2. (2006 Southeast Mathematical Olympiad) (1) Find the number of positive integer solutions $(m, n, r)$ to the indeterminate equation $m n+n r+m r=2(m+n+r)$;
(2) For a given integer $k>1$, prove: the indeterminate equation $m n+n r+m r=k(m+n+r)$ has at least $3 k+1$ sets of positive integer solutions $(m, n, r)$.
2. (1) If $m, n, r \geqslant 2$, from $m n \geqslant 2 m, n r \geqslant 2 n, m r \geqslant 2 r$, we get
$$
m n+n r+m r \geqslant 2(m+n+r).
$$

Therefore, the above inequalities all hold with equality, hence $m=n=r=2$.
If $1 \in\{m, n, r\}$, without loss of generality, let $m=1$, then $n r+n+r=2(1+n+r)$, thus $(n-1)(r-1)=3$, so $\{n-1, r-1\}=\{1,3\}$, hence $\{n, r\}=\{2,4\},\{m, n, r\}=\{1,2,4\}$, and there are 3! = 6 such solutions.
Therefore, the indeterminate equation $m n+n r+m r=2(m+n+r)$ has a total of 7 sets of positive integer solutions.
(2) Transform $m n+n r+m r=k(m+n+r)$ into
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$$
[n-(k-m)][r-(k-m)]=k^{2}-k m+m^{2}.
$$
$n=k-m+1, r=k^{2}-k m+m^{2}+k-m$ satisfies the above equation, and when $m=1,2, \cdots,\left[\frac{k}{2}\right]$, $0<m<n<r$.
When $k$ is even,
$$
\{m, n, r\}=\left\{l, k-l+1, k^{2}-k l+l^{2}+k-l\right\},
$$

where $l=1,2, \cdots, \frac{k}{2}$ gives $3 k$ sets of positive integer solutions to the indeterminate equation.
When $k$ is odd,
$$
\{m, n, r\}=\left\{l, k-l+1, k^{2}-k l+l^{2}+k-l\right\},
$$

where $l=1,2, \cdots, \frac{k-1}{2}$ gives $3(k-1)$ sets of positive integer solutions to the indeterminate equation, and when two of $m, n, r$ are $\frac{k+1}{2}$, the other is
$$
k^{2}-k \frac{k+1}{2}+\left(\frac{k+1}{2}\right)^{2}+k-\frac{k+1}{2}=\frac{(k+1)(3 k-1)}{4}
$$

which gives 3 sets of positive integer solutions to the indeterminate equation.
And $m=n=r=k$ is also a positive integer solution to the indeterminate equation.
Therefore, the indeterminate equation $m n+n r+m r=k(m+n+r)$ has at least $3 k+1$ sets of positive integer solutions.