Example 8 Try to find all natural number triples $(A, B, C)$, such that
$$
A^{2}+B-C=100, A+B^{2}-C=124 .
$$
(1990, Leningrad (now St. Petersburg) Mathematical Olympiad Third Round, Grade 8)
Solution: From the problem,
$A^{2}+B-C=100$,
$A+B^{2}-C=124$.
(2) - (1), $A+B^{2}-A^{2}-B=24$,

$\quad(B-A)(A+B-1)=24$.
Since $A, B, C$ are natural numbers,
we can obtain the following system of equations
(1) $\left\{\begin{array}{l}B-A=1, \\ A+B-1=24\end{array}\right.$
(2) $\left\{\begin{array}{l}B-A=24, \\ A+B-1=1\end{array}\right.$;
(3) $\left\{\begin{array}{l}B-A=2, \\ A+B-1=12 ;\end{array}\right.$
(4) $\left\{\begin{array}{l}B-A=12, \\ A+B-1=2 ;\end{array}\right.$
(5) $\left\{\begin{array}{l}B-A=3, \\ A+B-1=8 ;\end{array}\right.$
(6) $\left\{\begin{array}{l}B-A=8, \\ A+B-1=3\end{array}\right.$;
(7) $\left\{\begin{array}{l}B-A=4, \\ A+B-1=6\end{array}\right.$
(8) $\left\{\begin{array}{l}B-A=6 \\ A+B-1=4\end{array}\right.$
Solving (1) and (5) yields $\left\{\begin{array}{l}A=12, \\ B=13 ;\end{array}\left\{\begin{array}{l}A=3, \\ B=6 .\end{array}\right.\right.$
(2), (3), (4), (6), (7), (8) have no solutions in the natural number range.
Substituting $A=12, B=13$ into (1) gives $C=57$;
Substituting $A=3, B=6$ into (1) gives $C=-35$.
Therefore, the natural number solution set that meets the problem's requirements is $A=12, B=$ $13, C=57$.