8 at Thu, 20 Feb 2025 04:43:27 UTC No. 16592153

>>16592016
eight

Anonymous at Thu, 20 Feb 2025 05:00:53 UTC No. 16592163

>>16592016
8

Anonymous at Thu, 20 Feb 2025 05:24:13 UTC No. 16592179

4(x+y+z) = y(x^2-yz) + z(y^2-xz) + x(z^2-xy) = 0

24 = (x+y+z)^2 + 2(x^2-yz)+2(y^2-zx)+2(z^2-xy) = 3(x^2+y^2+z^2)

x^2+y^2+z^2 = 8

Anonymous at Thu, 20 Feb 2025 18:25:20 UTC No. 16593315

There is a parabola that describes the solutions to this equation. If the values of x and y are {x,y} ={2,-2), then the answer is
(2)^2+(-2)^2+(4+(2)(-2)) = 4 + 4 + (4-4) = 4 + 4 + 0 = 8

Anonymous at Fri, 21 Feb 2025 13:29:57 UTC No. 16594756

The OP's system is:
x^2 – y*z = 4
y^2 – z*x = 4
z^2 – x*y = 4

There are 6 integer solutions:
x = 0, y = –2, z = 2
x = –2, y = 0, z = 2
x = –2, y = 2, z = 0
and their negatives

In each case:
x^2 + y^2 + z^2 = 0 + 4 + 4 = 8

https://www.wolframalpha.com/input?i=%7Bx%5E2+-+y+z+%3D+4%2C+y%5E2+-+z+x+%3D+4%2C+z%5E2+-+x+y+%3D+4%7D

There are 2 real non-integer solutions:
x = –4/sqrt(3), y = 2/sqrt(3), z = 2/sqrt(3)
and its negative

In both cases:
x^2 + y^2 + z^2 = 16/3 + 4/3 + 4/3 = 24/3 = 8

Anonymous at Fri, 21 Feb 2025 16:21:29 UTC No. 16595008

a plot of the sum of the OP's system

https://www.wolframalpha.com/input?i=x%5E2+-+y+z+%2B+y%5E2+-+z+x+%2B+z%5E2+-+x+y+%3D+3%C3%974

Anonymous at Fri, 21 Feb 2025 16:30:36 UTC No. 16595020

3*(x^2 + y^2 + z^2) = (x + y + z)^2 + 24