RMO 2008

From Equation 2:

$(x+y)(y+z) + (y+z)(z+x) + (z+x)(x+y) = 1$

=> $(2-z)2-x) +(2-x)(2-y) + (2-y)(2-z) = 1$

With further solving we get,

$xy + yz + zx = -3....(i)$

Now, squaring Equation 1 : $x^2 + y^2 +z^2 + 2(xy+yz+zx) = 4$

Therefore, $x^2 + y^2 +z^2 = 4+6=10$

Now in equation (3)

$x^2(2-x) + y^2(2-y) + z^2(2-z) = -6$

Further solving we get,

$x^3+y^3+z^3 = 26$

Subtracting 3xyz from both sides,

$(x+y+z)(x^2 + y^2 +z^2) - (xy + yz + zx) = 26 - 3xyz$

=> $2(10+3) = 26 - 3xyz$

=> $xyz = 0$

$Case 1:$ when all 3 are 0

$x+y+z$ is not equal to 2 from equation 1.

$Case 2:$ when any 2 are 0

$x^2 + y^2 +z^2 = 4$ which is not equal to 10

From equation 1,2

$Case 3:$ when any one of them is 0

The following combination of $(x,y,z)$ arise : $(0,3,-1), (0,-1,3), (3,0,-1), (-1,0,3), (-1,3,0), (3,-1,0)$ or simply $3!$ possible permutations.

Hence there are $\boxed6$ integer solutions. :)

even SHORTER(1st part copied from samarpit's ans)

x+y+z=2..............let this be A

From Equation 2: (x+y)(y+z) + (y+z)(z+x) + (z+x)(x+y) = 1

=> (2-z)2-x) +(2-x)(2-y) + (2-y)(2-z) = 1

With further solving we get

xy + yz + zx = -3...........let this be B

Now, squaring Equation 1 : x^2 + y^2 +z^2 + 2(xy+yz+zx) = 4

Therefore, x^2 + y^2 +z^2 = 4+6=10

Now in equation (3)

x^2(2-x) + y^2(2-y) + z^2(2-z) = -6

Further solving we get,

x^3+y^3+z^3 = 26

Subtracting 3xyz from both sides,

(x+y+z){x^2 + y^2 +z^2 - (xy + yz + zx)} = 26 - 3xyz

=> 2(10+3) = 26 - 3xyz

=> xyz = 0 ........C

now look at A,B,C

asuume x,y,z are roots of a cubic eqn.then from viete's formulas cubic is

u^3-2u^2-3u=0 u(u-3)(u+1)=0

three distinct roots.from theory of permutation ANS is 3P3 = 3! = 6 ANS

(x + y)(y + z) + (y + z)(z + x) + (z + x)(x + y) = 1

(x + y)(2 - x) + (2 - x)(2 - y) + (2 - y)(x+y) = 1

(x + y)² + (x - 2)² + (y - 2)² = 14

1² + 2² + 3² = 14

x = 0, y = -1, z = 3

x = 0, y = 3, z = -1

x = -1, y = 0, z = 3

x = -1, y = 3, z = 0

x = 3, y = -1, z = 0

x = 3, y = 0, z = -1