Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $x_1$, $x_2$, $x_3,\ldots$ is a sequence such that $$x_n=\frac{x_{n-2}\space x_{n-1}}{2x_{n-2}-x_{n-1}},$$ where $x_i \in \mathbb R$ and $x_i \ne0 $ for all $i\in \mathbb N$ and $n=3$, $4$, $5,\ldots$

How can I establish necessary and sufficient conditions on $x_1$ and $x_2$ for $x_n$ to be an integer for infinitely many values of $n$?

I have been stuck on this problem for quite sometime now. I can't seem to find a pattern so that i could "make" $x_n$ an integer.

Any help is much appreciated!

Thanks in advance!

share|improve this question
    
Notice that if $x_1=x_2$, then $x_n = x_1$. –  user26872 Mar 25 '12 at 5:44
add comment

2 Answers 2

up vote 2 down vote accepted

Note that

$$ \frac{1}{x_n} = \frac{2}{x_{n-1}} - \frac{1}{x_{n-2}} $$

so $\tfrac{1}{x_n}$ satisfies a linear recursion. This can be solved explicitly: Let

$$ a = \frac{2}{x_1} - \frac{1}{x_2},\ b = \frac{1}{x_2} - \frac{1}{x_1} $$

Then $\tfrac{1}{x_n} = a + nb$, or

$$ x_n = \frac{1}{a+nb}. $$

Now if $b \neq 0$ then $x_n \rightarrow 0$ and in particular $x_n$ can only be an integer finitely many times. So $b=0$ and $a = \tfrac{1}{m}$ for some integer $m$.

share|improve this answer
add comment

Value of $x_n$ can be rewritten as, $$\frac{1}{x_n} = \frac{2}{x_{n-1}} - \frac{1}{x_{n-2}}.$$ $$\frac{1}{x_n} - \frac{1}{x_{n-1}} = \frac{1}{x_{n-1}} - \frac{1}{x_{n-2}}.$$ This means that difference of reciprocal remans same, hence sequence $1\over{x_n}$ is in arithmetic progression. This implies $${1\over{x_n}} = A + (n-1)D,$$ Where $A$ is first term and $D$ is difference of arithmetic progression. $${{x_n}} = {1\over{A + (n-1)D}},$$Only way $x_n$ can be integer for infinitely many values of n, is $D = 0$ and A is fraction whose numerator is 1. In such cases $x_n$ will be constant $1\over{A}$ for all value of n.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.