Subfield: Algebra


Question: 
Determine whether for every real number $t$ such that $0<t<\frac{1}{2}$ there exists an infinite set $S$ of positive integers such that

$$
|x-m y|>t y
$$

for every pair of different elements $x$ and $y$ of $S$ and every positive integer $m$ (i.e. $m>0$ ).

Solution:
We claim that an infinite set with the required property exists. To this end, we rewrite the required condition in the form

$$
\left|\frac{x}{y}-m\right|>t
$$

This is equivalent to saying that the distance between the ratio $x / y$ and the set of positive integers is greater than $t$.

Now we construct an increasing sequence $s_{n}$ of odd coprime positive integers satisfying

$$
\frac{1}{2}-\frac{1}{2 s_{n}}>t \quad \forall n \geq 1 \tag{Q6.2}
$$

and such that for every $j>i$ it turns out that

$$
\frac{s_{i}}{s_{j}}<\frac{1}{2} \quad \text { and } \quad t<\left\{\frac{s_{j}}{s_{i}}\right\}<\frac{1}{2} \tag{Q6.3}
$$

where $\{\alpha\}$ denotes the fractional part of $\alpha$. This is enough to show that the set $S:=\left\{s_{n}: n \geq 1\right\}$ has the required property.

To this end, we consider the sequence defined recursively by

$$
s_{n+1}=\frac{\left(s_{1} \cdot \ldots \cdot s_{n}\right)^{2}+1}{2}
$$

with $s_{1}$ large enough. An easy induction shows that this is an increasing sequence of odd positive integers. For every $i \in\{1, \ldots, n\}$ it turns out that

$$
\frac{s_{i}}{s_{n+1}} \leq \frac{2}{s_{i}} \leq \frac{2}{s_{1}}<\frac{1}{2}
$$

because $s_{1}$ is large enough, which proves the first relation in (Q6.3). Moreover, it turns out that

$$
\frac{s_{n+1}}{s_{i}}=\frac{\left(s_{1} \cdot \ldots \cdot s_{n}\right)^{2}}{2 s_{i}}+\frac{1}{2 s_{i}}
$$

The first term is a positive integer plus $1 / 2$, from which it follows that the distance of $s_{n+1} / s_{i}$ from the positive integers is greater than or equal to

$$
\frac{1}{2}-\frac{1}{2 s_{i}} \geq \frac{1}{2}-\frac{1}{2 s_{1}}
$$

which is greater than $t$ if $s_{1}$ is large enough. This proves the second relation in (Q6.3).
We claim that an infinite set with the required property exists. To this end, we rewrite the required condition in the form

$$
\left|\frac{x}{y}-m\right|>t
$$

This is equivalent to saying that the distance between the ratio $x / y$ and the set of positive integers is greater than $t$.

Now we construct an increasing sequence $s_{n}$ of odd coprime positive integers satisfying

$$
\frac{1}{2}-\frac{1}{2 s_{n}}>t \quad \forall n \geq 1 \tag{Q6.2}
$$

and such that for every $j>i$ it turns out that

$$
\frac{s_{i}}{s_{j}}<\frac{1}{2} \quad \text { and } \quad t<\left\{\frac{s_{j}}{s_{i}}\right\}<\frac{1}{2} \tag{Q6.3}
$$

where $\{\alpha\}$ denotes the fractional part of $\alpha$. This is enough to show that the set $S:=\left\{s_{n}: n \geq 1\right\}$ has the required property.

We produce an increasing sequence $s_{n}$ of odd and coprime positive integers that satisfies (Q6.3) every $j>i$. As in the previous solution, this is enough to conclude.

We argue by induction. To begin with, we choose $s_{1}$ to be any odd integer satisfying the inequality in (Q6.2). Let us assume now that $s_{1}, \ldots, s_{n}$ have already been chosen, and let us choose $s_{n+1}$ in such a way that

$$
s_{n+1} \equiv \frac{s_{i}-1}{2} \quad\left(\bmod s_{i}\right) \quad \forall i \in\{1, \ldots, n\}
$$



We can solve this system because the previously chosen integers are odd and coprime. Moreover, any solution of this system is coprime with $s_{1}, \ldots, s_{n}$. Indeed, for every $1 \leq i \leq n$ it turns out that

$$
s_{n+1}=\frac{s_{i}-1}{2}+k_{i} s_{i}
$$

for some positive integer $k_{i}$. Therefore, any prime $p$ that divides both $s_{n+1}$ and $s_{i}$ divides also $\left(2 k_{i}+1\right) s_{i}-2 s_{n+1}=1$, which is absurd. Finally, we observe that we can assume that $s_{n+1}$ is odd and large enough. In this way we can guarantee that

$$
\frac{s_{i}}{s_{n+1}}<\frac{1}{2} \quad \forall i \in\{1, \ldots, n\}
$$

which is the first requirement in (Q6.3), and

$$
k_{i}+t<k_{i}+\frac{1}{2}-\frac{1}{2 s_{i}}=\frac{s_{n+1}}{s_{i}}<k_{i}+\frac{1}{2} \quad \forall i \in\{1, \ldots, n\}
$$

which implies the second requirement in (Q6.3).
We claim that an infinite set with the required property exists. To this end, we rewrite the required condition in the form

$$
\left|\frac{x}{y}-m\right|>t
$$

This is equivalent to saying that the distance between the ratio $x / y$ and the set of positive integers is greater than $t$.

Now we construct an increasing sequence $s_{n}$ of odd coprime positive integers satisfying

$$
\frac{1}{2}-\frac{1}{2 s_{n}}>t \quad \forall n \geq 1 \tag{Q6.2}
$$

and such that for every $j>i$ it turns out that

$$
\frac{s_{i}}{s_{j}}<\frac{1}{2} \quad \text { and } \quad t<\left\{\frac{s_{j}}{s_{i}}\right\}<\frac{1}{2} \tag{Q6.3}
$$

where $\{\alpha\}$ denotes the fractional part of $\alpha$. This is enough to show that the set $S:=\left\{s_{n}: n \geq 1\right\}$ has the required property.

Again we produce an increasing sequence $s_{n}$ of positive integers that satisfies (Q6.3) every $j>i$.

To this end, for every positive integer $x$, we define its security region

$$
S(x):=\bigcup_{n \geq 1}\left((n+t) x,\left(n+\frac{1}{2}\right) x\right)
$$

The security region $S(x)$ is a periodic countable union of intervals of length $\left(\frac{1}{2}-t\right) x$, whose left-hand or right-hand endpoints form an arithmetic sequence. It has the property that

$$
t<\left\{\frac{y}{x}\right\}<\frac{1}{2} \quad \forall y \in S(x)
$$

Now we prove by induction that we can choose a sequence $s_{n}$ of positive integers satisfying (Q6.3) and in addition the fact that every interval of the security region $S\left(s_{n}\right)$ contains at least one interval of $S\left(s_{n-1}\right)$.

To begin with, we choose $s_{1}$ large enough so that the length of the intervals of $S\left(s_{1}\right)$ is larger than 1. This guarantees that any interval of $S\left(s_{1}\right)$ contains at least a positive integer. Now let us choose a positive integer $s_{2} \in S\left(s_{1}\right)$ that is large enough. This guarantees that $s_{1} / s_{2}$ is small enough, that the fractional part of $s_{2} / s_{1}$ is in $(t, 1 / 2)$, and that every interval of the security region $S\left(s_{2}\right)$ contains at least one interval of $S\left(s_{1}\right)$, and hence at least one positive integer.

Let us now assume that $s_{1}, \ldots, s_{n}$ have been already chosen with the required properties. We know that every interval of $S\left(s_{n}\right)$ contains at least one interval of $S\left(s_{n-1}\right)$, which in turn contains an interval in $S\left(s_{n-2}\right)$, and so on up to $S\left(s_{1}\right)$. As a consequence, we can choose a large enough positive integer $s_{n+1}$ that lies in $S\left(s_{k}\right)$ for every $k \in\{1, \ldots, n\}$. Since $s_{n+1}$ is large enough, we are sure that

$$
\frac{s_{k}}{s_{n+1}}<t \quad \forall k \in\{1, \ldots, n\}
$$

Moreover, we are sure also that all the intervals of $S\left(s_{n+1}\right)$ are large enough, and therefore they contain at least one interval of $S\left(s_{n}\right)$, which in turn contains at least one interval of $S\left(s_{n-1}\right)$, and so on. Finally, the condition

$$
t<\left\{\frac{s_{n-1}}{s_{n}}\right\}<\frac{1}{2}
$$

is guaranteed by the fact that $s_{n+1}$ was chosen in an interval that is contained in $S\left(s_{k}\right)$ for every $k \in\{1, \ldots, n\}$. This completes the induction.