8.33 Using the natural numbers in the interval $[87,91]$ as the hypotenuse of a right triangle, with the other two sides also being natural numbers, find all such right triangles with their side lengths.
(China Jiangxi Province Mathematics Competition, 1979)
[Solution] From $\left(m^{2}+n^{2}\right)^{2}=\left(m^{2}-n^{2}\right)^{2}+(2 m n)^{2}$, we can set the hypotenuse of a right-angled triangle to be $k\left(m^{2}+n^{2}\right)$, and the two legs to be $k\left(m^{2}-n^{2}\right)$ and $k(2 m n)$ (where $m > n$), with $m$ and $n$ being natural numbers.
(1) If the hypotenuse is 87, i.e.,
$$
k\left(m^{2}+n^{2}\right)=87.
$$

When $k=1$, $m^{2}+n^{2}=87$. Since 87 is a number of the form $4k+3$, and numbers of the form $4k+3$ cannot be expressed as the sum of two squares, there is no solution in this case.
When $k=3$, $m^{2}+n^{2}=29$. In this case, there is a solution $m=5, n=2$.
Such a triangle exists, with its sides being
$$
3\left(5^{2}+2^{2}\right), \quad 3\left(5^{2}-2^{2}\right), \quad 3 \cdot 2 \cdot 5 \cdot 2,
$$

i.e., the sides of the triangle are $87, 63, 60$.
When $k=29$, $m^{2}+n^{2}=3$, there is no solution in this case.
(2) If the hypotenuse is 88, i.e.,
$$
k\left(m^{2}+n^{2}\right)=88.
$$

When $k=1,2,4,8,11,22,44$,
$$
m^{2}+n^{2}=88,44,22,11,8,4,2.
$$

In these cases, there are no natural numbers $m, n$ that satisfy the equation.
(3) If the hypotenuse is 89, i.e.,
$$
k\left(m^{2}+n^{2}\right)=89.
$$

When $k=1$, $m^{2}+n^{2}=89=64+25$, then $m=8, n=5$.
In this case, the sides of the triangle are $89, 39, 80$.
(4) If the hypotenuse is 90, i.e.,
$$
k\left(m^{2}+n^{2}\right)=90.
$$

When $k=2$, $m^{2}+n^{2}=45=36+9$, then
$$
m=6, \quad n=3.
$$

When $k=9$, $m^{2}+n^{2}=10=9+1$, then
$$
m=3, \quad n=1.
$$

When $k=18$, $m^{2}+n^{2}=5=4+1$, then
$$
m=2, \quad n=1.
$$

In these three cases, the sides of the triangle are $90, 54, 72$.
When $k=1,3,5,6,10,15,30,45$, the triangle does not exist.
(5) If the hypotenuse is 91, i.e.,
$$
k\left(m^{2}+n^{2}\right)=91.
$$

When $k=7$, $m^{2}+n^{2}=13=3^{2}+2^{2}$, then
$$
m=3, n=2.
$$

The sides of the right-angled triangle are: $91, 35, 84$.
When $k=1,13$, the triangle does not exist.
Therefore, the sides of the right-angled triangles are as follows:
$$
(87,63,60),(89,39,80),(90,54,72),(91,35,84).
$$