smc

It never hurts to compute a few terms of the sequence in order to get a feel how it looks like. In our case, the definition is that $\forall$ (for all) $n>1:a_n=a_{n-1}{a}_{n+1}-1$. This can be rewritten as $a_{n+1}=\frac{a_n+1}{a_{n-1}}$. We have $a_1=x$ and $a_2=2000$, and we compute: $$\begin{array}{rl}{a}_3& =\frac{a_2+1}{a_1}=\frac{2001}{x}\\ a_4& =\frac{a_3+1}{a_2}=\frac{\frac{2001}{x}+1}{2000}=\frac{2001+x}{2000x}\\ a_5& =\frac{a_4+1}{a_3}=\frac{\frac{2001+x}{2000x}+1}{\frac{2001}{x}}=\frac{\frac{2001+2001x}{2000x}}{\frac{2001}{x}}=\frac{1+x}{2000}\\ a_6& =\frac{a_5+1}{a_4}=\frac{\frac{1+x}{2000}+1}{\frac{2001+x}{2000x}}=\frac{\frac{2001+x}{2000}}{\frac{2001+x}{2000x}}=x\\ a_7& =\frac{a_6+1}{a_5}=\frac{x+1}{\frac{1+x}{2000}}=2000\end{array}$$ At this point we see that the sequence will become periodic: we have $a_6=a_1$, $a_7=a_2$, and each subsequent term is uniquely determined by the previous two. Hence if $2001$ appears, it has to be one of $a_1$ to $a_5$. As $a_2=2000$, we only have four possibilities left. Clearly $a_1=2001$ for $x=2001$, and $a_3=2001$ for $x=1$. The equation $a_4=2001$ solves to $x=\frac{2001}{2000\cdot 2001-1}$, and the equation $a_5=2001$ to $x=2000\cdot 2001-1$. No two values of $x$ we just computed are equal, and therefore there are $\boxed{4}$ different values of $x$ for which the sequence contains the value $2001$.