13.2 設 $n$ 為整數。求 $n^{a}-n$ 除以 30 的稌值 $b$ 。
Let $n$ be an integer. Determine the remainder $b$ of $n^{a}-n$ divided by 30 .
$$
n^{5}-n=n\left(n^{4}-1\right)=n\left(n^{2}+1\right)\left(n^{2}-1\right)=(n-1) n(n+1)\left(n^{2}+1\right)
$$
$n-1, n$ and $n+1$ are three consecutive integers, the product of which must be divisible by 6 . If any one of $n-1, n$ or $n+1$ is divisible by 5 , then the product is divisible by 30 .
Otherwise, let $n-1=5 k+1, n=5 k+2, n+1=5 k+3, n^{2}+1=(5 k+2)^{2}+1=25 k^{2}+20 k+5$ which is a multiple of 5 , the product is divisible by 30 .
$$
\text { If } n-1=5 k+2, n=5 k+3, n+1=5 k+4, n^{2}+1=(5 k+3)^{2}+1=25 k^{2}+30 k+10
$$
which is a multiple of 5 , the product is divisible by 30 .
In all cases, $n^{5}-n$ is divisible by 30 . The remainder when $n^{5}-n$ divided by 30 is 0 .