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.

Find all positive integer pairs $(a, n)$ such that $\frac{(a+1)^{n}-a^{n}}{n}$ is an integer.
First, we 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: $(a+1)^{\varphi(n)} \equiv a^{\varphi(n)} \equiv 1(\bmod n)$.
Let $d=(n, \varphi(n))$, then by Bézout's theorem, there exist integers $\alpha$ and $\beta$ such that
$$d=\alpha n+\beta \varphi(n),$$

From $(a+1)^{n} \equiv a^{n}(\bmod n)$ and $(a+1)^{\varphi(n)} \equiv a^{\phi(n)} \equiv 1(\bmod n)$ (since $(a+1)^{n}-a^{n}=k n$, we have $(a+1)^{n} \equiv a^{n}(\bmod n)$), 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, the only 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).