(1) Find the number of positive integer solutions $(m, n, r)$ for the indeterminate equation
$$
m n + n r + m r = 2(m + n + r)
$$

(2) For a given integer $k (k > 1)$, prove that the indeterminate equation $m n + n r + m r = k(m + n + r)$ has at least $3k + 1$ sets of positive integer solutions $(m, n, r)$.
(Yuan Hanhui)
(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 $m n+n r+m r \geqslant 2(m+n+r)$.
Thus, all the inequalities above hold with equality.
Therefore, $m=n=r=2$.
If $1 \in\{m, n, r\}$, without loss of generality, let $m=1$, then $n r+n+r=2(1+n+r)$.
Thus, $(n-1)(r-1)=3$.
Therefore, $\{n-1, r-1\}=\{1,3\}$.
Hence, $\{n, r\}=\{2,4\},\{m, n, r\}=\{1,2,4\}$.
There are $3!=6$ such solutions.
Therefore, the indeterminate equation has 7 sets of positive integer solutions.
(2) Transform $m n+n r+m r=k(m+n+r)$ into $[n-(k-m)][r-(k-m)]=k^{2}-k m+m^{2}$.
It is easy to see that $n=k-m+1, r=k^{2}-k m+m^{2}+k-m$ 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 $l=1,2, \cdots, \frac{k}{2}$, provides $3 k$ 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 $l=1,2, \cdots, \frac{k-1}{2}$, provides $3(k-1)$ sets of positive integer solutions to the indeterminate equation, with two of $m, n, r$ being $\frac{k+1}{2}$ and the other being
$$
\begin{array}{l}
k^{2}-k \cdot \frac{k+1}{2}+\left(\frac{k+1}{2}\right)^{2}+k-\frac{k+1}{2} \\
=\frac{(k+1)(3 k-1)}{4}
\end{array}
$$

This situation provides 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 has at least $3 k+1$ sets of positive integer solutions.