用数学归纳法证明，对于所有正整数 $n$

$f(n) = 3^{4n-2} + 2^{6n-3}$

能被 17 整除。

先验证 $n=1$。归纳步骤中把 $f(k+1)$ 的两项分别写成 $81\cdot3^{4k-2}$ 与 $64\cdot2^{6k-3}$，再凑出 $81f(k)$；余下的一项含因子 $17$。

For $n=1$,

which is divisible by $17$.

Assume that the result is true for $n=k$, where $k\in\mathbb{Z}^+$. Thus

is divisible by $17$.

Now

By the inductive hypothesis, $81f(k)$ is divisible by $17$, and $17\cdot2^{6k-3}$ is also divisible by $17$. Therefore, $f(k+1)$ is divisible by $17$.

Hence, if the result is true for $n=k$, it is true for $n=k+1$. Since it is true for $n=1$, $f(n)$ is divisible by $17$ for every positive integer $n$ by mathematical induction.