18. Find the largest integer $n$ such that $n$ is a divisor of $a^{5}-a$ for all integers $a$.
18. Ans: 30

Note that
$$
a^{5}-a=a(a-1)(a+1)\left(a^{2}+1\right) .
$$

It is clear that $2 \mid a^{5}-a$ and $3 \mid a^{5}-a$. We can show that $5 \mid a^{5}-a$ by considering the five cases: $a \equiv i \quad(\bmod 5), i=0,1,2,3,4$. Thus $30 \mid a^{5}-a$.
When $a=2$, we have $a^{5}-a=30$. Thus the maximum $n$ is 30 .