# Weird infimum of fractional part expression

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?

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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
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.
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