8. Suppose that $\left\{a_{n}\right\}_{n \geq 1}$ is an increasing arithmetic sequence of integers such that $a_{a_{20}}=17$ (where the subscript is $a_{20}$ ). Determine the value of $a_{2017}$.
Answer: 4013
Solution: Let $a_{1}=a$ be the first term of such arithmetic sequence and $d>0$ be its common difference. Then the condition $a_{a_{20}}=17$ is equivalent to $17=a+\left(a_{20}-1\right) d=a+(a+19 d-1) d=$ $a(1+d)+19 d^{2}-d$. Solving for $a$, we get
$$
a=\frac{-19 d^{2}+d+17}{d+1}=-19 d+20-\frac{3}{d+1} .
$$

Since $a$ and $d$ are integers, $d+1$ must divide 3 . With $d>0$, this forces $d+1=3$ or $d=2$, so $a=-19(2)+20-1=-19$. Hence, $a_{2017}=a+2016 d=-19+2016(2)=4013$.