1 Introduction
By Hurwitz’s theorem, for each irrational number
$\alpha>0$
, there are infinitely many pairs of positive integers
$(m,n)$
such that
$$ \begin{align}|m\alpha-n|< \frac{1}{\sqrt{5} m}\end{align} $$
(see, for example, [Reference Coppel4, page 189] or [Reference Schmidt16]). In particular, (1.1) implies that if
$\alpha>0$
is irrational, then, for any
$\varepsilon>0$
, there exist
$m,n \in {\mathbb N}$
for which
$$ \begin{align}|m\alpha-n|< \varepsilon.\end{align} $$
For some infinite subsets S of
${\mathbb N}$
, the inequality (1.2) also holds for infinitely many pairs
$(m,n)$
, where
$m \in S$
and
$n \in {\mathbb N}$
. In [Reference Iyer10], such a set S is called a Heilbronn set. For example, by Furstenberg’s theorem (see [Reference Boshernitzan2, Reference Furstenberg7]), the inequality (1.2) with any
$\varepsilon>0$
holds for some
$m \in S$
and
$n \in {\mathbb N}$
, where
$S \subseteq {\mathbb N}$
is a multiplicative semigroup with at least two multiplicatively independent integers, for instance,
${S=\{p^u q^v \mid u, v \in {\mathbb N} \cup \{0\}\}}$
, where
$p<q$
are two fixed prime numbers. (See [Reference Katz11, Reference Kra12, Reference Urban17, Reference Urban18] for some generalisations of Furstenberg’s theorem.) Also, there are some interesting sets S for which the inequality weaker than (1.1) but stronger than (1.2), namely,
$|m\alpha -n|<m^{-\tau }$
, has been derived for some
$\tau $
in the range
$0<\tau <1$
. These are, for example, the set of squares
$S=\{n^2 \mid n \in {\mathbb N}\}$
(see [Reference Zaharescu19]) and the set of prime numbers
${S=P=\{p_1<p_2<p_3<\cdots \}}$
(see [Reference Balog, Daboussi, Liardet and Rauzy1, Reference Heath-Brown and Jia8, Reference Matomäki14]), so they are Heilbronn sets.
In this paper, we are interested in obtaining inequality (1.2) for each irrational
$\alpha>0$
when not only just m but both m and n belong to a subset S of
${\mathbb N}$
. For an irrational
$\alpha>0$
it is clear that, for each
$\varepsilon>0$
, the inequality (1.2) holds with some
$m,n \in S$
if and only if
$\liminf _{m,n \in S} |m\alpha -n|=0.$
For a subset E of the set of real numbers
${\mathbb R}$
, we define
$$ \begin{align}\Delta(E):= \liminf_{x, y \in E, \> x \ne y} |x-y|.\end{align} $$
It is clear that
$\Delta (S) \geq 1$
for
$S \subseteq {\mathbb N}$
. With the notation as in (1.3), the problem we are interested in can be rephrased as follows: for a given
$S \subseteq {\mathbb N}$
, determine whether or not, for each irrational
$\alpha>0$
,
$$ \begin{align} \Delta(S \cup \alpha S)=0 \end{align} $$
or, alternatively, whether or not there exists an irrational
$\alpha>0$
for which
$$ \begin{align} \Delta(S \cup \alpha S)>0. \end{align} $$
For the set of squares
$S=\{n^2 \mid n \in {\mathbb N}\}$
, we have option (1.5). Indeed, the distance between any two distinct elements of S is at least
$3$
, while the distance between any two distinct elements of
$\alpha S$
is at least
$3\alpha $
. Recall that the number
$\beta>0$
is badly approximable if there exists a constant
$c=c(\beta )>0$
such that
$|m\beta -n|>{c}/{m}$
for all
$m,n \in {\mathbb N}$
. (A number is badly approximable if and only if the partial quotients of its continued fraction are bounded [Reference Coppel4, page 190]. For example, all quadratic algebraic numbers
$\beta $
are badly approximable [Reference Coppel4, page 194].) For
$\alpha =\beta ^2$
, where
$\beta>0$
is a badly approximable number, the distance between
$\alpha m^2 \in \alpha S$
and
$n^2 \in S$
is
$$ \begin{align*}|m^2\beta^2-n^2|=|(m\beta -n)(m \beta+n))| \geq \frac{c}{m}|m\beta+n|=\frac{c}{m}(m\beta+n)>c\beta=c \sqrt{\alpha}\end{align*} $$
for some
$c>0$
. Hence,
$$ \begin{align*}\Delta(S \cup \alpha S) \geq \min(3,3\alpha,c\sqrt{\alpha})>0\end{align*} $$
for each such
$\alpha $
, which proves (1.5). This example appears in Ruzsa’s paper [Reference Ruzsa15] in a slightly different context. (We will also use another idea from the proof of [Reference Ruzsa15, Theorem 1] in the proof of our own Theorem 1.2.)
On the other hand, for the set of primes
$S=P$
, the problem of determining whether we have option (1.4) or (1.5) seems to be out of reach. Option (1.4) takes place if and only if, for each irrational
$\alpha>0$
and any
$\varepsilon>0$
, there are prime numbers
$p_i, p_j$
satisfying
$|p_i \alpha -p_j|<\varepsilon $
. This is true if and only if there is an infinite sequence of primes
$q_1<q_2<q_3<\cdots $
such that
$$ \begin{align} \|q_j \alpha\| \to 0 \quad \text{as } j \to \infty \end{align} $$
and the nearest integer to
$\alpha q_j$
, namely,
$$ \begin{align} \lfloor q_j \alpha + 1/2\rfloor, \end{align} $$
is a prime number. In particular, condition (1.7) alone, without condition (1.6), is satisfied if and only if there are infinitely many primes p for which
$\lfloor p \alpha + 1/2\rfloor $
is also a prime number. For any
$\alpha>0$
, which is not an integer, this problem is completely out of reach (even for rational numbers
$\alpha $
). For example, for
$\alpha =1/2$
, this problem is equivalent to the following. Are there infinitely many primes p for which
$2p-1$
is also a prime?
As for the problem described in (1.2), in general, it is natural to expect that (1.4) is true when the set S is ‘large’ whereas (1.5) is true when S is ‘small’. However, we show that the answer to the problem does not depend just on the size of S. Recall that the lower and the upper density of the set
$E \subseteq {\mathbb N}$
are defined by
$$ \begin{align*}\underline{d}(E)=\liminf_{x \to \infty} \frac{\#\{E \cap [1,x]\}}{x} \quad \text{and}\quad \overline{d}(E)=\limsup_{x \to \infty} \frac{\#\{E \cap [1,x]\}}{x}, \end{align*} $$
respectively. Clearly,
$0\leq \underline {d}(E) \leq \overline {d}(E) \leq 1$
. In the case when
$\underline {d}(E) = \overline {d}(E)$
, their common value
$d(E)=\underline {d}(E) = \overline {d}(E)$
is called the density of E.
First, observe that, for any
$\delta>0$
, there is a set of positive integers S with density at most
$\delta $
such that, for each irrational
$\alpha>0$
, we have
$\Delta (S \cup \alpha S)=0$
. To see this, we can take, for example, an integer
$b>{1}/{\delta }$
and
$S=\{bk \mid k \in {\mathbb N}\}$
. Then the set S has density
$d(S)=1/b<\delta $
. Also, by (1.1), for each irrational number
$\alpha>0$
there are infinitely many pairs
$(m,n) \in {\mathbb N}^2$
for which
$$ \begin{align*}|bm\alpha-bn|<\frac{b}{\sqrt{5}m}.\end{align*} $$
For any
$\varepsilon>0$
, selecting
$m>{b}/{\varepsilon \sqrt {5}}$
, we see that
$0<|bm\alpha -bn|<\varepsilon $
with
$bm, bn \in S$
. Hence,
$\Delta (S \cup \alpha S)=0$
, as claimed. In this direction, it would be of interest to determine whether or not there is a set
$S \subseteq {\mathbb N}$
with density zero such that
$\Delta (S \cup \alpha S)=0$
for each irrational
$\alpha $
.
In this paper, we investigate the problem in the opposite direction. First, we show that there is a ‘large’ set S (much greater than the set of squares
$\{n^2 \mid n \in {\mathbb N}\}$
with density zero) for which we have option (1.5).
Theorem 1.1. For each
$\delta>0$
and each sufficiently large real number
$\alpha $
, there is a set of positive integers S with lower density greater than
$1-\delta $
such that
$$ \begin{align} \Delta(S \cup \alpha S) \geq \Delta\bigg(\bigcup_{k=0}^{\infty} \alpha^k S \bigg) \geq 1. \end{align} $$
Second, we prove that every set
$S \subseteq {\mathbb N}$
with density
$1$
satisfies option (1.4).
Theorem 1.2. If S is a set of positive integers with density
$1$
, then, for each irrational number
$\alpha>0$
, we have
$\Delta (S \cup \alpha S)=0$
.
One can also consider approximation weaker than that in (1.2), namely, for a given
$S \subseteq {\mathbb N}$
, investigate whether or not, for each
$\alpha>0$
and any
$\varepsilon>0$
, there are
$m,n \in S$
for which
$$ \begin{align}\bigg|\alpha-\frac{n}{m}\bigg|<\varepsilon.\end{align} $$
For example, for the set of primes
$S=P$
, this problem has been considered in [Reference Hobby and Silberger9]. It was shown there that the quotients of primes are everywhere dense in
$[0,\infty )$
, so each
$\alpha>0$
can be approximated as in (1.9) by a quotient of two primes
$n/m$
. The density of the sequence of rational numbers of the form
$b^m/m$
modulo one, where
$b \geq 2$
is a fixed integer and m runs through the set
${\mathbb N}$
, and similar sequences, have been considered in [Reference Cilleruelo, Kumchev, Luca, Rué and Shparlinski3, Reference Dubickas5, Reference Dubickas6, Reference Lind13].
The proofs of Theorems 1.1 and 1.2 will be given in Sections 2 and 3, respectively. In fact, the irrationality of
$\alpha $
is not relevant in Theorem 1.2. We show that if
$S \subseteq {\mathbb N}$
is a set with density
$1$
, then, for each rational
$\alpha>0$
,
$$ \begin{align} m\alpha-n=0 \end{align} $$
for infinitely many pairs
$(m,n) \in S^2$
(see the end of Section 3).
2 Proof of Theorem 1.1
By the definition of
$\Delta $
in (1.3), it is clear that
$\Delta (E) \geq \Delta (F)$
whenever
$E \subseteq F$
. Since
$S \cup \alpha S$
is a subset of
$\bigcup _{k=0}^{\infty } \alpha ^k S$
, this immediately implies the first inequality in (1.8).
In order to prove the second inequality in (1.8), we fix
$\delta $
in the interval
$(0,1)$
and a real number
$\alpha $
satisfying
$$ \begin{align} \alpha> \frac{3}{\delta}+1. \end{align} $$
We begin the construction of an infinite set
$S=\{s_1<s_2<s_3<\cdots \}$
depending on
$\alpha $
by selecting
$s_1=1$
. Assume that, for some
$m \in {\mathbb N}$
, we have already chosen the first m elements
$s_1<s_2<\cdots <s_m$
of S. The next element
$s_{m+1}$
is always taken as the least positive integer that is greater than
$s_{m}$
and is not equal to any of the numbers
$$ \begin{align} \lfloor \alpha^k s_j \rfloor, \quad \lceil \alpha^k s_j \rceil, \quad \text{where} \>\> k \in {\mathbb N} \>\> \text{and} \>\> j=1,\ldots,m. \end{align} $$
To see that the integers in (2.2) do not occupy all integers greater than
$s_m$
and that such an
$s_{m+1}>s_m$
always exists, we choose
$t=t(m) \in {\mathbb N}$
so large that
$\alpha ^t>s_m+2tm+1$
. (This is possible because
$\alpha>1$
.) Then, for
$k \geq t$
, the numbers in (2.2) are all greater than or equal to
$$ \begin{align*}\lfloor \alpha^k \rfloor>\alpha^k-1 \geq \alpha^t-1>s_m+2tm,\end{align*} $$
while for k in the range
$1 \leq k \leq t-1$
, there are at most
$2m(t-1)<2mt$
integers of the form (2.2). So, for each
$m \in {\mathbb N}$
, it is always possible to choose the required integer
$s_{m+1}$
in the interval
$[s_m+1,s_m+2tm]$
; therefore, the set S is infinite.
We claim that, for this set S, the distance between any two distinct elements of the set
$$ \begin{align*}S_{\alpha}:=\bigcup_{k=0}^{\infty} \alpha^k S\end{align*} $$
is at least
$1$
. Indeed, take
$x=\alpha ^u s_i \in S_{\alpha }$
and
$y=\alpha ^v s_j \in S_{\alpha }$
, where
$u,v \in {\mathbb N} \cup \{0\}$
and
$i,j \in {\mathbb N}$
. Assume that
$x \ne y$
. Then
$|x-y| \geq 1$
in the case when
$u=v$
, since
$i \ne j$
and
${|x-y|=\alpha ^u|s_i-s_j|}$
. Assume that
$u \ne v$
. Without restriction of generality, we may assume that
$u<v$
. Setting
$w:=v-u \in {\mathbb N}$
, we find that
$$ \begin{align*}|x-y|=|\alpha^u s_i-\alpha^v s_j|=\alpha^u|s_i-\alpha^{w}s_j| \geq |\alpha^w s_j-s_i|.\end{align*} $$
Now, in the case when
$i \leq j$
, using (2.1) and
$s_j \geq s_i$
, we deduce that
$$ \begin{align*}|\alpha^w s_j-s_i| = \alpha^w s_j -s_i \geq \alpha^w s_j - s_j \geq \alpha^w-1 \geq \alpha-1>\frac{3}{\delta}>3,\end{align*} $$
so
$|x-y|>3$
. In the case when
$i>j$
, by (2.2),
$s_i$
is neither
$\lfloor \alpha ^w s_j \rfloor $
nor
$\lceil \alpha ^w s_j \rceil $
. Thus, the distance between
$\alpha ^w s_j$
and
$s_i \in {\mathbb N}$
is greater than or equal to
$1$
, that is,
${|\alpha ^w s_j-s_i| \geq 1}$
. This yields
$|x-y| \geq 1$
and implies that
$\Delta (S_{\alpha }) \geq 1$
, which is the second inequality in (1.8).
It remains to show that the lower density of S is greater than
$1-\delta $
. Let
$x \geq \alpha $
be a real number. Choose the unique
$\ell \in {\mathbb N}$
for which
$\alpha ^{\ell } \leq x+1 < \alpha ^{\ell +1}$
. We derive a lower bound for the number of elements of S in the interval
$(x/\alpha ,x]$
. By (2.2), an integer in this interval belongs to S if and only if it is not of the form
$\lfloor \alpha ^k s_j \rfloor $
or
$\lceil \alpha ^k s_j \rceil $
for some
$k \in {\mathbb N}$
and some
$j \in {\mathbb N}$
. Note that it is sufficient to consider k in the range
$ 1 \leq k \leq \ell $
, since, otherwise, when
$k>\ell $
,
$$ \begin{align*} \lceil \alpha^k s_j \rceil \geq \lfloor \alpha^k s_j \rfloor \geq \lfloor \alpha^{k} \rfloor \geq \lfloor \alpha^{\ell+1} \rfloor> \alpha^{\ell+1}-1>x.\end{align*} $$
Fix
$k \in \{1,\ldots ,\ell \}$
. For this k, at least one of the numbers
$\lfloor \alpha ^k s_j \rfloor $
,
$\lceil \alpha ^k s_j \rceil $
belongs to the interval
$(x/\alpha ,x]$
only if j is such that
$x/\alpha <\lceil \alpha ^k s_j \rceil $
or j is such that
$\lfloor \alpha ^k s_j \rfloor \leq x$
. The first inequality does not hold if
$$ \begin{align*}x \geq \alpha \lceil \alpha^k s_j \rceil \geq \alpha^{k+1} s_j,\end{align*} $$
while the second inequality does not hold if
$$ \begin{align*}x< \lfloor \alpha^k s_j \rfloor \leq \alpha^k s_j.\end{align*} $$
Consequently, at least one of the inequalities
$x/\alpha <\lceil \alpha ^k s_j \rceil $
or
$\lfloor \alpha ^k s_j \rfloor \leq x$
holds only if j is such that
$$ \begin{align}\frac{x}{\alpha^{k+1}}<s_j \leq \frac{x}{\alpha^k}.\end{align} $$
Fix a pair of positive integers
$(k,j)$
for which (2.3) is true. Recall that
$1 \leq k \leq \ell $
. The pair
$(k,j)$
prevents at most two integers
$\lfloor \alpha ^k s_j \rfloor $
,
$\lceil \alpha ^k s_j \rceil $
in the interval
$(x/\alpha ,x]$
from belonging to the set S. Evidently, for each
$k \in \{1,\ldots ,\ell \}$
, there are at most
$x/\alpha ^k$
indices j satisfying (2.3). So, the collection of all relevant pairs
$(k,j)$
, where
$k=1,\ldots ,\ell $
and j satisfies (2.3), prevents at most
$$ \begin{align*}2\sum_{k=1}^{\ell} \frac{x}{\alpha^k}<2\sum_{k=1}^{\infty} \frac{x}{\alpha^k}=\frac{2x}{\alpha-1}\end{align*} $$
integers of the interval
$(x/\alpha ,x]$
from being in S. It follows that the intersection
${S \cap (x/\alpha ,x]}$
contains at least
$$ \begin{align*}\lfloor x \rfloor - \lfloor x/\alpha \rfloor -1 -\frac{2x}{\alpha-1}> x-2- \frac{x}{\alpha}-\frac{2x}{\alpha-1}=x\bigg(1-\frac{1}{\alpha}-\frac{2}{\alpha-1}\bigg)-2\end{align*} $$
elements. Therefore,
$$ \begin{align*}\underline{d}(S) =\liminf_{x \to \infty} \frac{\#\{S \cap [1,x]\}}{x} &\geq \liminf_{x \to \infty} \frac{\#\{S \cap (x/\alpha,x]\}}{x}\\ &\geq 1-\frac{1}{\alpha}-\frac{2}{\alpha-1}>1-\frac{3}{\alpha-1},\end{align*} $$
which is greater than
$1-\delta $
in view of (2.1).
3 Proof of Theorem 1.2
Let S be a set of positive integers with density
$1$
and let
$\alpha>0$
be an irrational number. It is sufficient to prove that
$$ \begin{align}\liminf_{m,n \in S} |m\alpha-n|=0\end{align} $$
for each irrational
$\alpha $
in the range
$0<\alpha <1$
. Indeed, for irrational
$\alpha>1$
, applying (3.1) to the number
$\alpha ^{-1} \in (0,1)$
, by
$|m\alpha ^{-1}-n|=\alpha ^{-1}|m-n\alpha |$
, we deduce that
$$ \begin{align*}\liminf_{m,n \in S} |m-n\alpha|=0,\end{align*} $$
and hence
$\Delta (S \cup \alpha S)=0$
.
So, from now on, we assume that
$0<\alpha <1$
. Let
$\varepsilon $
be in the range
$$ \begin{align*} 0<\varepsilon<\tfrac{1}{9}. \end{align*} $$
Throughout, we consider positive integers n satisfying
$$ \begin{align} n> \frac{3}{\varepsilon} \quad \text{and} \quad n > \frac{1}{1-\alpha}. \end{align} $$
Assume that the nth and the
$(n+1)$
st convergents of the continued fraction of
$\alpha $
are
$h_n/k_n$
and
$h_{n+1}/k_{n+1}$
(here
$h_n,k_n,h_{n+1},k_{n+1} \in {\mathbb N}$
), which means that
$$ \begin{align}\bigg|\alpha-\frac{h_n}{k_n}\bigg|<\frac{1}{k_n k_{n+1}} \quad \text{and} \quad \bigg|\alpha-\frac{h_{n+1}}{k_{n+1}}\bigg|<\frac{1}{k_{n+1} k_{n+2}}\end{align} $$
(see [Reference Coppel4, page 181]). Let
$u,v$
be positive integers satisfying
$$ \begin{align} u \leq \varepsilon k_{n+1} \quad \text{and} \quad v \leq \varepsilon k_n.\end{align} $$
(Such integers exist, because
$\varepsilon k_{n+1} \geq \varepsilon (k_n+k_{n-1})> \varepsilon k_n \geq \varepsilon n > 3$
by the first inequality in (3.2).) Consider the rational number
$$ \begin{align*}\mu:=\frac{uh_n+vh_{n+1}}{uk_n+vk_{n+1}}.\end{align*} $$
It is well known that
${h_n}/{k_n}<\alpha <{h_{n+1}}/{k_{n+1}}$
for even n and
${h_{n+1}}/{k_{n+1}}<\alpha <{h_{n}}/{k_{n}}$
for odd n (see [Reference Coppel4, page 181]). In both cases, the numbers
$\alpha $
and
$\mu $
are between the fractions
${h_n}/{k_n}$
and
${h_{n+1}}/{k_{n+1}}$
. Therefore, by the identity
$$ \begin{align} h_{n+1}k_n-h_nk_{n+1}=(-1)^n \end{align} $$
(see [Reference Coppel4, page 180]), we derive
$$ \begin{align*}\bigg|\alpha-\frac{uh_n+vh_{n+1}}{uk_n+vk_{n+1}}\bigg|=|\alpha-\mu| < \bigg|\frac{h_n}{k_n}-\frac{h_{n+1}}{k_{n+1}}\bigg|=\frac{1}{k_n k_{n+1}}.\end{align*} $$
This, combined with (3.4), implies that, for
$$ \begin{align}s(u,v,n):=uk_n+vk_{n+1} \in {\mathbb N} \quad \text{and} \quad t(u,v,n):=uh_n+vh_{n+1} \in {\mathbb N}, \end{align} $$
we have
$$ \begin{align}|s(u,v,n)\alpha-t(u,v,n))|<\frac{uk_n+vk_{n+1}}{k_n k_{n+1}} =\frac{u}{k_{n+1}}+\frac{v}{k_n}\leq 2\varepsilon.\end{align} $$
Now, to complete the proof of the theorem it suffices to show that there are
${u,v, n \in {\mathbb N}}$
satisfying (3.2) and (3.4) such that the integers
$s(u,v,n)$
,
$t(u,v,n)$
as defined in (3.6) both belong to the set S.
Put
$$ \begin{align} L_n:=\lfloor 2\varepsilon k_n k_{n+1} \rfloor. \end{align} $$
We first show that, for infinitely many
$n \in {\mathbb N}$
,
$$ \begin{align} \# \{j \notin S, \> 1 \leq j \leq L_n\} \leq \varepsilon^2 L_n. \end{align} $$
Indeed, if the inequality opposite to (3.9) holds for all sufficiently large
$n \in {\mathbb N}$
, then
$$ \begin{align*}\frac{\# \{j \in S, \> 1\leq j \leq L_n\}}{L_n}<\frac{L_n-\varepsilon^2 L_n }{L_n}=1-\varepsilon^2,\end{align*} $$
and hence
$$ \begin{align*}\underline{d}(S) =\liminf_{x \to \infty} \frac{\#\{S \cap [1,x]\}}{x} \leq \liminf_{n \to \infty} \frac{\# \{j \in S, \> 1 \leq j \leq L_n\}}{L_n} \leq 1-\varepsilon^2,\end{align*} $$
which is contrary to
$d(S)=\underline {d}(S)=1$
.
We want to show that there are n satisfying (3.2) and
$u,v \in {\mathbb N}$
satisfying (3.4) such that
$s(u,v,n)$
and
$t(u,v,n)$
both belong to S. Take any
$n \in {\mathbb N}$
for which the inequalities (3.2) and (3.9) hold. Note that, by (3.4), (3.6) and (3.8), it follows that
$ s(u,v,n) \leq L_n$
. We claim that
$$ \begin{align} t(u,v,n)< s(u,v,n) \leq L_n. \end{align} $$
Indeed, by the first inequality in (3.3), we find that
$|k_n \alpha -h_n|<{1}/{k_{n+1}}$
. Hence,
${h_n<k_n \alpha +{1}/{k_{n+1}}}$
. By the second inequality in (3.2), we obtain
$$ \begin{align*}1 < (1-\alpha)n \leq (1-\alpha) k_n \leq (1-\alpha) k_n^2.\end{align*} $$
It follows that
$k_n\alpha +{1}/{k_{n+1}}<{1}/{k_n}+k_n\alpha < k_n$
, and hence
$h_n<k_n$
. By the same argument, from the second inequality in (3.3), we get
$h_{n+1}<k_{n+1}$
. Thus,
$$ \begin{align*}t(u,v,n)=uh_n+vh_{n+1}<uk_n+vh_{n+1}=s(u,v,n),\end{align*} $$
which completes the proof of (3.10) because
$s(u,v,n) \leq L_n$
.
By (3.10), the integers
$s(u,v,n)$
and
$t(u,v,n)$
are distinct. Assume that, for some two pairs of positive integers
$(u,v) \ne (u',v')$
satisfying (3.4), we have
${s(u,v,n)=s(u',v',n)}$
. This implies that
$uk_n+vk_{n+1}=u'k_n+v'k_{n+1}$
by (3.6). Hence,
$(u-u')k_n=(v'-v)k_{n+1}$
. By (3.5), the numbers
$k_n$
and
$k_{n+1}$
are coprime, which implies that
$k_n \mid (v'-v)$
. However, by (3.4),
$1 \leq v,v' \leq \varepsilon k_n<k_n$
, so this is only possible if
$v=v'$
. This forces
$u=u'$
, which is a contradiction. Therefore,
$s(u,v,n) \ne s(u',v',n)$
. By the same argument, we conclude that
$t(u,v,n) \ne t(u',v',n)$
.
We call a positive integer bad if it does not belong to the set S. Similarly, we call a pair of distinct integers
$(s(u,v,n),t(u,v,n))$
bad if at least one of those integers is bad. Let us consider all bad integers not exceeding
$L_n$
. Because of (3.9), there are at most
$\varepsilon ^2 L_n$
of them. By what we have just shown above, each of them occurs in at most two pairs
$(s(u,v,n),t(u,v,n))$
. (It may happen that
$s(u,v,n)$
is equal to
$t(u',v',n)$
for
$(u,v) \ne (u',v')$
.) So, by (3.8), at most
$$ \begin{align*}2\varepsilon^2 L_n \leq 4\varepsilon^3 k_n k_{n+1}\end{align*} $$
among the pairs under consideration are bad. Note that, by (3.4), there are exactly
$\lfloor \varepsilon k_{n+1}\rfloor \lfloor \varepsilon k_{n}\rfloor $
pairs
$(s(u,v,n),t(u,v,n))$
with
$u,v$
satisfying (3.4). Using
$\varepsilon k_{n+1}>\varepsilon k_n>3$
and
$0<\varepsilon <{1}/{9}$
, we deduce that
$$ \begin{align*} \lfloor \varepsilon k_{n+1}\rfloor\lfloor \varepsilon k_{n}\rfloor> (\varepsilon k_{n+1}-1)(\varepsilon k_n-1) >\frac{2\varepsilon k_{n+1}}{3} \cdot \frac{2\varepsilon k_{n}}{3} >4\varepsilon^3 k_n k_{n+1}.\end{align*} $$
Consequently, there is a pair
$(s(u,v,n),t(u,v,n))$
, where
$u,v$
satisfy (3.4), which is not bad. This means that these two positive integers
$s(u,v,n)$
,
$t(u,v,n)$
for which (3.7) is true are both in S, which is the desired conclusion. This completes the proof of Theorem 1.2.
Finally, to prove (1.10), we write
$\alpha ={u}/{v}$
, where
$u,v \in {\mathbb N}$
are coprime. The result is trivial for
$u=v=1$
, so assume that
$u \ne v$
. Take
$N \in {\mathbb N}$
and consider the N pairs
$(m,n)=(kv,ku)$
with
$k=1,\ldots ,N$
. As above, a positive integer is called bad if it does not belong to the set S. Since the density of S is
$1$
, for infinitely many
$N \in {\mathbb N}$
, the set
$\{1,2,\ldots ,N\max (u,v)\}$
contains at most
$N/4$
bad integers. Each of those bad integers can appear in at most two pairs
$(kv,ku)$
for
$k=1,2,\ldots ,N$
. So, for at least
${N-2N/4=N/2}$
indices k in the range
$1 \leq k \leq N$
, we must have
$m=kv \in S$
and
${n=ku \in S}$
. For each of those k and
$(m,n)=(kv,ku) \in S^2$
, we get
${m\alpha -n=kv({u}/{v})-ku=0}$
, as claimed in (1.10).
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