20. Find the total number of positive integers $n$ not more than 2013 such that $n^{4}+5 n^{2}+9$ is divisible by 5 .
20. Answer: 1611
Solution. Note that $n^{4}+5 n^{2}+9=n^{4}-1+5 n^{2}+10=(n-1)(n+1)\left(n^{2}+1\right)+5\left(n^{2}+2\right)$. If $n \equiv 1$ or $4(\bmod 5)$, then 5 divides $n-1$ or $n+1$.
If $n \equiv 2$ or $3(\bmod 5)$, then 5 divides $n^{2}+1$.
If $n \equiv 0(\bmod 5)$, then 5 does not divide $(n-1) n\left(n^{2}+1\right)$ but divides $5\left(n^{2}+2\right)$, hence does not divide $n^{4}+5 n^{2}+9$.
Thus, there are $2010 \div 5=402$ multiples of 5 from 1 to 2013 . The number of integers thus required is $2013-402=1611$.