用数学归纳法证明，对于所有正整数 $n \in \mathbb{Z}^+$， $f(n) = 7^{n-1} + 8^{2n+1}$

能被 $57$ 整除。

先验证 $n=1$。归纳步骤中，把 $f(k+1)$ 的两项分别写成 $7$ 倍与 $64$ 倍，再凑出 $7f(k)$；余下部分恰好含因数 $57$。

For $n=1$,

so the result is true for $n=1$.

Assume that the result is true for $n=k$, where $k\in\mathbb Z^+$. Thus $f(k)$ is divisible by 57.

Now,

By the induction hypothesis, $7f(k)$ is divisible by 57, and $57\big(8^{2k+1}\big)$ is also divisible by 57. Hence $f(k+1)$ is divisible by 57.

Therefore, if the result is true for $n=k$, it is true for $n=k+1$. Since it is true for $n=1$, by mathematical induction,

另一种官方认可的重排方式是凑出 $64f(k)$。设 $f(k)=57\ell$，则 $f(k+1)$ 可以整体提出因数 57，归纳结论随即成立。

For $n=1$,

so the result is true for $n=1$.

Assume that the result is true for $n=k$. Then, for some integer $\ell$,

Also,

Since $64\ell-7^{k-1}$ is an integer, $f(k+1)$ is divisible by 57.

Therefore, if the result is true for $n=k$, it is true for $n=k+1$. Since it is true for $n=1$, by mathematical induction,