No-three-in-line problem on a torus: periodicity

TL;DR

This paper investigates the maximum number of points on a discrete torus with no three collinear points, proving periodicity of certain sequences and providing bounds on their periods, extending known results in combinatorial geometry.

Contribution

It generalizes tools for determining u_{m,n} and proves the periodicity of the sequence u_{z,n} for fixed z > 1, with bounds on the period when z is a prime power.

Findings

u_{m,n} is known when d(m,n) is prime.

The sequence (u_{z,n}) is periodic for fixed z > 1.

Bounds on the period are established for prime power cases.

Abstract

Let $\tau_{m,n}$ denote the maximal number of points on the discrete torus (discrete toric grid) of sizes $m \times n$ with no three collinear points. The value $\tau_{m,n}$ is known for the case where $\gcd(m,n)$ is prime. It is also known that $\tau_{m,n} \leq 2\gcd(m,n)$. In this paper we generalize some of the known tools for determining $\tau_{m,n}$ and also show some new. Using these tools we prove that the sequence $(\tau_{z,n})_{n \in \mathbb{N}}$ is periodic for all fixed $z > 1$. In general, we do not know the period; however, if $z = p^a$ for $p$ prime, then we can bound it. We prove that $\tau_{p^a,p^{(a-1)p+2}} = 2p^a$ which implies that the period for the sequence is $p^b$ where $b$ is at most $(a-1)p+2$.

Figures

Tables

Table 1. Table 1: Initial values of τ z , n subscript 𝜏 𝑧 𝑛 \tau_{z,n} .

$z$	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20 …
2	2	4	2	4	2	4	2	4	2	4	2	4	2	4	2	4	2	4	2	4 …
3	2	2	4	2	2	4	2	2	6	2	2	4	2	2	4	2	2	6	2	2 …
4	2	4	2	6	2	4	2	8	2	4	2	6	2	4	2	8	2	4	2	6 …
5	2	2	2	2	6	2	2	2	2	6	2	2	2	2	6	2	2	2	2	6 …
6	2	4	4	4	2	8	2	4	6	4	2	8	2	4	4	4	2	10	2	4 …

Table 2. Table 2: Values of σ 6 ​ ( n ) subscript 𝜎 6 𝑛 \sigma_{6}(n) according to gcd ⁡ ( n , 2 2 ⋅ 3 3 ) = 2 i ⋅ 3 j 𝑛 ⋅ superscript 2 2 superscript 3 3 ⋅ superscript 2 𝑖 superscript 3 𝑗 \gcd(n,2^{2}\cdot 3^{3})=2^{i}\cdot 3^{j} .

$\gcd (n,2^2\cdot 3^3)$	$2^0$	$2^1$	$2^2$
$3^0$	2	4	4
$3^1$	4	8	8
$3^2$	6	10	12
$3^3$	6	12	12

Equations

$$x$$

$$u^{\ast }k$$

$$v^{\ast }k$$

$$u^{\ast }$$

$$v^{\ast }$$

$$b_1$$

$$b_2$$

$$b_1$$

$$b_2$$

$$v^{\ast }{b}_1$$

$$u^{\ast }{b}_2$$

$$\gcd \left(v^{\ast }m,u^{\ast }n\right)$$

$$=g_{u,m}{g}_{v,n}\cdot \gcd \left(\frac{m}{g_{u,m}},\frac{n}{g_{v,n}}\right)=g_{u,m}{g}_{v,n}\cdot \gcd \left(x\frac{m}{g_{u,m}},y\frac{n}{g_{v,n}}\right)$$

$$=\gcd \left(g_{v,n}{p}_{v}x{g}_{u,m}\frac{m}{g_{u,m}},g_{u,m}{p}_{u}y{g}_{v,n}\frac{n}{g_{v,n}}\right)=\gcd \left(v^{\ast }xm,u^{\ast }yn\right).$$

$$s$$

$$u^{\ast }k$$

$$v^{\ast }k$$

$$\forall J\subseteq I:{\sigma }_z\left(\prod _{i\in \overline{J}}{p}_i^b\prod _{i\in J}{p}_i^{c_i}\right)={\sigma }_z\left(\prod _{i\in \overline{J}}{p}_i^{d_i}\prod _{i\in J}{p}_i^{c_i}\right)$$

$${\mathcal{C}}^{(b)}=\left(J^{(b)},C^{(b)},{\delta }^{(b)}\right),\text{where}{J}^{(b)}\subseteq I,C^{(b)}=(c_i^{(b)}{)}_{i\in J^{(b)}},{\delta }^{(b)}\ge b,$$

$${\sigma }_z\left(\prod _{i\in \overline{J^{(b)}}}{p}_i^b\prod _{i\in J^{(b)}}{p}_i^{c_i^{(b)}}\right)<{\sigma }_z\left(\prod _{i\in \overline{J^{(b)}}}{p}_i^{{\delta }^{(b)}}\prod _{i\in J^{(b)}}{p}_i^{c_i^{(b)}}\right).$$

$${\sigma }_z\left(\prod _{i\in \overline{J}}{p}_i^{t_b}\prod _{i\in J}{p}_i^{c_i^{(t_b)}}\right)<{\sigma }_z\left(\prod _{i\in \overline{J}}{p}_i^{{\delta }^{(t_b)}}\prod _{i\in J}{p}_i^{c_i^{(t_b)}}\right)=S$$

$$x+\alpha m_z=\prod _{i\in J}{p}_i^{c_i}\prod _{i\in K\cup L}{p}_i^b\left(r\prod _{i\in L}{p}_i^{c_i-b}+\alpha \prod _{i\in J}{p}_i^{b-c_i}\right).$$

$$x+\alpha m_z=h\left(\prod _{i\in J}{p}_i^{c_i}\prod _{i\in K\cup L}{p}_i^{d_i}\right)$$

$${\sigma }_z\left(x+\alpha m_z\right)$$

$$\displaystyle\stackrel{{\scriptstyle\mathclap{\mbox{$\triangle$}}}}{{=}}\sigma_{z}\left(\prod_{i\in J}p_{i}^{c_{i}}\prod_{i\in K\cup L}p_{i}^{c_{i}}\right)=\sigma_{z}\left(\prod_{i\in I}p_{i}^{c_{i}}\right)=\sigma_{z}(x),$$

$$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ 1& 1& 1\end{array}\right|.$$

$${\text{ord}}_{{\mathbb{Z}}_m\times {\mathbb{Z}}_n}(u,v)=\text{lcm}({\text{ord}}_{{\mathbb{Z}}_m}(u),{\text{ord}}_{{\mathbb{Z}}_n}(v)).$$

$${\text{ord}}_{{\mathbb{Z}}_x}(h)=\frac{\mathrm{lcm}(x,h)}{h}=\frac{\frac{xh}{\gcd (x,h)}}{h}=\frac{x}{\gcd (x,h)}.$$

$$k{b}_1$$

$$k{b}_2$$

$$k{v}_1$$

$$k{v}_2$$

$$g^{-1}k{v}_1$$

$$g^{-1}k{v}_2$$

$$\mathrm{lcm}\left(\frac{p^a}{\gcd (p^a,v)},\frac{p^b}{\gcd (p^b,w{p}^r)}\right)=\mathrm{lcm}\left(p^a,\frac{p^b}{p^r}\right)=\max \left\{p^a,\frac{p^b}{p^r}\right\}=\max \left\{p^a,\frac{p^b}{\gcd (p^b,a_2-b_2)}\right\}.$$

$$w_1$$

Full text

In this paper we generalize some of the known tools for determining $\tau_{m,n}$ and also show some new. Using these tools we prove that the sequence $(\tau_{z,n})_{n\in\mathbb{N}}$ is periodic for all fixed $z>1$. In general, we do not know the period; however, if $z=p^{a}$ for $p$ prime, then we can bound it. We prove that $\tau_{p^{a},p^{(a-1)p+2}}=2p^{a}$ which implies that the period for the sequence is $p^{b}$, where $b$ is at most $(a-1)p+2$.

1 Introduction

In 1917, Dudeney introduced the No-three-in-line-problem (see problem 317 in [Dud17]). The goal of this problem is to place as many points as possible on an $n\times n$ grid so that no three points are collinear. This problem is still not solved for all $n\in\operatorname{\mathbb{N}}$.

In 2012, Fowler, Groot, Pandya and Snapp introduced a modification of this problem: No-three-in-line-problem on a torus (see [FGPS12]). The purpose is to place as many points as possible on a discrete torus of size $m\times n$, where $m,n\in\operatorname{\mathbb{N}}$ so that no three of these points are collinear.

We consider the discrete torus as a Cartesian product $\{0,\ldots,m-1\}\times\{0,\ldots,n-1\}\subset\operatorname{\mathbb{Z}}^{2}$ and we denote it $T_{{m}\times{n}}$. A line on this torus is an image of a line $\ell$ in $\operatorname{\mathbb{Z}}^{2}$ under a mapping $\pi_{m,n}$ which maps a point $(x,y)\in\operatorname{\mathbb{Z}}^{2}$ to the point $(x\mod{m},y\mod{n})$.111 As a line in $\operatorname{\mathbb{Z}}^{2}$ we consider $\ell_{\operatorname{\mathbb{R}}^{2}}\cap\operatorname{\mathbb{Z}}^{2}$, where $\ell_{\operatorname{\mathbb{R}}^{2}}$ is a line in $\operatorname{\mathbb{R}}^{2}$ which contains at least two points of integer coordinates.

We are interested in estimating the quantity $\tau_{m,n}$ which denotes the maximal number of points that can be placed on the torus $T_{{m}\times{n}}$ so that no three of these points are collinear.222From another point of view, we may identify $T_{{m}\times{n}}$ with the group $\operatorname{\mathbb{Z}}_{m}\times\operatorname{\mathbb{Z}}_{n}$. Then lines are cosets of cyclic subgroups $\operatorname{\mathbb{Z}}_{m}\times\operatorname{\mathbb{Z}}_{n}$ generated by $(x,y)$ where $x,y$ are relatively prime.

Fowler et al. in [FGPS12] proved that $\tau_{p,p}=p+1$ and $\tau_{p,p^{2}}=2p$ where $p$ is an odd prime. They also showed $\tau_{m,n}=2$ whenever $\gcd(m,n)=1$, where $\gcd$ denotes the greatest common divisor. This fact follows directly from the Chinese remainder theorem. These results were generalized by Misiak, Stȩpień, A. Szymaszkiewicz, L. Szymaszkiewicz and Zwierzchowski [MSS*+*16] who determined $\tau_{m,n}$ whenever $\gcd(m,n)=p$ is a prime. More concretely, they have shown in this case that $\tau_{m,n}=2p$ if $m$ or $n$ is divisible by $p^{2}$ or if $p=2$ and that $\tau_{m,n}=p+1$ if neither $m$ nor $n$ is divisible by $p^{2}$ and $p$ is an odd prime; see Theorem 1.2. in [MSS*+*16].

Misiak et al. also provided a useful general upper bound for $\tau_{m,n}$.

Theorem 1** (Theorem 1.1 in [MSS*+*16]).**

Let $m,n\in\operatorname{\mathbb{N}}$. Then $\tau_{m,n}\leq 2\gcd(m,n)$.

Results of the paper: Periodicity.

We contribute to the study of $\tau_{m,n}$ by showing that once we fix one of the coordinates, then we get a periodic sequence. Namely, for a positive integer $z$, we define the sequence $\sigma_{z}=(\sigma_{z}(n))_{n=1}^{\infty}$ by setting $\sigma_{z}(n):=\tau_{z,n}$. Note that $\sigma_{z}(n)\leq 2n$ for arbitrary $n$ by Theorem 1. In particular, $\sigma_{z}$ attains only finitely many values. We get following:

Theorem 2**.**

The sequence $\sigma_{z}$ is periodic for all positive integers $z$ greater than 1.

Theorem 2 in principle allows to determine the values $\tau_{z,n}$ for arbitrary big $n$ from several initial values of $n$ (the number of initial values for $n$ depends on $z$, of course). Unfortunately, for a general $z$ we do not know what is the period. Our proof is purely existential. However, if $z$ is a power of a prime, we can say more. We show that in such case the sequence reaches its potential maximum $2z=2p^{a}$ for $a\in\operatorname{\mathbb{N}}$ and a prime $p$.

Theorem 3**.**

Let $T_{{p^{a}}\times{p^{(a-1)p+2}}}$ be a torus where $p$ is a prime and $a\in\operatorname{\mathbb{N}}$. Then $\tau_{p^{a},p^{(a-1)p+2}}=2p^{a}$.

We can also determine the period of $\sigma_{p^{a}}$.

Theorem 4**.**

Let $p$ be a prime, $a\in\operatorname{\mathbb{N}}$. Let us denote $m:=\min\{x;\sigma_{p^{a}}(x)=2p^{a}\}$. Then $m=p^{b}$ for some $b\geq a$ and the sequence $\sigma_{p^{a}}$ is periodic with the period $m$.

Note that Theorem 3 implies that such an $m$ exists. In fact, we show that each $x$ such that $\sigma_{p^{a}}(x)=2p^{a}$ is a period for $\sigma_{p^{a}}$. Therefore, from these two theorems above we get that $\sigma_{p^{a}}$ is periodic with the (not necessarily least) period $p^{(a-1)p+2}$.

Tools.

A useful tool for determining $\tau_{m,n}$ is to find out for which values $m,n,x,y\in\operatorname{\mathbb{N}}$ we have $\tau_{m,n}=\tau_{xm,yn}$. Misiak et al. showed $\tau_{p,p}=\tau_{xp,yp}$ if $p$ is odd prime, $x,y$ are relatively prime and neither $x$ nor $y$ is divisible by $p$. In other words $\tau_{p,p}=\tau_{a,b}$ if $\gcd(a,b)=p$ and neither $a$ nor $b$ is divisible by $p^{2}$; see Theorem 4.5 in [MSS*+*16].

We generalize this result to the following one.

Theorem 5**.**

Let $m,n,x,y$ be positive integers such that $m,n$ are not both 1 and $\gcd(x,y)=\gcd(m,y)=\gcd(n,x)=1$. Then $\tau_{xm,yn}=\tau_{m,n}$.

Misiak et al. used the following idea in [MSS*+*16]. Let us consider tori $T_{{xm}\times{yn}}$ and $T_{{m}\times{n}}$, where $m,n,x,y\in\operatorname{\mathbb{N}}$, a mapping $f:T_{{xm}\times{yn}}\rightarrow T_{{m}\times{n}}$ defined by $f((a_{1},a_{2}))=(a_{1}\mod{m},a_{2}\mod{n})$, and the mappings $\pi_{xm,yn}$, $\pi_{m,n}$ defined above from $\operatorname{\mathbb{Z}}^{2}$ to $T_{{xm}\times{yn}}$, $T_{{m}\times{n}}$, respectively. Then $\pi_{m,n}=f\circ\pi_{xm,yn}$. Hence the image of every line on $T_{{xm}\times{yn}}$ under the mapping $f$ is a line on $T_{{m}\times{n}}$. Consequently, a set of points on $T_{{m}\times{n}}$ such that no three are collinear can be also used on $T_{{xm}\times{yn}}$. Indeed, if such points were collinear on $T_{{xm}\times{yn}}$ they would be collinear also on $T_{{m}\times{n}}$ since the image of every line on $T_{{xm}\times{yn}}$ is a line on $T_{{m}\times{n}}$. This leads to the following result.

Lemma 6** (Lemma 4.1(1) in [MSS*+*16]).**

Let $m,n,x,y$ be positive integers. Then $\tau_{xm,yn}\geq\tau_{m,n}$.

Theorem 5 and Lemma 6 are our main tools for proving Theorem 2 and Theorem 4.

Small values of the sequence $\sigma_{z}(n)$.

For sake of example, we present here few initial values of $\tau_{z,n}$, our table is analogous to the table in [FGPS12]. See Table 1.

The values $\sigma_{p}(n)$ for a prime number $p$ can be fully determined from Theorem 1.2 in [MSS*+*16]. In general, it follows from their result that the sequence $\sigma_{p}(n)$ is periodic with the period $p^{2}$ for a prime number $p$.

Considering $\sigma_{4}(n)$, we can determine the values $\sigma_{4}(n)$ again by Theorem 1.2 in [MSS*+*16] for $n\in\{1,2,3,5,6,7\}$. We also have a computer assisted proof that $\sigma_{4}(4)=6$.333[FGPS12] also claim this value without a detailed proof; with a moderate effort it is also possible to get a computer-free proof. In particular, a configuration of points showing $\sigma_{4}(4)\geq 6$ is shown in Figure 1. We can also easily deduce that $\sigma_{4}(8)=8$: due to Theorem 1, it is sufficient to show that $\sigma_{4}(8)\geq 8$ which follows from the example in Figure 1. Given that $\sigma_{4}(n)=8$, Theorem 4 implies that $\sigma_{4}(n)$ is periodic with the period 8.

Regarding the values presented in the table (Table 1), we can determine exactly $\sigma_{6}(n)$ for $n\not\equiv 0\pmod{6}$ due to Misiak et al. (see [MSS*+*16]). For other values we rely on the computer proof. It also gives that $\sigma_{6}(36)=\sigma_{6}(54)=12$, which is the maximum of $\sigma_{6}(n)$ by Theorem 1, and $\sigma_{6}(24)=8$.

Unfortunately, we are not able to determine the least period for $\sigma_{6}(n)$. However, we conjecture $\sigma_{6}(2^{k}\cdot 3)=8$ for all $k\in\operatorname{\mathbb{N}}$.444We are able to prove it is true for $k\leq 5$ using computer. If it is true then we can determine all values of $\sigma_{6}(n)$ and the least period is $2^{2}\cdot 3^{3}=108$; See Table 2. Indeed, by Theorem 5, Lemma 6 and by known initial values we get $\sigma_{6}(n)=2$ whenever $\gcd(n,2^{2}\cdot 3^{3})=1$, $\sigma_{6}(n)=4$ whenever $\gcd(n,2^{2}\cdot 3^{3})=3^{1},2^{1},2^{2}$ or $2^{3}$, $\sigma_{6}(n)=6$ whenever $\gcd(n,2^{2}\cdot 3^{3})=3^{2}$ or $3^{3}$, $\sigma_{6}(n)=10$ whenever $\gcd(n,2^{2}\cdot 3^{3})=2\cdot 3^{2}$, $\sigma_{6}(n)=12$ whenever $\gcd(n,2^{2}\cdot 3^{3})=2^{2}\cdot 3^{2},2\cdot 3^{3}$ or $2^{2}\cdot 3^{3}$. Finally, $\sigma_{6}(n)=8$ whenever $\gcd(n,2^{2}\cdot 3^{3})=2\cdot 3,2^{2}\cdot 3$ by our conjecture.

2 Proof of Theorem 5

In this section, we prove Theorem 5. For this purpose we need the following two well-known theorems.

Theorem 7** (Chinese Remainder Theorem, see [How02]).**

Let $m,n$ be positive integers. Then two simultaneous congruences

[TABLE]

are solvable if and only if $a\equiv b\pmod{\gcd(m,n)}$. Moreover, the solution is unique modulo $\operatorname{\textup{lcm}}(m,n)$, where $\operatorname{\textup{lcm}}$ denotes the least common multiple.

Theorem 8** (Dirichlet’s Theorem, see [Sel49]).**

Let $a,b$ be positive relatively prime integers. Then there are infinitely many primes of the form $a+nb$, where $n$ is a non-negative integer.

We again use the idea of Misiak et al.

Lemma 9** (essentially Lemma 4.1(2) in [MSS*+*16]).**

Let $T_{{xm}\times{yn}}$, $T_{{m}\times{n}}$ be tori, where $m,n,x,y\in\operatorname{\mathbb{N}}$ and $m,n$ are not both 1. Let $f\colon T_{{xm}\times{yn}}\to T_{{m}\times{n}}$ be a mapping defined by $f((a_{1},a_{2}))=(a_{1}\mod{m},a_{2}\mod{n})$. If the preimage of every line on $T_{{m}\times{n}}$ is a line on $T_{{xm}\times{yn}}$, then $\tau_{xm,yn}=\tau_{m,n}$.

Proof.

The Lemma follows from Lemma 6 if $\tau_{xm,yn}=2$. Suppose $\tau_{xm,yn}>2$. Let $M\subset T_{{xm}\times{yn}}$ be a maximal set such that no three points from $M$ are collinear. We show that the image $f(M)$ on $T_{{m}\times{n}}$ also satisfies the no-three-in-line condition.

First of all, note that $|M|=|f(M)|$. Indeed, if $f(A)=f(B)$ for two distinct $A,B\in M$, then $A,B$ and $C$ would be collinear for some point $C\in M\setminus\{A,B\}$.

Now, let us consider three distinct points from $f(M)$. If they were collinear, their preimages would be also collinear since the preimage of every line on $T_{{m}\times{n}}$ is a line on $T_{{xm}\times{yn}}$. Consequently, we have $\tau_{xm,yn}\leq\tau_{m,n}$ and by Lemma 6 we have $\tau_{xm,yn}\geq\tau_{m,n}$. ∎

Now, we generalize Lemmas 4.3 and 4.4 from [MSS*+*16].

Lemma 10**.**

Let $x,y,m,n\in\operatorname{\mathbb{N}}$ such that $\gcd(x,y)=\gcd(m,y)=\gcd(n,x)=1$. Let $T_{{xm}\times{yn}}$, $T_{{m}\times{n}}$ be tori and let $f\colon T_{{xm}\times{yn}}\to T_{{m}\times{n}}$ be a mapping defined by $f((a_{1},a_{2}))=(a_{1}\mod{m},a_{2}\mod{n})$. Then the preimage of every line on $T_{{m}\times{n}}$ is a line on $T_{{xm}\times{yn}}$.

Proof.

First, note that it is sufficient to consider only lines which contain the origin. The other lines are just translations of such lines.

Let $\ell$ be a line on $T_{{m}\times{n}}$ which contains the origin. We can express it as $\ell=\{\pi_{m,n}\left(k\left(u,v\right)\right);k\in\operatorname{\mathbb{Z}}\}$ where $\gcd(u,v)=1$. We have to find $u^{\ast},v^{\ast}\in\operatorname{\mathbb{N}}_{0}$ such that $\gcd(u^{\ast},v^{\ast})=1$ defining the line $\ell^{\ast}=\{\pi_{xm,yn}\left(k(u^{\ast},v^{\ast})\right);k\in\operatorname{\mathbb{Z}}\}$ on $T_{{xm}\times{yn}}$ which it is the preimage of $\ell$. In other words, for every point $(a_{1},a_{2}$) from $\ell^{\ast}$ its image $f((a_{1},a_{2}))$ belongs to $\ell$ and for every point $(b_{1},b_{2})$ from $f^{-1}(\ell)$ the following equations have solution.

[TABLE]

Let us denote $g_{u,m}:=\gcd(u,m)$ and $g_{v,n}:=\gcd(v,n)$. Since $\frac{u}{g_{u,m}}$ and $\frac{m}{g_{u,m}}$ are relatively prime, by the Dirichlet’s Theorem (Theorem 8) there is $i$ defining a prime $p_{u}=\frac{u}{g_{u,m}}+i\frac{m}{g_{u,m}}$ which is greater than $mnxy$. At the same way, we can get $j$ defining another prime $p_{v}=\frac{v}{g_{v,n}}+j\frac{n}{g_{v,n}}$ greater than $mnxy$ such that $p_{v}\not=p_{u}$. This does not work if either $u$ or $v$ is zero (one of them is always non-zero). In such case, we set $i:=1$ or $j:=1$, respectively, and hence $p_{u}=1$ or $p_{v}=1$, respectively. Now, we set

[TABLE]

Note that $\gcd(g_{u,m},g_{v,n})=1$ and, therefore, $\gcd(u^{\ast},v^{\ast})=1$. Indeed, $g_{u,m}$ divides $u$, $g_{v,n}$ divides $v$ and $\gcd(u,v)=1$.

Since $u^{\ast}=u+im$ and $v^{\ast}+jn$, the image $f((a_{1},a_{2}))$ belongs to $\ell$ for every $(a_{1},a_{2})$ from $\ell^{\ast}$. It remains to check whether equalities (1), (2) have solution for every point $(b_{1},b_{2})$ from $f^{-1}(\ell)$.

Since $(b_{1},b_{2})\in f^{-1}(\ell)$ there is $t\in\operatorname{\mathbb{N}}_{0}$ such that

[TABLE]

By the definition of $u^{\ast}$ and $v^{\ast}$ (see (3), (4)) we have

[TABLE]

Therefore, $v^{\ast}b_{1}\equiv u^{\ast}b_{2}\pmod{\text{GCD}(v^{\ast}m,u^{\ast}n)}$.

Again, by the definition of $u^{\ast}$ and $v^{\ast}$ (see (3), (4)) we have

[TABLE]

Hence $v^{\ast}b_{1}\equiv u^{\ast}b_{2}\pmod{\gcd(v^{\ast}xm,u^{\ast}yn)}$.

Now, let us consider the following equations.

[TABLE]

Since $v^{\ast}b_{1}\equiv u^{\ast}b_{2}\pmod{\gcd(v^{\ast}xm,u^{\ast}yn)}$, these equations have solution by the Chinese Remainder Theorem (Theorem 7). Since $\gcd(u^{\ast},v^{\ast})=1$, $s=u^{\ast}v^{\ast}k$ for some $k\in\operatorname{\mathbb{Z}}$. Therefore,

[TABLE]

Consequently, equations (1), (2) have solution and the point $(b_{1},b_{2})\in f^{-1}(\ell)$ lies on $\ell^{\ast}$. We are done. ∎

Now, we are able to prove Theorem 5 using Lemmas 9 and 10.

Proof of Theorem 5.

Considering a mapping $f\colon T_{{xm}\times{yn}}\to T_{{m}\times{n}}$ defined by $f((a_{1},a_{2}))=(a_{1}\mod{m},a_{2}\mod{n})$, the preimage of every line on $T_{{m}\times{n}}$ is a line on $T_{{xm}\times{yn}}$ by Lemma 10. Therefore, $\tau_{xm,yn}=\tau_{m,n}$ by Lemma 9. ∎

3 Periodicity

In this section, we prove that the sequence $\sigma_{z}$ is periodic for all $z\in\operatorname{\mathbb{N}}$ (Theorem 4 and Theorem 2).

Before we start, we mention two lemmas which are used in the proofs of Theorem 4 and 2. They are special cases of Theorem 5 and Lemma 6, respectively.

Lemma 11**.**

Let $z,x,m\in\operatorname{\mathbb{N}}$ such that $z>1$ and $\gcd(x,z)=1$. Then $\sigma_{z}(m)=\sigma_{z}(xm)$.

Proof.

We may use Theorem 5 for the tori $T_{{m}\times{z}}$ and $T_{{xm}\times{z}}$. ∎

Lemma 12**.**

Let $z,x,m\in\operatorname{\mathbb{N}}$. Then $\sigma_{z}(m)\leq\sigma_{z}(xm)$.

Proof.

We use Lemma 6 for the tori $T_{{m}\times{z}}$ and $T_{{xm}\times{z}}$. ∎

First, we prove Theorem 4, that is, the sequence $\sigma_{p}$ is periodic with the period $m:=\min\{x;\sigma_{p^{a}}(x)=2p^{a}\}$ for $p$ prime.

Proof of Theorem 4.

Note that the existence of $m=\min\{x;\sigma_{p^{a}}(x)=2p^{a}\}$ follows from Theorem 3 which is proven in the following section (Section 4). More precisely, $m\leq p^{(a-1)p+2}$ by Theorem 3.

First, we observe that $m=p^{b}$ for some $b\in\operatorname{\mathbb{N}}$. Indeed, if $m$ were $hp^{b}$ for some $h>1$ such that $p$ does not divide $h$, then $\sigma_{p^{a}}(p^{b})=\sigma_{p^{a}}(hp^{b})$ by Lemma 11. This would contradict the minimality of $m$.

Let us consider an arbitrary $x\in\{1,\ldots,p^{b}\}$. We show that $\sigma_{p^{a}}(x)=\sigma_{p^{a}}(x+\alpha p^{b})$ for any $\alpha\in\operatorname{\mathbb{N}}$. Since $x\leq p^{b}$ we can express it as $x=rp^{l}$, where $0\leq l\leq b$ and $\gcd(r,p)=1$. By Lemma 11 $\sigma_{p^{a}}(x)=\sigma_{p^{a}}(p^{l})$. We consider two cases:

$x<p^{b}$.

In this case $x+\alpha p^{b}=p^{l}(r+\alpha p^{b-l})$. Since $b-l\geq 1$, we get $\gcd(r+\alpha p^{b-l},p)=1$. Therefore, $\sigma_{p^{a}}(p^{l}(r+\alpha p^{b-l}))=\sigma_{p^{a}}(p^{l})=\sigma_{p^{a}}(x)$ by Lemma 11. 2. 2.

$x=p^{b}$.

In this case $x+\alpha p^{b}=p^{b}(1+\alpha)$. Since $\sigma_{p^{a}}(p^{b})=2p^{a}=\max\sigma_{p^{a}}$, Lemma 12 implies $\sigma_{p^{a}}(p^{b})=\sigma_{p^{a}}(hp^{b})$ for any $h>0$ and hence also for $h=(1+\alpha)$.

We proved that $\sigma_{p^{a}}(x)=\sigma_{p^{a}}(x+\alpha p^{b})$ for any $x\in\{1,\ldots,p^{b}\}$. ∎

Now, we prove that the sequence $\sigma_{z}(x)$ is periodic for all $z\in N$. For this purpose, we use the following observation and lemma.

Observation 13**.**

Every infinite sequence $\{C^{(i)}\}_{i\in\operatorname{\mathbb{N}}}$ of $n-$tuples of natural numbers $C^{(i)}\in\operatorname{\mathbb{N}}^{n}$ contains an infinite non-decreasing subsequence, that is, $\{C^{(t_{i})}\}_{i\in\operatorname{\mathbb{N}}}$ such that $C^{(t_{i})}_{j}\leq C^{(t_{i+1})}_{j}$ for all $i,j\in\operatorname{\mathbb{N}}$.

Proof.

It holds trivially for $n=1$. For $n>1$ we may proceed by induction. ∎

Lemma 14**.**

Let $z\in\operatorname{\mathbb{N}}$ and $z=\prod_{i\in I}p_{i}^{a_{i}}$ be its prime factorization. There exists $m_{z}=\prod_{i\in I}p_{i}^{b}$, where $b\geq a_{i}$ for each $i\in I$ which satisfies the following condition.

[TABLE]

for arbitrary $0\leq c_{i}<b_{i}$, $d_{i}\geq b$ and where $\overline{J}:=I\setminus J$.

Proof.

First, note that the left expression in (5) is always less than or equal to the right expression by Lemma 12. The same lemma also implies that if (5) does not hold for some $J\subseteq I,(c_{i})_{i\in J},(d_{i})_{i\in\overline{J}}$ (that is, the right side of (5) is strictly greater than the left side) then it does not hold also for $J\subseteq I,(c_{i})_{i\in J},(\delta_{i})_{i\in\overline{J}}$ where all $\delta_{i}=\max\{d_{j};j\in\overline{J}\}$.

For a contradiction, let us assume there is no $m_{z}$ which satisfies condition (5). This implies there is an infinite sequence $\mathfrak{C}$ of counterexamples

[TABLE]

for all $b\in\operatorname{\mathbb{N}}$ such that

[TABLE]

Since $I$ is finite, there exists an infinite subsequence $\mathfrak{C}^{\prime}$ of counterexamples whose index sets $J^{(b)}\subseteq I$ are equal to some $J\subseteq I$. Since $\sigma_{z}$ attains only finitely many values, there exists an infinite subsequence $\mathfrak{C}^{\prime\prime}$ of counterexamples from $\mathfrak{C}^{\prime}$ such that the right sides of (6) for them are equal to some $S\in\operatorname{\mathbb{N}}$. Finally, Observation 13 implies there exists an infinite subsequence $\mathfrak{C}^{\prime\prime\prime}=\{\mathcal{C}^{(t_{b})}\}_{b\in\operatorname{\mathbb{N}}}$ of counterexamples from $\mathfrak{C^{\prime\prime}}$ such that $\{C^{(t_{b})}\}_{b\in\operatorname{\mathbb{N}}}$ is a non-decreasing sequence.

Now, let $t_{b}\geq\delta^{(t_{1})}$. Then we get

[TABLE]

A contradiction. Note that the first inequality holds by (6). The second inequality holds by Lemma 12 since $t_{b}\geq\delta^{(t_{1})}$ and $c_{i}^{(t_{1})}\leq c_{i}^{(t_{k})}$ for all $i\in J$. ∎

Finally, we prove Theorem 2 using the lemma above.

Proof of Theorem 2.

Let $z=\prod_{i\in I}p_{i}^{a_{i}}$ be the prime factorization of $z$. We show that $\sigma_{z}$ is periodic with the period $m_{z}=\prod_{i\in I}p_{i}^{b}$ given by Lemma 14 satisfying condition (5).

We aim to show that $\sigma_{z}(x)=\sigma_{z}(x+\alpha m_{z})$ for all $x\leq m_{z}$ and for all $\alpha\in\operatorname{\mathbb{N}}_{0}$. We can express $x$ as $x=r\prod_{i\in I}p_{i}^{c_{i}}$ such that $\gcd(r,p_{i})=1$ for all $i\in I$. Lemma 11 implies $\sigma_{z}(x)=\sigma_{z}(\prod_{i\in I}p_{i}^{c_{i}})$. We consider two cases:

$c_{i}<b$ for all $i\in I$.

In this case $x+\alpha m_{z}=\prod_{i\in I}p_{i}^{c_{i}}(r+\alpha\prod_{i\in I}p_{i}^{b-c_{i}})$. Therefore, $\gcd(r+\alpha\prod_{i\in I}p_{i}^{b-c_{i}},p_{j})=1$ for all $j\in I$ since $\gcd(r,p_{j})=1$. Lemma 11 implies $\sigma_{z}(x+\alpha m_{z})=\sigma_{z}(\prod_{i\in I}p_{i}^{c_{i}})=\sigma_{z}(x)$. 2. 2.

There is $i\in I$ such that $c_{i}\geq b$.

Let $J:=\{i\in I;c_{i}<b\}$, $K:=\{i\in I;c_{i}=b\}$ and $L:=\{i\in I;c_{i}>b\}$. Then we get

[TABLE]

The expression $\left(r\prod_{i\in L}p_{i}^{c_{i}-b}+\alpha\prod_{i\in J}p_{i}^{b-c_{i}}\right)$ is not divisible by $p_{i}$ for $i\in J\cup L$. However, it could be divisible by $p_{i}$ for $i\in K$. Therefore,

[TABLE]

for some $d_{i}\geq b$ where $i\in K\cup L$555In fact, $d_{i}=b$ for $i\in L$. and $h\in\operatorname{\mathbb{N}}$ such that $\gcd(h,p_{i})=1$ for all $i\in I=J\cup K\cup L$. Now, we use the property of $m_{z}$ (see (5)) and Lemma 11:

[TABLE]

where $\ast$ holds by Lemma 11, $\star$ and $\triangle$ hold by the property of $m_{z}$ (see (5)) since $c_{i}<b$ for $i\in J$, $d_{i}\geq b$ for $i\in K\cup L=\overline{J}$ ($\star$) and $c_{i}\geq b$ for $i\in K\cup L=\overline{J}$ ($\triangle$).

That is, $\sigma_{z}$ is periodic for all $z>1$ with the period $m_{z}$. ∎

Note, that it is Lemma 14 what makes the proof of this theorem existence.

4 Proof of Theorem 3

In the last section, we prove Theorem 3.

First of all, we need a tool for determining whether three points are collinear. Let $A=(a_{1},a_{2}),B=(b_{1},b_{2}),C=(c_{1},c_{2})$ be the points on the torus $T_{{m}\times{n}}$. Let $D(A,B,C)$ denote the following determinant.

[TABLE]

Lemma 15** (see Lemma 3.2. in [MSS*+*16]).**

Let $m,n$ be positive integers and let $A=(a_{1},a_{2}),B=(b_{1},b_{2}),C=(c_{1},c_{2})$ be the points on the torus $T_{{m}\times{n}}$.

$A,B,C$* are not collinear if and only if there exist $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2},\delta_{1},\delta_{2},\gamma_{1},\gamma_{2}\in\operatorname{\mathbb{Z}}$ such that $D((a_{1}+\alpha_{1}m,a_{2}+\alpha_{2}n),(b_{1}+\beta_{1}m,b_{2}+\beta_{2}n),(c_{1}+\gamma_{1}m,c_{2}+\gamma_{2}n))=0$.* 2. 2.

If $A,B,C$ are collinear then $D((a_{1},a_{2}),(b_{1},b_{2}),(c_{1},c_{2}))\equiv 0\pmod{\gcd(m,n)}$.

In our proof of Theorem 3 we will also need to be able to determine the length of a line on a torus.

Lemma 16**.**

Let $m,n$ be positive integers and $\ell=\{\pi_{m,n}(A+k(u,v));k\in\mathbb{Z}\}$ be a line on $T_{{m}\times{n}}$, where $A,(u,v)\in T_{m\times n}$ such that $\gcd(u,v)=1$. Then the length of $\ell$ is $\operatorname{\textup{lcm}}\left(\frac{m}{\gcd(m,u)},\frac{n}{\gcd(n,v)}\right)$.

Proof.

The line $\ell$ is given by the vector $(u,v)$ and its length is equal to the order of the element $(u,v)$ in $\mathbb{Z}_{m}\times\mathbb{Z}_{n}$. Which is

[TABLE]

The order of an element $h$ in the group $\operatorname{\mathbb{Z}}_{x}$ is

[TABLE]

∎

We also need lemmas about properties of lines on a torus.

Lemma 17**.**

Let $m,n$ be positive integers such that $m$ is divisible by $p$ if and only if $n$ is divisible by $p$ for every prime $p$. Let $(a_{1},a_{2})$ be a point on $T_{{m}\times{n}}$ such that $\gcd(a_{1},a_{2})=1$. Then there is exactly one line containing the origin and $(a_{1},a_{2})$.

Proof.

Let $\ell$ denote the line $\{\pi_{m,n}(k(a_{1},a_{2}));k\in\operatorname{\mathbb{Z}}\}$. This line contains the origin and $(a_{1},a_{2})$. Let us assume that there is another line $\ell^{\prime}=\{\pi_{m,n}(k(b_{1},b_{2}));k\in\operatorname{\mathbb{Z}}\}$ containing the origin and $(a_{1},a_{2})$. That means there exists $k\in\operatorname{\mathbb{Z}}$ such that

[TABLE]

Therefore, $\ell\subseteq\ell^{\prime}$. Moreover, $a_{1}=kb_{1}-s_{1}m$ and $a_{2}=kb_{2}-s_{2}n$ for some $s_{1},s_{2}\in\operatorname{\mathbb{Z}}$. Since $\gcd\left(m,kb_{1}-s_{1}m\right)=\gcd\left(m,kb_{1}\right)$ and $\gcd\left(n,kb_{2}-s_{2}n\right)=\gcd\left(n,kb_{2}\right)$, the length of the line $\ell$ is $\operatorname{\textup{lcm}}\left(\frac{m}{\gcd\left(m,kb_{1}\right)},\frac{n}{\gcd\left(n,kb_{2}\right)}\right)$ by Lemma 16. By the same lemma the length of $\ell^{\prime}$ is $\operatorname{\textup{lcm}}\left(\frac{m}{\gcd\left(m,b_{1}\right)},\frac{n}{\gcd\left(n,b_{2}\right)}\right)$.

Now, we show that $\gcd\left(m,kb_{1}\right)=\gcd\left(m,b_{1}\right)$. For a contradiction, let us assume $\gcd\left(m,kb_{1}\right)>\gcd\left(m,b_{1}\right)$. Then there is a prime $p$ which divides both $k$ and $m$. By the assumption $p$ also divides $n$ and thus $\gcd(a_{1},a_{2})=\gcd(kb_{1}-s_{1}m,kb_{2}-s_{2}n)\geq p\neq 1$. A contradiction. Analogously we get $\gcd\left(n,kb_{2}\right)=\gcd\left(n,b_{2}\right)$. Consequently, $\ell$ has the same length as $\ell^{\prime}$. Moreover, $\ell=\ell^{\prime}$ since $\ell\subseteq\ell^{\prime}$. ∎

Lemma 18**.**

Let $T_{{p^{a}}\times{p^{b}}}$ be a torus where $p$ is a prime, $a\leq b$, and let $(x,y)$ be a point on $T_{{p^{a}}\times{p^{b}}}$. Let $g$ denote $\gcd(x,y)$. If $x$ or $y$ is not divisible by $p$, then there is exactly one line between the origin and $(x,y)$.

Proof.

Let $\ell=\{\pi_{p^{a},p^{b}}(k(v_{1},v_{2}));k\in\operatorname{\mathbb{Z}}\}$ be a line which contains $(x,y)$. Then for some $k\in\operatorname{\mathbb{Z}}$

[TABLE]

Let $g$ denote $\gcd(x,y)$. Since $\gcd(g,p)=1$, $g$ has an inverse element modulo $p^{b}$. Let us denote it $g^{-1}$. Then

[TABLE]

Hence every line on $T_{{p^{a}}\times{p^{b}}}$ contains $(x,y)$ if and only if it contains $(\frac{x}{g},\frac{y}{g})$. Since $\gcd(\frac{x}{g},\frac{y}{g})=1$, Lemma 17 implies that there is exactly one line containing the origin and $(\frac{x}{g},\frac{y}{g})$. Consequently, there is exactly one line containing the origin and the point $(x,y)$. ∎

Corollary 19**.**

Let $T_{{p^{a}}\times{p^{b}}}$ be a torus where $p$ is a prime and let $A=(a_{1},a_{2}),B=(b_{1},b_{2})\in T_{{p^{a}}\times{p^{b}}}$ be points on that torus such that $a_{1}-b_{1}$ is not divisible by $p$. Then there is exactly one line between $A,B$ and its length is $\max\left\{p^{a},\frac{p^{b}}{\gcd(p^{b},a_{2}-b_{2})}\right\}$.

Proof.

The lines between $(a_{1},a_{2}),(b_{1},b_{2})$ are in one-to-one correspondence with the lines between the origin and $P:=\pi_{p^{a},p^{b}}(a_{1}-b_{1},a_{2}-b_{2})$. Since $a_{1}-b_{1}$ is not divisible by $p$ there is exactly one line between the origin and $P$ by Lemma 18.

Such line is given by the vector $\pi_{p^{a},p^{b}}(\frac{a_{1}-b_{1}}{\gcd(a_{1}-b_{1},a_{2}-b_{2})},\frac{a_{2}-b_{2}}{\gcd(a_{1}-b_{2},a_{2}-b_{2})})$. We can express it as $(v,wp^{r})$, where $v,w$ is not divisible by $p$ and $r<b$. Moreover, $p^{r}=\gcd(p^{b},\frac{a_{2}-b_{2}}{\gcd(a_{1}-b_{1},a_{2}-b_{2})})=\gcd(p^{b},a_{2}-b_{2})$ since $a_{1}-b_{1}$ is not divisible by $p$. Lemma 16 implies the length of the line between the origin and $P$ is

[TABLE]

∎

Lemma 20**.**

Let $T_{{p^{a}}\times{p^{b}}}$ be a torus where $p$ is a prime and let $A,B,C$ be points on it such that $\pi_{p^{a-1},p^{b}}(A)=\pi_{p^{a-1},p^{b}}(B)$. If each line $\ell$ containing the points $A,C$ on $T_{{p^{a}}\times{p^{b}}}$ has the same length as its image $\ell^{\prime}:=\pi_{p^{a-1},p^{b}}(\ell)$ then $A,B,C$ are not collinear. (See Figure 2 for an example.)

Proof.

If there is a line $\ell$ which contains $A,B,C$ then its image $\ell^{\prime}$ has smaller length since $\pi_{p^{a-1},p^{b}}(A)=\pi_{p^{a-1},p^{b}}(B)$. However, it is impossible by our assumption. ∎

Lemma 21**.**

Let $T_{{p^{a}}\times{p^{b}}}$ be a torus where $p$ is a prime and $a\leq b$. Let $(v_{1}p^{c},v_{2}p^{d})$ be a point on $T_{{p^{a}}\times{p^{b}}}$ such that $p$ does not divide $v_{1}$ or $v_{2}$, $c<a$, $d<b$ and $c\leq d$. Then each line which contains the origin and $(v_{1}p^{c},v_{2}p^{d})$ also contains some point $(w_{1},w_{2})$ such that $w_{1},w_{2}$ satisfy the following equations.

[TABLE]

Proof.

Let $\ell$ be a line which contains the origin and $(v_{1}p^{c},v_{2}p^{d})$. Then we can express it as $\ell=\{\pi_{p^{a},p^{b}}(k(u_{1},u_{2}));k\in\operatorname{\mathbb{Z}}\}$ such that $\gcd(u_{1},u_{2})=1$ and there exists $k\in\operatorname{\mathbb{Z}}$ which satisfies the following equations.

[TABLE]

Since $\gcd(u_{1},u_{2})=1$, $p$ does not divide $u_{1}$ or $u_{2}$. Therefore, $p^{c}$ has to divide $k$ since $c\leq d$. Consequently,

[TABLE]

∎

Now, we are able to prove Theorem 3.

Proof of Theorem 3.

Theorem 1 implies $\tau_{p^{a},p^{(a-1)p+2}}\leq 2p^{a}$. Therefore, it is sufficient to show a construction of $2p^{a}$ points on $T_{{p^{a}}\times{p^{(a-1)p+2}}}$ where no three of them are collinear.

Let $X:=\{(i,i^{2}p);i\in P\}$ and $Y:=\{(i,i^{2}p+1);i\in P\}$, where $P=\{0,\ldots,p-1\}$. Misiak et al. (see Theorem 1.2(3a) in [MSS*+*16]) proved that no three points from the set $\pi_{p,p^{2}}(X\cup Y)$ are collinear on the torus $T_{{p}\times{p^{2}}}$. In this proof, we inductively construct a set $X_{a}\cup Y_{a}\subset\operatorname{\mathbb{Z}}^{2}$ by taking multiple copies of $X\cup Y$ and prove by induction that no three points from $\pi_{p^{a},p^{(a-1)p+2}}(X_{a}\cup Y_{a})$ are collinear on the torus $T_{{p^{a}}\times{p^{(a-1)p+2}}}$.

First, let us define the set $X_{a}\cup Y_{a}$.

For $a=1$ we define $X_{a}:=X$ and $Y_{a}:=Y$. 2. 2.

For $a>1$ we take $p$ copies of the set $X_{a-1}\cup Y_{a-1}$ on $\operatorname{\mathbb{Z}}^{2}$ so that

[TABLE]

As an example of the construction see Figures 3 and 4. For a point $A\in\pi_{p^{a},p^{(a-1)p+2}}(X_{a}\cup Y_{a})$ on $T_{{p^{a}}\times{p^{(a-1)p+2}}}$ we call the point $\widehat{A}\in X_{a}\cup Y_{a}$ such that $\pi_{p^{a},p^{(a-1)p+2}}(\widehat{A})=A$ the preimage of $A$ (see Figure 4). Now, we inductively prove that $\pi_{p^{a},p^{(a-1)p+2}}(X_{a}\cup Y_{a})$ does not contain three collinear points. Case $a=1$ was proved by Misiak et al. (see Theorem 1.2(3a) in [MSS*+*16]).

Let us assume $a>1$ and let $A,B,C\in\pi_{p^{a},p^{(a-1)p+2}}(X_{a}\cup Y_{a})$. We have to consider several cases.

First, let us assume that $\pi_{p^{a-1},p^{(a-2)p+2}}(A),\pi_{p^{a-1},p^{(a-2)p+2}}(B),\pi_{p^{a-1},p^{(a-2)p+2}}(C)\in X_{a-1}\cup Y_{a-1}$ are three distinct points. In other words, the preimages of $A,B,C$ in $X_{a}\cup Y_{a}$ are copies of three distinct points from $X_{a-1}\cup Y_{a-1}$. If $A,B,C$ are collinear on $T_{{p^{a}}\times{p^{(a-1)p+2}}}$, then $\pi_{p^{a-1},p^{(a-2)p+2}}(A)$, $\pi_{p^{a-1},p^{(a-2)p+2}}(B)$, $\pi_{p^{a-1},p^{(a-2)p+2}}(C)$ are collinear on $T_{{p^{a-1}}\times{p^{(a-2)p+2}}}$ as well which contradicts the induction hypothesis.

Now, we assume $A$, $B$, $C$ satisfy $\pi_{p^{a-1},p^{(a-2)p+2}}(A)=\pi_{p^{a-1},p^{(a-2)p+2}}(B)$. In other words, the preimages of $A$ and $B$ are copies of the same point from $X_{a-1}\cup Y_{a-1}$.

In the first case, we assume that the preimages of $A,B$ are from $X_{a}$ and the preimage of $C$ is from $Y_{a}$. Therefore,

[TABLE]

for some $k,m\in\{0,\ldots,p-1\}$, $u\in\{1,\ldots,p-1\}$, for suitable $t_{1},t_{2},v_{1},v_{2},x\in\operatorname{\mathbb{N}}_{0}$, $q\geq(a-2)p+4\geq a$ and where $p$ does not divide $x$. Note that $u>0$; otherwise $A=B$. We may use the translation given by the vector $(-t_{1}p,-t_{2}p)$ and we get

[TABLE]

where $s_{1}=v_{1}-t_{1}\mod p^{a}$ and $s_{2}=v_{2}-t_{2}\mod p^{(a-1)p+2}$. The points $A_{t},B_{t},C_{t}$ are collinear if and only if $A,B,C$ are collinear. We compute $D(A_{t},B_{t},C_{t})$.

[TABLE]

for some $R\in\operatorname{\mathbb{Z}}$. Therefore, $D(A_{t},B_{t},C_{t})\not\equiv 0\pmod{p^{a}}$ since $u>0$. Consequently, $A,B,C$ are not collinear on $T_{{p^{a}}\times{p^{(a-1)p+2}}}$ by Lemma 15 (2).

Similarly, we check the case when the preimages of $A,B$ are from $Y_{a}$ and the preimage of $C$ from $X_{a}$. Hence

[TABLE]

We use the translation given by the vector $(-t_{1}p,-t_{2}p)$ as we did in the previous case:

[TABLE]

and we get

[TABLE]

Therefore, $A,B,C$ are not collinear on $T_{{p^{a}}\times{p^{(a-1)p+2}}}$ by the same lemma.

Now, we check the case when the preimages of all three points are from $X_{a}$. The case where they are all from $Y_{a}$ is just a translation. We can express the points as follows:

[TABLE]

where $p$ divides $s_{1},s_{3}$ and $p^{4}$ divides $s_{2},s_{4}$ by the definition of the construction (the first iteration for $a=2$), $q\in\{p^{(a-2)p+4},\ldots,p^{(a-1)p+2}\}$, $u\geq 1$ is not divisible by $p$, $x\in\{0,1\}$ and $k,m\in\{0,\ldots,p-1\}$.

First, we assume $m\neq k$. Let us map these points to the smaller torus $T_{{p^{a}}\times{p^{(a-2)p+4}}}$ using $\pi_{p^{a},p^{(a-2)p+4}}$ and let $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ be the images. Then

[TABLE]

where $p^{4}$ divides $s_{2}^{\prime},s_{4}^{\prime}$. Let us check the line between $A^{\prime}$ and $C^{\prime}$. Since $((k-m)+s_{1}-s_{3})$ is not divisible by $p$, Corollary 19 implies there is exactly one line between them and its length is $d=\max\{p^{a},\frac{p^{(a-2)p+4}}{\gcd(p^{(a-2)p+4},p(k-m)(k+m)+s_{2}^{\prime}-s_{4}^{\prime})}\}=\max\{p^{a},p^{(a-2)p+4-h}\}$, where $h\in\{1,2\}$ (the sum of $k$ and $m$ may be $p$).

Now, let us map $A^{\prime},B^{\prime},C^{\prime}$ to the torus $T_{{p^{a-1}}\times{p^{(a-2)p+4}}}$ and let $A^{\prime\prime}$, $B^{\prime\prime}$, $C^{\prime\prime}$ be the images. We can express them as

[TABLE]

where $p$ divides $s_{1}^{\prime}$ a $s_{3}^{\prime}$. Again, by Corollary 19 there is exactly one line between $A^{\prime\prime}$ and $C^{\prime\prime}$ and its length is $d^{\prime}=\max\{p^{a-1},p^{(a-2)p+4-h}\}$. Since $(a-2)p+4-h\geq a$ for $a>1$, $d^{\prime}=d$ and $A^{\prime},B^{\prime},C^{\prime}$ are not collinear by Lemma 20. Therefore, $A,B,C$ are not collinear on $T_{{p^{a}}\times{p^{(a-1)p+2}}}$. 2. 2.

Let us check the case when $k=m$. That means $\pi_{p,p^{2}}(A)=\pi_{p,p^{2}}(C)$.

First, let us assume that $\pi_{p^{a-1},p^{(a-2)p+2}}(A)\neq\pi_{p^{a-1},p^{(a-2)p+2}}(C)$. In other words, the preimages of all three points are copies of the same point from $X$ but they are not copies of the same point from $X_{a-1}$. We have

[TABLE]

where $p$ divides $s_{1},s_{3}$ and $p^{4}$ divides $s_{2},s_{4}$ by the definition of the construction (the first iteration for $a=2$), $q\in\{p^{(a-2)p+4},\ldots,p^{(a-1)p+2}\}$, $u\geq 1$ is not divisible by $p$, $x\in\{0,1\}$ and $k,m\in\{0,\ldots,p-1\}$. We can see that $p^{a-1}$ does not divide $s_{1}$ and $p^{(a-2)p+4}$ does not divide $s_{2}$ or $p^{a-1}$ does not divide $s_{3}$ and $p^{(a-2)p+4}$ does not divide $s_{4}$; otherwise $\pi_{p^{a-1},p^{(a-2)p+2}}(A)=\pi_{p^{a-1},p^{(a-2)p+2}}(C)$. Now, we use the translation given by the vector $(-k-s_{1},-k^{2}p-s_{2})$ and we get

[TABLE]

where $u,x,v,y$ are not divisible by $p$, $t\leq a-2$, $r\leq(a-2)p+2$. Let us have a look at the images of these points on the torus $T_{{p^{a}}\times{p^{(a-2)p+4}}}$. We get

[TABLE]

for suitable $y^{\prime}$ which is not divisible by $p$. Let us compute the length of the lines between $A^{\prime}$ a $C^{\prime}$. By Lemma 21, every such line passing through some point $(w_{1},w_{2})$ satisfies

[TABLE]

By Corollary 19, there is exactly one line between the origin and such $(w_{1},w_{2})$ and its length is $\max\{p^{a},\frac{p^{(a-2)p+4}}{p^{r-t}}\}$. Therefore, each line between $A^{\prime}$ and $C^{\prime}$ has the length $d=\max\{p^{a},\frac{p^{(a-2)p+4}}{p^{r-t}}\}$.

Let us map $A^{\prime},B^{\prime},C^{\prime}$ to the torus $T_{{p^{a-2}}\times{p^{(a-2)p+4}}}$ and let $A^{\prime\prime},B^{\prime\prime},C^{\prime\prime}$ be the images. We get

[TABLE]

for suitable $v^{\prime}$ which is not divisible by $p$. Similarly, the length $d^{\prime}$ of each line between $A^{\prime\prime}$ and $C^{\prime\prime}$ is $d^{\prime}=\max\{p^{a-1},\frac{p^{(a-2)p+4}}{p^{r-t}}\}$ by Corollary 19. Note that the maximal $r-t$ satisfying the definition of the construction equals $\max_{x\in\{1,\ldots,a-2\}}(xp+2-x)=(a-2)(p-1)+2$. Hence $r-t\leq(a-2)p-a+4$ and $(a-2)p+4-(r-t)\geq a$. Consequently, $d=d^{\prime}$ and Lemma 20 implies that $A^{\prime},B^{\prime},C^{\prime}$ are not collinear and, therefore, $A,B,C$ are not collinear on $T_{{p^{a}}\times{p^{(a-1)p+2}}}$.

Finally, we check the case when all three $A,B,C$ points are copies of one point from $X_{a-1}$ which is without loss of generality the origin. Otherwise, we use a translation similarly to the previous cases. Therefore,

[TABLE]

for distinct $q,r,s\in\{p^{(a-2)p+4},\ldots,p^{(a-1)p+2}\}$ and for distinct $u,v,w\in\{0,\ldots,p-1\}$ satisfying the definition of the construction. Let us denote $b:=(a-1)p+2$ and $D:=D((up^{a-1}+\alpha_{1}p^{a},p^{q}+\alpha_{2}p^{b}),(vp^{a-1}+\beta_{1}p^{a},p^{r}+\beta_{2}p^{b}),(wp^{a-1}+\gamma_{1}p^{a},p^{s}+\gamma_{1}p^{b}))$ the determinant for $A,B,C$ from Lemma 15 (1), where $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2},\gamma_{1},\gamma_{2}\in\operatorname{\mathbb{Z}}$. Then

[TABLE]

where $Q=D((u,\alpha_{2}),(v,\beta_{2}),(w,\gamma_{2}))$ and $W=D((\alpha_{1},\alpha_{2}),(\beta_{1},\beta_{2}),(\gamma_{1},\gamma_{2}))$. Without loss of generality let $q$ be the smallest number of $\{q,r,s\}$. Then indeed $q<b$ and thus we get $D=p^{a+q-1}((w-v)+pH)$ for suitable $H\in\operatorname{\mathbb{Z}}$. Therefore, $D\neq 0$ and $A,B,C$ are not collinear on $T_{{p^{a}}\times{p^{(a-1)p+2}}}$ by Lemma 15 (1) and we are done.

∎

Acknowledgement

I would like to thank my supervisor Martin Tancer for all his support and advice. I would also like to thank the anonymous reviewers for valuable advice and remarks.