RMO Part -2 !

$\begin{cases} x^2z+y^2z+4xy = 40 & ...(1) \\ x^2+y^2 +xyz = 20 &...(2) \end{cases}$

$\begin{aligned} (2) \times z: \quad x^2z+y^2z +xyz^2 = 20z \quad \quad ...(3) \end{aligned}$

$\begin{aligned} (1) - (3): \quad 4xy - xyz^2 & = 40-20z \\ xy(4-z^2) & = 20(2-z) \\ xy(z+2)(z-2) & = 20(z-2) \\ xy(z+2) & = 20 \quad \quad \quad \quad ...(4) \end{aligned}$

$\begin{aligned} (2)=(4): \quad x^2+y^2+xyz & = xy(z+2) \\ x^2 -2xy +y^2 & = 0 \\ (x-y)^2 & = 0 \\ x & = y \end{aligned}$

$\begin{aligned} (4): \quad xy(z+2) & = x^2(z+2) = 20 \\ \Rightarrow z & = \frac{20}{x^2} - 2 \end{aligned}$

For integer $z$ the possible $x=\pm 1, \pm 2$ and we have:

$\begin{cases} x = \pm 1 & \Rightarrow z = \dfrac{20}{1} -2 = 18 \\ x = \pm 2 & \Rightarrow z = \dfrac{20}{4} -2 = 3\end{cases}$

Therefore, the sum of all $x$ , $y$ and $z$ is $-2-2+3-1-1+18+1+1+18+2+2+3 = \boxed{42}$

x²z+y²z+4xy=40 .......(1)

x²+y²+xyz=20 ....(2)

Multiply (2) by 2 and subtract it from (1)

x²z+y²z+4xy-2x²-2y²-2xyz=0

x²z+y²z-2xyz-2(x²+y²-2xy)=0

z(x-y)²-2(x-y)²=0

(x-y)²(z-2)=0

So, z=2 or x=y .

But if we substitute z=2 in (1) we see that it does not satisfy the given conditions. Thus neglect z=2

Now, x=y substitute this in (1) or (2) we get ,

(x,y,z)=(2,2,3);(-2,-2,3);(1,1,18);(-1,-1,18)

Thus ,sum of all values of x,y,z =2+2+3-2-2+3+1+1-1-1+18+18=42