TITLE: Beautiful problem on a progression
QUESTION [6 upvotes]: $\{x_n\}$ is a sequence defined as follows:
$x_1=20,\quad x_2=14,\quad x_{n+2}=x_n - \frac{1}{x_{n+1}}$.
Prove that $0$ is among the members of this sequence. Find its number. 
I tried some stuff for quite some time, I think it's solved using limits but I can't solve it.

REPLY [16 votes]: $$\require{cancel}\begin{align}
x_{n+2}&=x_n-\frac 1{x_{n+1}}\\
x_{n+1}x_{n+2}&=x_n x_{n+1}-1\\
\end{align}$$
Given that $x_1=20$ and $x_2=14$, we have
$$\begin{align}
x_1 x_2&=280\\
x_2 x_3&=279\\
x_3 x_4&=278\\
\vdots &= \vdots\\
x_{280} x_{281}&=1\\
x_{281} x_{282}&=0\\
\end{align}$$
Hence $$x_{282}=0$$
i.e. $x_n=0$ when $n=282$.
NB - technically the series is not defined for $n>282$ because $x_{283}$ has a $\frac 10$ term; however if it is accepted that $\frac 10=\infty$ and $\frac 1\infty=0$, then every even term after the $282^{\text{nd}}$ is zero. An interesting problem though.

Further to my solution above, this is in response to the request for expressing $x_n$ explicitly.
We can see that 
$$\begin{align}
x_1=20&, 
x_2=14\\
x_3=\frac {279}{14}&,
x_4=14\cdot \frac{278}{279}\\
x_5=\frac{277}{278}\cdot \frac{279}{14}&,
x_6=14\cdot \frac{278\cdot 276}{279\cdot 277}\\
x_7=\frac{275\cdot 277}{276\cdot 278}\cdot \frac{279}{14}&,
x_8=14\cdot \frac{278\cdot 276\cdot 274}{279\cdot 277\cdot 275}\\
\vdots &,\vdots \end{align}$$
hence, $x_n$ can be explicitly defined depending on whether $n$ is odd  or even, i.e.
$$x_{2m}=14\cdot \frac{\prod_{r=0}^{m-2}278-2r}{\prod_{r=0}^{m-2}279-2r}=
14\cdot \frac{278^\underline{\underline{m-1}}}{279^\underline{\underline{m-1}}}
=\frac 1{20}\cdot \frac{280^\underline{\underline{m}}\;\;\;}{279^\underline{\underline{m-1}}}$$
and 
$$x_{2m+1}=\frac{279}{14}\cdot \frac{\prod_{r=0}^{m-2}277-2r}{\prod_{r=0}^{m-2}278-2r}
=\frac 1{14}\cdot \frac{279^\underline{\underline{m}}\;\;\;}{278^\underline{\underline{m-1}}}
=20\cdot \frac{279^\underline{\underline{m}}}{280^\underline{\underline{m}}}$$
where $m\geq 2$, and $x^\underline{\underline{n}}=\overbrace{x(x-2)(x-4)\cdots (x-2n+2)}^{n \text{ terms}}$