About the sequence satisfying $a_n=a_{n-1}a_{n+1}-1$ "Consider sequences of positive real numbers of the form x,2000,y,..., in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of x does the term 2001 appear somewhere in the sequence?
(A) 1 (B) 2 (C) 3 (D) 4 (E) More than 4"
Can anyone suggest a systematic way to solve this problem? Thanks!

(This is basically EuYu's answer with the details of periodicity added; took a while to type up.)
Suppose that $a_0 , a_1 , \ldots$ is a generalised sequence of the type described, so that $a_i = a_{i-1} a_{i+1} - 1$ for all $i > 0$.  Note that this condition is equivalent to demanding that $$a_{i+1} = \frac{ a_i + 1 }{a_{i-1}}.$$
Using this we find the following recurrences:
$$ a_2 = \frac{ a_1 + 1}{a_0}; \\
a_3 = \frac{ a_2 + 1}{a_1} = \frac{ \frac{ a_1 + 1}{a_0} }{a_1} = \frac{ a_0 + a_1 + 1 }{ a_0a_1 }; \\
a_4 = \frac{ a_3 + 1 }{a_2} = \frac{\frac{ a_0 + a_1 + 1 }{ a_0a_1 } + 1}{\frac{ a_1 + 1}{a_0}} = \frac{ ( a_0 + 1 )( a_1 + 1) }{ a_1 ( a_1 + 1 ) } = \frac{a_0 + 1}{a_1};\\
a_5 = \frac{ a_4 + 1 }{ a_3 } = \frac{ \frac{a_0 + 1}{a_1} + 1}{\frac{ a_0 + a_1 + 1 }{ a_0a_1 }} = \frac{ \left( \frac{a_0 + a_1 + 1}{a_1} \right) }{ \left( \frac{a_0+a_1+1}{a_0a_1} \right) } = a_0 \\
a_6 = \frac{ a_5 + 1 }{a_4} = \frac{ a_0 + 1}{ \left( \frac{ a_0 + 1 }{a_1} \right) } = a_1.
$$
Thus every such sequence is periodic with period 5, so if 2001 appears, it must appear as either $a_0, a_1, a_2, a_3, a_4$.


*

*Clearly if $a_0 = 2001$, we're done.

*As we stipulate that $a_1 = 2000$, it is impossible for $a_1 = 2001$.

*If $a_2 = 2001$, then it must be that $2001 = \frac{ 2000 + 1 }{a_0}$ and so $a_0 = 1$.

*If $a_3 = 2001$, then it must be that $2001 = \frac{a_0 + 2000 + 1}{a_0 \cdot 2000}$, and it follows that $a_0 = \frac{2001}{2000 \cdot 2001 - 1}$.

*If $a_4 = 2001$, then it must be that $2001 = \frac{ a_0 + 1 }{2000}$, and so $a_0 = 2001 \cdot 2000 - 1$.


There are thus exactly four values of $a_0$ such that 2001 appears in the sequence.
