Solving a Cubic System for Real Solutions

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[XJWill]

It is interesting to extend this system of equations to
(x + y)^n = z
(x + z)^n = y
(y + z)^n = x
For integers n > 0
It turns out that for all such n, the only real-valued solutions occur when x = y = z, and so the
real solutions for x (and thus y and z) are
x = 0
x = 1 / 2^(n / (n - 1)) and when n is odd there is the additional solution :
x = -1 / 2^(n / (n - 1))

It was not easy to demonstrate that the only real-valued solutions satisfy x = y = z for all integer n > 0.
It is easy enough for n = 1, 2, 3, and in fact it is not difficult to demonstrate it for all even n > 1 since
an even exponent means x >= 0 and y >= 0 and z >= 0, and so the factorization is only equal to zero
for x = y = z.
For n = 3 :
a^2 + b^2 + a*b = -1 and this can be factored as
(1/2)*( a^2 + b^2 + (a + b)^2 ) = -1
Similarly, for n = 5 :
a^4 + b^4 + a*b*(a^2 + b^2) + (a*b)^2 = -1
= (1/4)*(2*a^4 + 2*b^4 + (a^2 - b^2)^2 + (a + b)^4)
Also, for n = 7 :
a^6 + b^6 + a*b*(a^4 + b^4) + (a*b)^2 * (a^2 + b^2) + (a*b)^3 = -1
= (1/4)*( ( a^4 + b^4 + (a^2 - b^2)^2 ) * (a^2 + b^2) +
( a^4 + b^4 + (a^2 + b^2)^2 ) * (a + b)^2 )
But at this point the factorizations became to difficult. But I found that a^n - b^n can be factored
as the roots of unity, and taking them in complex-conjugate pairs, it factors as
a^2 + b^2 - 2*a*b*cos(2*pi * k/n)
and since cos() is between -1 and 1, each term can be factored by completing the square and
and therefore each term is non-negative.

[elkr]

x1=0; y1=0; z1=0 is the first evident solution that EVERYONE needs to recognize.

[SergeySvotin]

So, russian equations won in this battle, since you couldn't solve it -_-