Find all the side lengths of triangles that satisfy the following conditions:
(1) The side lengths of the triangle are integers;
(2) The inradius of the triangle is 2.
Let the three sides of the triangle be $a, b, c$, the semi-perimeter be $p$, the inradius be $r$, and $S$ represent the area of the triangle. By Heron's formula $S=\sqrt{p(p-a)(p-b)(p-c)}$ and $S=pr=2p$, we have
$$
2_{p}=\sqrt{p(p-a)(p-b)(p-c)}.
$$

Therefore, $4 p=(p-a)(p-b)(p-c)$, which means
$$
4 \cdot \frac{a+b+c}{2}=\frac{b+c-a}{2} \cdot \frac{c+a-b}{2} \cdot \frac{a+b-c}{2}
$$

From equation (1), we know that among $a, b, c$, there must be two odd and one even or all three even, so $b+c-a, c+a-b, a+b-c$ are all even numbers, hence $\frac{b+c-a}{2}, \frac{c+a-b}{2}, \frac{a+b-c}{2}$ must be integers. Let $\frac{b+c-a}{2}=x, \frac{c+a-b}{2}=y, \frac{a+b-c}{2}=z$.
Then $x, y, z$ are integers, and $a+b+c=2(x+y+z)$. Therefore, equation (1) can be transformed into
$$
4(x+y+z)=x y z.
$$

Since equation (2) is symmetric with respect to $x, y, z$, without loss of generality, assume $x \geqslant y \geqslant z$. Then
$$
\frac{1}{4}=\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \leqslant \frac{1}{z^{2}}+\frac{1}{z^{2}}+\frac{1}{z^{2}}=\frac{3}{z^{2}}.
$$

Therefore, $z^{2} \leqslant 12, z \leqslant 2 \sqrt{3}<4$.
Thus, $z$ can only be $1, 2, 3$.
(i) When $z=1$, from (2) we get $4(x+y+1)=x y$, which means
$$
(x-4)(y-4)=20=20 \times 1=10 \times 2=5 \times 4
$$
(It is easy to verify that 20 decomposed into negative factors does not satisfy the problem).
Since $x \geqslant y$, then $x-4 \geqslant y-4$. Thus,
we get $\left\{\begin{array}{l}x=24, \\ y=5 ;\end{array}\left\{\begin{array}{l}x=14 \\ y=6 ;\end{array},\left\{\begin{array}{l}x=9, \\ y=8 .\end{array}\right.\right.\right.$
At this point, $(x, y, z)=(24,5,1),(14,6,1),(9,8,1)$.
From the transformation we know $a=y+z, b=z+x, c=x+y$.
Thus, the three sides of the triangle $(a, b, c)$ are $(6,25,29)$, $(7,15,20),(9,10,17)$.
(ii) When $z=2$, we can similarly solve to get
$$
(x, y, z)=(10,3,2),(6,4,2).
$$

Thus, the three sides of the triangle $(a, b, c)$ are $(5,12,13)$, $(6,8,10)$.
(iii) When $z=3$, from (2) we get
$4(x+y+3)=3 x y$,

which means $\frac{1}{4}=\frac{1}{3 x}+\frac{1}{3 y}+\frac{1}{x y}$.
Since $x \geqslant y$, we have $\frac{1}{4} \leqslant \frac{1}{3 y}+\frac{1}{3 y}+\frac{1}{y^{2}}=\frac{2}{3 y}+\frac{1}{y^{2}}$, which means $3 y^{2}-8 y-12 \leqslant 0$.
Solving this, we get $\frac{4-2 \sqrt{13}}{3} \leqslant y \leqslant \frac{4+2 \sqrt{13}}{3}$.
Since $y \geqslant_{z}=3$, thus, $3 \leqslant y \leqslant \frac{4+2 \sqrt{13}}{3}$.
But $y \leqslant \frac{4+2 \sqrt{13}}{3}<\frac{4+8}{3}=4$.
Therefore, $y$ can only be 3.
Substituting $y=3$ into equation (3), we get $x=4.8$, which is not an integer.
Hence, when $z=3$, equation (2) has no solution.
Combining (i), (ii), and (iii), we know that there are 5 triangles that satisfy the conditions, and their three sides $(a, b, c)$ are $(6,25,29),(7,15,20),(9,10,17),(5,12,13),(6,8,10)$.
(Department of Shouwen, Anhui Province, Nangang County, Gongshan No.2 Middle School, 242418)