Example 5 Calculate the coefficient of $x^{100}$ in the expansion of $\left(1+x+x^{2}+\cdots+x^{100}\right)^{3}$ after combining like terms.
(7th All-Russian High School Mathematics Olympiad, Third Round Competition)
Solve the product of three polynomials, $\left(1+x+x^{2}+\cdots+x^{100}\right)\left(1+x+x^{2}+\right.$ $\left.\cdots+x^{100}\right)\left(1+x+x^{2}+\cdots+x^{100}\right)$. Without combining like terms, we can obtain a sum of products of the form $x^{p} x^{q} x^{r} = x^{p+q+r}$, where the integers $p, q, r$ satisfy the following inequalities:
$$
0 \leqslant p \leqslant 100,0 \leqslant q \leqslant 100,0 \leqslant r \leqslant 100
$$

We are interested in the terms that satisfy $p+q+r=100$. Clearly, the number of such terms is equal to the number of integer solutions $(p, q, r)$ of the equation $p+q+r=100$ that satisfy the inequalities (11).

We will find the number of such solutions. If $p=0$, then $q+r=100$, and this equation has 101 solutions $(q, r)$, which are the pairs $(0,100),(1,99),(2,98), \cdots,(100,0)$; if $p=1$, then $q+r=99$, and it has 100 solutions $(0,99),(1,98),(2,97)$, $\cdots,(99,0)$; if $p=2$, then $q+r=98$, and it has 99 solutions, and so on. Therefore, the total number of solutions that meet the requirements is
$$
101+100+99+\cdots+2+1=\frac{(1+101) \times 101}{2}=5151 \text {. }
$$

This is also the coefficient of $x^{100}$ in the expanded form.
In fact, we can generalize the above two examples as follows:
Theorem 2 If the number of positive integer solutions of $x_{1}+x_{2}+\cdots+x_{n}=m(n, m \in \mathbf{N}$ and $n \leqslant m)$ is denoted by $N_{n}(m)$, then $N_{n}(m)=\mathrm{C}_{m n-1}^{n-1}$.

Proof Clearly, we should have $N_{1}(m)=1$. We will now calculate $N_{2}(m)$, which is the number of positive integer solutions of the equation
$$
x_{1}+x_{2}=m .
$$

In fact, in (12), let $x_{1}=1,2,3, \cdots, m-1$, then the number of solutions of (12) is $(1, m-1),(2, m-2)(3, m-3), \cdots,(m-1,1)$, thus
$$
N_{2}(m)=m-1 .
$$

Next, we find $N_{3}(m)$, which is the number of positive integer solutions of the equation
$$
x_{1}+x_{2}+x_{3}=m
$$

At this point, we only need to let $x_{3}=1,2,3, \cdots, m-2$. Then, by the previous situation, we know
$$
N_{3}(m)=N_{2}(m-1)+N_{2}(m-2)+N_{2}(m-3)+\cdots+N_{2}(2) .
$$

If we also consider (2), we know
$$
\begin{aligned}
N_{3}(m) & =(m-2)+(m-3)+) m-4)+\cdots+2+1 \\
& =\frac{(m-1)(m-2)}{1 \cdot 2}=\mathrm{C}_{m-1}^{2} .
\end{aligned}
$$

Now, we generalize this for dimension $n$ using mathematical induction. It is easy to prove
$$
N_{n}(m)=\mathrm{C}_{m-1}^{n-1} .
$$

In fact, let $x_{n}=1,2,3, \cdots,[m-(n-1)]$, then we know
$$
\begin{aligned}
N_{n}(m)= & N_{n-1}(m-1)+N_{n-1}(m-2)+N_{n-1}(m-3) \\
& +\cdots+N_{n-1}(n-1),
\end{aligned}
$$

By the induction hypothesis, we know $\quad N_{n-1}(m)=C_{m-1}^{n-2}$,
Thus $\quad N_{n}(m)=\mathrm{C}_{m-2}^{n-2}+\mathrm{C}_{m-3}^{n-2}+\mathrm{C}_{m-3}^{n-2}+\mathrm{C}_{m-4}^{n-2}+\cdots+\mathrm{C}_{n-2}^{n-2}$.
If we consider Pascal's identity
$$
\mathrm{C}_{m}^{n}+\mathrm{C}_{m}^{n-1}=\mathrm{C}_{m+1}^{n},
$$

Then we know
$$
\begin{aligned}
\mathrm{C}_{m-1}^{n-1} & =\mathrm{C}_{m-2}^{n-2}+\mathrm{C}_{m-2}^{n-1}=\mathrm{C}_{m-2}^{n-2}+\mathrm{C}_{m-3}^{n-2}+\mathrm{C}_{m-3}^{n-1} \\
& =\mathrm{C}_{m-2}^{n-2}+\mathrm{C}_{m-3}^{n-2}+\mathrm{C}_{m-4}^{n-2}+\mathrm{C}_{m-4}^{n-1} \\
& =\mathrm{C}_{m-2}^{n-2}+\mathrm{C}_{m-3}^{n-2}+\mathrm{C}_{m-4}^{n-2}+\cdots+\mathrm{C}_{m-2}^{n-2}
\end{aligned}
$$

Thus
$$
N_{n}(m)=\mathrm{C}_{m-1}^{n-1} \text {. }
$$

If we use a more advanced method of power series, we can also obtain the following famous results:
(1) Hardy's Theorem: The number of non-negative integer solutions of the Diophantine equation $x+2 y+3 z=n$ is $A_{n}=\left[\frac{(n+3)^{2}}{12}\right]$.
(2) Catalan's Theorem: The total number of non-negative integer solutions of the equations $x+2 y=n, 2 x+3 y=n-1,3 x+4 y=n-2, \cdots,(n-1) x+(n+2) y=0$ is $n+1$.
(3) Cesàro's Theorem: The total number of non-negative integer solutions of the Diophantine equations $x+4 y=3 n-1,4 x+9 y=5 n-4,9 x+16 y=7 n-9, \cdots, k^{2} x+(k+1)^{2} y=(2 k+1) n-k^{2}, \cdots$ is $n$.
(4) Laguerre's Theorem: Let $a_{1}, a_{2}, \cdots, a_{k}$ be $k$ relatively prime positive integers, and let $A_{n}$ be the number of non-negative integer solutions of the equation $a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{k} x_{k}=n$, then
$$
\lim _{n \rightarrow \infty} \frac{A_{n}}{n^{k-1}}=\frac{1}{(k-1) a_{1} a_{2} \cdots a_{n}} .
$$

Of course, Diophantine equations come in various forms, and the methods for finding the number of their solutions are also diverse, which requires adapting to the specific problem.