INMO - 2000

The answer given is wrong to the problem, actually it is $\infty$ answers possible.

$x^3+y^3+z^2=1$

$x^3+y^3 + \underbrace{ (1-x-y)^2}_\mathrm{= 1-(x+y)^2} =1$

$(x+y)(x^2-xy+y^2) + 1+ (x+y)^2 -2(x+y) =1$

$(x+y)(x^2-xy+y^2) + (x+y) (x+y-2)=0$

$(x+y)(x^2+y^2-xy+x+y-2)=0$

Thus if we have $x=k$ and $y=-k$ then we get $z=1$ and here we get INFINITE solutions $(x,y,z)=(k,-k,1)$

But because we can't enter the answer as infinite, most probably Ameya wanted us to find solutions for $z\neq 1$

Then we are left with the only case that

$x^2+y^2-xy+x+y-2=0$

To get this in solvable integer solutions we make it

$4x^2+4y^2 - 4xy+4x+4y=8$

$4x^2+y^2 - 4xy+1 +4x-2y +3y^2+6y 3=8+1+3$

$(2x-y+1)^2 + 3(y+1)^2=12$

And this has integer solutions at $(2x-y+1)^2=0$ and $(y+1)^2 = 1$

or $(2x-y+1)^2=0$ and $(y+1)^2=4$

This gives 6 pairs of $(x,y)$ namely $(0,1) , (1,0) , (0,-2) , (-2,0), (-3,-2) , (-2,-3)$

And for each of these pairs, there will be some value of $z$ , we needn't find it as we eliminated $z$ to get these solutions. Hence there are $\boxed{6}$ triples.