China Western Mathematical Olympiad

According to the assumption we have $$a_{n+2}{a}_{n+1}-a_{n+1}{a}_n=1.$$ Thus, $\{a_{n+1}{a}_n\}$ is an arithmetic progression with first term $1$ and common difference $1$. Hence $$a_{n+1}{a}_n=n,n=1,2,\dots$$ So $a_{n+2}=\frac{n+1}{a_{n+1}}=\frac{n+1}{n}=\frac{n+1}{n}{a}_n,n=1,2,\dots$. Consequently, $$\begin{array}{rl}{a}_{2004}& =\frac{2003}{2002}{a}_{2002}=\frac{2003}{2002}\cdot \frac{2001}{2000}{a}_{2000}\\ & =\cdots =\frac{2003}{2002}\cdot \frac{2001}{2000}\cdot \cdots \cdot \frac{3}{2}{a}_2\\ & =\frac{3\cdot 5\cdot \cdots \cdot 2003}{2\cdot 4\cdot \cdots \cdot 2002}\end{array}$$