Example 6 (2006 National Training Team Test) Find all positive integer pairs $(a, n)$ such that $\frac{(a+1)^{n}-a^{n}}{n}$ is an integer.
First, point out that $(a, 1)$ is clearly a solution to the original problem (here $a$ is any positive integer). Below, we prove that the original problem has no other solutions.
Assume $(a, n)(n \geqslant 2)$ is a solution to the original problem, then there must exist a positive integer $k$ such that $(a+1)^{n}-a^{n}=k n$.
Since $a+1$ and $a$ are coprime, it follows from the above equation that $n$ is coprime with both $a$ and $a+1$. By Euler's theorem, we have:
$$
(a+1)^{\varphi(n)} \equiv a^{\varphi(n)} \equiv 1(\bmod n) \text {. }
$$

Let $d=(n, \varphi(n))$, then by Bézout's theorem, there exist integers $\alpha$ and $\beta$ such that
$$
d=\alpha n+\beta \varphi(n) \text {, }
$$

By $(a+1)^{n} \equiv a^{n}(\bmod n)$ and $(a+1)^{\varphi(n)} \equiv a^{\varphi(n)} \equiv 1(\bmod n)$ (from $(a+1)^{n}-a^{n}=k n$, we know $(a+$ $\left.1)^{n} \equiv a^{n}(\bmod n)\right)$, we can deduce
$$
(a+1)^{d} \equiv(a+1)^{\alpha n+\beta \varphi(n)} \equiv a^{\alpha n+\beta \varphi(n)} \equiv a^{d}(\bmod n) .
$$

Clearly, $d>1$ (otherwise $a+1 \equiv a(\bmod n)$, then $n=1$, which contradicts the assumption), and note that $\varphi(n) > 1$ for $n > 1$.
Therefore, all positive integer pairs $(a, n)$ that make $\frac{(a+1)^{n}-a^{n}}{n}$ an integer are $(a, 1)$ (where $a$ can be any positive integer).