a bit of cleanup:
5*[-(x+y) + 2*(x+y)^2 - 2*(x+y)^3 + (x+y)^4] = (x+y)^5 - x^5 - y^5
Now lets expand the right side:
5*[-(x+y) + 2*(x+y)^2 - 2*(x+y)^3 + (x+y)^4] = x^5 + 5x^4*y + 10*x^3*y^2 + 10*x^2*y^3 + 5*x*y^4 + y^5 - x^5 - y^5
Some more cleanup:
5*[-(x+y) + 2*(x+y)^2 - 2*(x+y)^3 + (x+y)^4] = 5*[x^4*y + 2*x^3*y^2 + 2*x^2*y^3 + x*y^4]
Now pull 5*(x+y) out of each side:
5*(x+y) * [-1 + 2*(x+y) - 2*(x+y)^2 + (x+y)^3] = 5*(x+y) * [x^3*y + x^2*y^2 + x*y^3]
5 can be canceled on each side and x+y=0 forms one case of solutions, so lets pull that out and move everything to one side:
x^3*y + x^2*y^2 + x*y^3 - (x+y)^3 + 2*(x+y)^2 - 2*(x+y) + 1 = 0
Now this is a lot easier for some math program (such as wolfram online) to factor into:
(x-1)*(y-1)*(x^2+xy+y^2-x-y+1) = 0
So x=1 and y=1 are two other solution case, but from x+y+z=1 then x=1 implies y+z=0 and y=1 implies x+z=0.
x+y=0 from earlier is the z=1 a solution case.
So solutions thus far are {1,b,-b}, {c,1,-c}, {d,-d,1} for parameters b,c,d.
But there is still x^2+xy+y^2-x-y+1 = 0 to consider.
A bit of playing with square polynomials eventually yields:
(3/4)*(x+y-2/3)^2 + (1/8)*(x-y+2/3)^2 + (1/8)*(-x+y+2/3)^2 = -5/9.
The left side consists solely of nonnegative multiples of squares while the right side is negative, so no real solutions exist in this case.
re: Second look - third look
