A number theory problem by Ilham Saiful Fauzi

Consider the following:

$\begin{aligned} a_n & = a_{n-1}a_{n+1}-1 \\ \implies \color{#3D99F6} a_{n-1} & = \color{#3D99F6}\frac {a_n+1}{a_{n+1}} \\ \implies a_{n+1} & = \frac {a_n+1}{a_{n-1}} \\ \implies a_{n+4} & = \frac {a_{n+3}+1}{a_{n+2}} \\ & = \left(\frac {a_{n+2}+1}{a_{n+1}} +1 \right) \cdot \frac 1{a_{n+2}} \\ & = \frac 1{a_{n+1}} + \frac {1+a_{n+1}}{a_{n+1}a_{n+2}} \\ & = \frac 1{a_{n+1}} + \frac {1+a_{n+1}}{a_{n+1}} \cdot \frac {a_n}{a_{n+1} + 1} \\ & = \color{#3D99F6} \frac {1+a_n}{a_{n+1}} \\ & = \color{#3D99F6} a_{n-1} & \small \color{#3D99F6} \text{Putting }k = n-1 \\ \implies a_{k+5} & = a_k \end{aligned}$

Therefore, $a_n$ has a period of 5. And $a_{2017} = a_{2017 \text{ mod 5}} = a_2 = \boxed{2}$ .

We can write

$a_n = a_{n-1} a_{n+1} - 1 \implies a_{n+1} = \dfrac{a_n +1}{a_{n-1}}$

The first few terms of the sequence are

$\begin{aligned} a_1 && &=& 1 \\ a_2 && &=& 2 \\ a_3 &=& \dfrac{a_2 +1}{a_1} &=& 3 \\ a_4 &=& \dfrac{a_3 +1}{a_2} &=& 2 \\ a_5 &=& \dfrac{a_4 +1}{a_3} &=& 1 \\ a_6 &=& \dfrac{a_5 +1}{a_4} &=& 1 \\ a_7 &=& \dfrac{a_6 +1}{a_5} &=& 2 \\ a_8 &=& \dfrac{a_7 +1}{a_6} &=& 3 \\ a_9 &=& \dfrac{a_8 +1}{a_7} &=& 2 \\ a_{10} &=& \dfrac{a_9 +1}{a_8} &=& 1 \\ a_{11} &=& \dfrac{a_{10} +1}{a_9} &=& 1 \end{aligned}$

We observe the pattern

$a_k := \begin{cases} 1 & \text{if } k \equiv 0 \text{ or } k \equiv 1 \pmod{5} \\ 2 & \text{if } k \equiv 2 \text{ or } k \equiv 4 \pmod{5} \\ 3 & \text{if } k \equiv 3 \pmod{5} \end{cases}$

Since $2017 \equiv 2 \pmod{5}$ , so we have $a_{2017} = \boxed{2}$ .