RMO 2008

Algebra Level 4

{x+y+z=2(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)=1x2(y+z)+y2(z+x)+z2(x+y)=6 \begin{cases}{x+y+z = 2} \\ {(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y) = 1} \\ {x^2(y+z)+y^2(z+x)+z^2(x+y) = -6}\end{cases}

How many integer solutions exist for that satisfy the equations above?

Details and Assumptions:

  • Permutations are allowed, which means the values for x,y,zx, y,z are interchangeable & are considered as different solutions.
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