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.

I found this problem a while ago on AoPS.

Let $k$ be a squarefree positive integer.

Find $\inf_{n \in \Bbb{Z}_+^*} n\{n\sqrt{k}\,\}$, where $\{\cdot\}$ denotes the fractional part.

I tried to find a solution, but it is still unsolved. My guess for the infimum is $0$, and I tried proving the existence of a sequence $(n_i)$ such that $\displaystyle \{n_i \sqrt{k}\} < \frac{1}{n_i^2}$. Denote $p_i$ the integer part of $n_i\sqrt{k}$. Then we have $\displaystyle p_i<n_i\sqrt{k}<p_i+\frac{1}{n_i^2}$. Squaring we get $\displaystyle p_i^2<n_i^2k<p_i^2+2p_i\frac{1}{n_i^2}+\frac{1}{n_i^4}$.

The first inequality must be strict, and the best case would be $p_i^2-kn_i^2=-1$, which is a Pell equation, solvable for some squarefree $k$, but not for all squarefree $k$.

Do you have some ideas of what to do next?

share|improve this question
1  
The infimum is not $0$. The reason roughly speaking is that quadratic irrationals are poorly approximable by rationals. The best you can do is to use (alternate) convergents of the continued fraction. –  André Nicolas May 25 '11 at 23:24
    
Comment continued. As you probably noticed, if the Pell Equation $x^2-dy^2=-1$ has a solution, then as $n \to\infty$, the minimum value you are interested in approaches $1/(2\sqrt{d})$. I do not know any result this exact in other cases. –  André Nicolas May 26 '11 at 2:49
add comment

1 Answer

up vote 3 down vote accepted

Expanding on the comment of user6312; for every squarefree positive integer $k$, there's a constant $C_k$ such that for all integers $m$, $n$, $|\sqrt k-(m/n)|\gt C_kn^{-2}$. This leads to $n\lbrace n\sqrt k\rbrace\gt C_k$. In turn, $C_k$ is related to the partial quotients in the continued fraction expansion of $\sqrt k$; roughly speaking, big partial quotients give small $C_k$. Any good intro Number Theory text will go into continued fraction expansions of quadratic irrationals and the relation to diophantine approximation, and there are probably good web resources that you can find by searching for some of the keyphrases in what I've written.

share|improve this answer
    
Thank you. I'll look into it. –  Beni Bogosel May 26 '11 at 9:18
    
I would like to know a reference for the inequality you posted, namely $|\sqrt{k}-(m/n)|>C_k n^{-2}$. –  Beni Bogosel May 26 '11 at 11:54
    
@Beni, as I said, any good intro Number Theory text; or, search the web for keyphrases you can pick out of my answer. –  Gerry Myerson May 26 '11 at 12:31
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.