Problem 1300

7. Suppose positive integers $m, n$ satisfy
$f(m, n)=2018$.
For the integer solutions $(x, y, z)$ of the system of equations (1):
if $x=y=z$, then it is called the 1st type solution;
if exactly two of $x, y, z$ are equal, then it is called the 2nd type solution;
if $x, y, z$ are all distinct, then it is called the 3rd type solution.
Let the number of the $i$-th type solution be $a_{i}$.
Since the integer solutions $(x, y, z)$ of the system of equations (1) remain solutions after permuting the components, we have
$3\left|a_{2}, 6\right| a_{3}$.
Thus, $a_{1}=a_{1}+a_{2}+a_{3}=2018 \equiv 2(\bmod 3)$.
This indicates that the number of integer roots of the equation
$x^{3}-3 x-m=0$

satisfying $|x| \leqslant n$ is congruent to 2 modulo 3. Hence, it has at least two integer roots, denoted as $\alpha, \beta$.

By the properties of cubic equations, the third root of equation (2) is a real number, denoted as $\gamma$. According to Vieta's formulas, we have
${ }^{\alpha+\beta+\gamma=0,}$
$\left\{\begin{array}{l}\alpha \beta+\beta \gamma+\gamma \alpha=-3, \\ \alpha \beta \gamma=m .\end{array}\right.$
Thus, $\gamma=-\alpha-\beta \in \mathbf{Z}$.
Assume $\alpha \leqslant \beta \leqslant \gamma$.
From $\alpha \beta \gamma=m>0$, we know $\alpha \leqslant \beta<0<\gamma$.
Then $-3=\alpha \beta+(\alpha+\beta)(-\alpha-\beta)$
$=-\alpha^{2}-\alpha \beta-\beta^{2} \leqslant-3$.
This implies $\alpha=\beta=-1$.
Thus, $\gamma=2, m=2$.
Next, consider the system of equations (1) under $m=2$.
Notice that, $(2 n+1)^{3} \geqslant 2018$ (there are only $(2 n+1)^{3}$

integer triples $(x, y, z)$ satisfying $|x| 、|y| 、|z|$ are all no greater than $n$).
Thus, $n \geqslant 6$.
Next, calculate $a_{1} 、 a_{2} 、 a_{3}$.
(1) Since $-1,2$ are all integer solutions of $x^{3}-3 x-2=0$, the triples
$(-1,-1,-1) 、(2,2,2)$ are all the 1st type solutions of the system of equations (1), i.e., $a_{1}=2$.
(2) For the 2nd type solutions, assume $y=z \neq x$, substituting into

the system of equations (1) gives
$x y^{2}=x+2 y+2$
$\Rightarrow(y+1)(x(y-1)-2)=0$.
When $y=-1$, $(x,-1,-1)(-n \leqslant x \leqslant n$,
$x \neq-1)$ are $2 n$ 2nd type solutions;
When $y \neq-1$, from $x(y-1)=2$, we have
$\left\{\begin{array}{l}x= \pm 2, \pm 1, \\ y-1= \pm 1, \pm 2\end{array}\right.$
Noting that $x \neq y$, we have
$(x, y)=(-2,0),(1,3)$

satisfying the conditions.
Considering the permutations of $x, y, z$, the number of 2nd type solutions
$a_{2}=3(2 n+2)$.
(3) For the 3rd type solutions $(x, y, z)$, consider the number of the corresponding 3-element sets $\{x, y, z\}$.

First, $-1 \notin\{x, y, z\}$ (otherwise, assume $z=-1$, substituting into the system of equations (1) and rearranging gives
$(x+1)(y+1)=0 \Rightarrow-1 \in\{x, y\}$,

contradicting that $x, y, z$ are all distinct).
If $0 \in\{x, y, z\}$, assume $z=0$, substituting into the system of equations (1) gives $x+y=-2$. Then
$\{x, y, z\}=\{k,-2-k, 0\}(k=1,2, \cdots, n-2)$

are $n-2$ 3-element sets.
If $0 \notin\{x, y, z\}$, assume $y 、 z$ have the same sign, and
$|y| \geqslant|z|$.
When $|z|=1$, we must have $z=1$, substituting into the system of equations (1) and rearranging gives
$(x-1)(y-1)=4$.
Noting that $x \neq y, x y \neq 0$, we have
$(x, y)=(5,2),(2,5)$

satisfying the conditions. Thus,
$\{x, y, z\}=\{1,2,5\}$.
When $|z| \geqslant 2$, from $x=\frac{y+z+2}{y z-1}$, we have
$|x| \leqslant \frac{|y|+|z|+2}{|y||z|-1} \leqslant \frac{2|y|+2}{2|y|-1}$
$=1+\frac{3}{2|y|-1} \leqslant 2$.
The equality cannot hold simultaneously (otherwise, we must have $|x|=|y|=|z|=2$, and $(x, y, z)$ would not be a 3rd type solution), thus, $|x|<2, x=1$.
This implies, $\{x, y, z\}=\{1,2,5\}$.
Therefore, the 3-element sets $\{x, y, z\}$ corresponding to the 3rd type solutions of the system of equations (1) are $n-1$.
Thus, $a_{3}=6(n-1)$.
Combining (1), (2), and (3), we have for any $n \geqslant 6$,
$a_{1}+a_{2}+a_{3}$
$=2+3(2 n+2)+6(n-1)$
$=12 n+2$.
Since $a_{1}+a_{2}+a_{3}=f(m, n)=2018$, we have
$n=168$.

In summary, when and only when $m=2, n=168$,