NZMO Round Two

Solution:
First we exclude $-1$ and $2$:
$x = 2$ implies $y = 2$ implies $z = 2$ implies $x = 2$
$x = -1$ implies $y = -1$ implies $z = -1$ implies $x = -1$
In both these cases we have $x + y + z$ being an integer. So henceforth we assume none of $x,y,z$ are $2$ nor $-1$. Now let $x,y,z$ be the roots of the following cubic equation.
$$(\lambda - x)(\lambda - y)(\lambda - z) = \lambda^{3} - A\lambda^{2} + B\lambda - C.$$
Applying Viete's formula to this cubic gives us $A = x + y + z$, $B = xy + yz + zx$ and $C = xyz$. This means that $x^{2} + y^{2} + z^{2} = (x + y + z)^{2} - 2(xy + yz + zx) = A^{2} - 2B$. Now sum the three given equations ($x^{2} = y + 2$ and $y^{2} = z + 2$ and $z^{2} = x + 2$) to get
$$A^{2} - 2B = x^{2} + y^{2} + z^{2} = (y + 2) + (z + 2) + (x + 2) = A + 6.$$
$$A^{2} - A - 2B = 6. \quad (1)$$
Next rearrange the equations to be $x^{2} - 1 = y + 1$ and $y^{2} - 1 = z + 1$ and $z^{2} - 1 = x + 1$. These can then be multiplied to get
$$(x^{2} - 1)(y^{2} - 1)(z^{2} - 1) = (y + 1)(z + 1)(x + 1)$$
$$(x - 1)(x + 1)(y - 1)(y + 1)(z - 1)(z + 1) = (y + 1)(z + 1)(x + 1)$$
$$(x - 1)(y - 1)(z - 1) = 1$$
$$xyz - (xy + yz + zx) + (x + y + z) - 1 = 1$$
$$B - A + 2 = C \quad (2)$$
In the above algebraic manipulation, we are allowed to cancel the $(x + 1)(y + 1)(z + 1)$ factor because none of $x,y,z$ are equal to $-1$. Finally rearrange the equations to be $x^{2} - 4 = y - 2$ and $y^{2} - 4 = z - 2$ and $z^{2} - 4 = x - 2$. These can then be multiplied to get
$$(x^{2} - 4)(y^{2} - 4)(z^{2} - 4) = (y - 2)(z - 2)(x - 2)$$
$$(x - 2)(x + 2)(y - 2)(y + 2)(z - 2)(z + 2) = (y - 2)(z - 2)(x - 2)$$
$$(x + 2)(y + 2)(z + 2) = 1$$
$$xyz + 2(xy + yz + zx) + 4(x + y + z) + 8 = 1$$
$$C + 2B + 4A = -7$$
$$C = -4A - 2B - 7 \quad (3)$$
In the above algebraic manipulation, we are allowed to cancel the $(x - 2)(y - 2)(z - 2)$ factor because none of $x,y,z$ are equal to $2$. Combining equations (2) and (3) gives us $B - A + 2 = C = -4A - 2B - 7$ which rearranges to give us $B = -A - 3$. Substituting this into 1 gives us.
$$A^{2} - A - 2B = 6$$
$$A^{2} - A - 2(-A - 3) = 6$$
$$A(A + 1) = 0.$$
Therefore $A = 0$ or $A = -1$ both of which are integers. Since $A = x + y + z$ we are done.

Consider the polynomial $P$ defined by
$$P(\lambda) = \lambda^{8} - 8\lambda^{6} + 20\lambda^{4} - 16\lambda^{2} - \lambda + 2$$
$$\qquad = (\lambda + 1)(\lambda - 2)(\lambda^{3} - 3\lambda + 1)(\lambda^{3} + \lambda^{2} - 2\lambda - 1).$$
If we substitute $z = y^{2} - 2$ into $z^{2} = x + 2$ gives us $(y^{2} - 2)^{2} = x + 2$. Then substitute $y = x^{2} - 2$ to get $\left((x^{2} - 2)^{2} - 2\right)^{2} = x + 2$. Expanding gives
$$x^{8} - 8x^{6} + 20x^{4} - 16x^{2} - x + 2 = 0.$$
Therefore $x$ is a root of the polynomial $P$. By symmetry we must have all of $x,y,z$ being roots of $P$. Now we consider cases:
Case 1: at least one of $x,y,z$ is equal to $-1$. Wlog assume $x = -1$. Using $y = x^{2} - 2$ we get $y = -1$. Then using $z = y^{2} - 2$ we get $z = -1$. In this case we get $(x,y,z) = (-1, -1, -1)$ which has sum $-3$ which is an integer.
Case 2: at least one of $x,y,z$ is equal to $2$. Wlog assume $x = 2$. Using $y = x^{2} - 2$ we get $y = 2$. Then using $z = y^{2} - 2$ we get $z = 2$. In this case we get $(x,y,z) = (2,2,2)$ which has sum $6$ which is an integer.
Case 3: at least one of $x,y,z$ is a root of $(\lambda^{3} - 3\lambda + 1)$. Wlog assume $x^{3} - 3x + 1 = 0$. Note that $z = y^{2} - 2 = (x^{2} - 2)^{2} - 2 = x^{4} - 4x + 2$. Now consider the sum of $x$ and $y = x^{2} - 2$ and $z = x^{4} - 4x^{2} + 2$,
$$x + y + z = x + (x^{2} - 2) + (x^{4} - 4x^{2} + 2) = x^{4} - 3x^{2} + x = (x^{3} - 3x + 1)x.$$
But since $x^{3} - 3x + 1 = 0$ this means $x + y + z = 0$ in this case.
Case 4: some two of $x,y,z$ are the same. Wlog assume $x = y$ therefore $x = y = x^{2} - 2$. Hence
$$0 = x^{2} - x - 2 = (x - 2)(x + 1).$$
and so $x = -1$ or $x = 2$, and this was covered in cases 1 and 2.
Case 5: Since $x,y,z$ are all roots of $P$, the only remaining possibility is that $x,y$ and $z$ are distinct roots of $\lambda^{3} + \lambda^{2} - 2\lambda - 1$. By Viete's formula this means that the sum of the roots is $-1$ in this case.
In all cases we conclude that $x + y + z$ is an integer.