Let $a$ and $b$ be two square-free, distinct natural numbers. Show that there exist $c>0$ such that \[ \left | \{n\sqrt{a}\}-\{n\sqrt{b}\} \right |>\frac{c}{n^3}\] for every positive integer $n$.
Problem
Source: Romania TST 2013 Day 3 Problem 1
Tags: inequalities, algebra, polynomial, absolute value, triangle inequality, number theory unsolved, number theory
randomusername
22.01.2015 22:42
We know that $n\sqrt a-\{n\sqrt a\}$ and $n\sqrt b-\{n\sqrt b\}$ are integers, hence for some integer $k$, we have
\[\{n\sqrt a\}-\{n\sqrt b\}=n\sqrt a-n\sqrt b+k.\]
First, suppose $k=0$. Trivially, $|\{n\sqrt a\}-\{n\sqrt b\}|\le 1$, hence this means $n(\sqrt a-\sqrt b)+k\in[-1;1]$. But since $a\neq b$, this isn't true for large enough $n$, it isn't true for all $n>n_0$. Let's consider
\[c_0:=\min_{1\le n\le n_0}|\{n\sqrt a\}-\{n\sqrt b\}|,\]
then for all $n\le n_0$, we have
\[|\{n\sqrt a\}-\{n\sqrt b\}|\ge c_0\ge \frac{c_0}{n^3}.\]
Now let $n>n_0$ be arbitrary. Then $k\neq 0$.
Note the following identity:
\[(n\sqrt a-n\sqrt b+k)(n\sqrt a+n\sqrt b+k)=(n\sqrt a+k)^2-(n\sqrt b)^2=\]
\[=2kn\sqrt a+(n^2a+k^2-n^2b),\]
and its 'conjugate',
\[(-n\sqrt a-n\sqrt b+k)(-n\sqrt a+n\sqrt b+k)=-2kn\sqrt a+(n^2a+k^2-n^2b).\]
Multiplying these two together, it is immediate to see
\[\prod(\pm n\sqrt a\pm n\sqrt b+k)\in\mathbb{Z}.\]
Now if this product is zero, then one of $\pm 2kn\sqrt a+(n^2a+k^2-n^2b)$ is zero, which implies that $2kn=0$, because $\sqrt a$ is irrational. Since $n$ is a positive integer and $k\neq 0$, this is impossible, hence the product is nonzero. So we can surely say that its absolute value is $\ge 1$. Therefore:
\[|\{n\sqrt a\}-\{n\sqrt b\}|=|n\sqrt a-n\sqrt b+k|\ge\]
\[\ge \frac1{|n\sqrt a+n\sqrt b+k|\cdot |-n\sqrt a-n\sqrt b+k|\cdot |-n\sqrt a+n\sqrt b+k|}.\]
To finish, we note that
\[|n\sqrt a-n\sqrt b+k|=|\{n\sqrt a\}-\{n\sqrt b\}|\le 1,\]
hence by the triangle inequality,
\[X:=|n\sqrt a+n\sqrt b+k|\le |n\sqrt a-n\sqrt b+k|+|2n\sqrt b|\le 1+2n\sqrt b,\]
\[Y:=|-n\sqrt a-n\sqrt b+k|\le 1+2n\sqrt a,\]
\[Z:=|-n\sqrt a+n\sqrt b+k|\le 1+2n(\sqrt a+\sqrt b).\]
Denote $M=4(\sqrt a+\sqrt b)$, then $X,Y,Z\le M\cdot n$, yielding
\[|\{n\sqrt a\}-\{n\sqrt b\}|\ge \frac1{XYZ}\ge \frac1{M^3n^3}.\]
Therefore if $c=\min\{M^{-3},c_0\}$, then $\forall n\ge 1$
\[|\{n\sqrt a\}-\{n\sqrt b\}|\ge \frac{c}{n^3}.\]
mavropnevma
22.01.2015 23:43
The number $\lambda = \sqrt{a} - \sqrt{b}$ is an algebraic integer of degree $4$; its minimal polynomial is \[\prod (x \pm \sqrt{a} \pm \sqrt{b}) = x^4 - 2(a+b)x^2 + (a-b)^2 \in \mathbb{Z}[x].\] The condition on $a,b$ to be square-free is present only in order to ensure that $\sqrt{ab}$ is irrational, thus the degree of $\lambda$ is $4$. The Liouville approximation theorem states that for an algebraic integer $\lambda$ of degree $d$ there exists a real positive constant $c$ such that \[\displaystyle \left | \lambda - \dfrac {m} {n} \right | > \dfrac {c} {n^d},\] for any positive integers $m,n$, thus in our case $\displaystyle \left | n\lambda - m \right | > \dfrac {c} {n^3}$, i.e. $\min \{\{n\lambda\}, 1 - \{n\lambda\}\} > \dfrac {c} {n^3}$. But this is clearly equivalent with $ \left | \{n\sqrt{a}\} - \{n\sqrt{b}\} \right | > \dfrac {c} {n^3}. \quad \square$ From the Thue-Siegel-Roth approximation theorem follows even more, namely that for any $\varepsilon > 0$ there exists a real positive constant $c(\varepsilon)$ such that \[\displaystyle \left | \lambda - \dfrac {m} {n} \right | > \dfrac {c(\varepsilon)} {n^{2+\varepsilon}},\] and then $\displaystyle \left | n\lambda - m \right | > \dfrac {c(\varepsilon)} {n^{1+\varepsilon}}$. On the other hand, the Dirichlet approximation theorem states that the equation \[\displaystyle \left | \lambda - \dfrac {m} {n} \right | < \dfrac {1} {n^2}\] is satisfied for infinitely many positive integers $m,n$, thus the exponent $2+\varepsilon$ is best possible. I wondered what was the purpose of this question ... Although several people tried to convince me of its benefit, I'm still reserved upon it.
lazizbek42
25.12.2021 07:58
hard problem for p1.