8.

用数学归纳法证明，对于所有正整数 $n$

能被 17 整除。

先验证 $n=1$。归纳时两项的倍率分别是 $81$ 与 $64$，相差 $17$，因此凑出 $81f(k)$ 后，余下的一项含因子 $17$。

For $n=1$,

which is divisible by $17$.

Assume that $f(k)$ is divisible by $17$ for some positive integer $k$. Write

Then $A,B$ are integers and $f(k)=A+B$. Since

we have

Both terms are divisible by $17$, so $f(k+1)$ is divisible by $17$.

Thus the statement for $n=k$ implies the statement for $n=k+1$. Together with the base case, mathematical induction proves the result for every positive integer $n$.

同样可以先作差，再加回 $f(k)$；在同一归纳假设下：

若把第二项写成 $8^{2n-1}$，因 $8^2=64$，归纳步骤完全相同。

改为凑出 $64f(k)$，剩余部分就成为第一项的 $17$ 倍；归纳的起始与结论保持不变。

For $n=1$, $f(1)=17$, so the statement is true.

Assume that $f(k)$ is divisible by $17$ for a positive integer $k$. Let

Then

The inductive hypothesis makes $64f(k)$ divisible by $17$, and $17A$ is also divisible by $17$. Hence $f(k+1)$ is divisible by $17$.

The statement therefore holds for all positive integers $n$ by mathematical induction.

先作差时，相应的写法为

再加回 $f(k)$，便得到上面的递推式。

更一般地，先减去整数倍 $mf(k)$。选择使 $81-m$ 与 $64-m$ 都是 $17$ 的倍数的 $m$，就能直接完成归纳步骤。例如可取 $m=47$。

The base case is $f(1)=17$.

Assume that $f(k)$ is divisible by $17$ for a positive integer $k$, and put

For any integer $m$, let

Then

Taking $m=47$ gives

Therefore,

Both terms are divisible by $17$, so the inductive step holds. Hence mathematical induction proves the result for every positive integer $n$.

选取其他合适整数也得到同样结论。例如：

将 $G_m$ 加回 $mf(k)$ 就得到 $f(k+1)$，因此这些写法同样完成归纳步骤。