Diophantine Tuples over $\mathbb{Z}_p$

TL;DR

This paper investigates the structure and measure of Diophantine tuples over the ring of p-adic integers and their residue fields, providing new insights into their distribution and properties.

Contribution

It introduces methods to compute and estimate the measures of Diophantine m-tuples over _p and _p, advancing understanding of their distribution in p-adic contexts.

Findings

Computed measures of Diophantine tuples in _p and _p.

Estimated the distribution of these tuples in p-adic integers.

Provided bounds and asymptotic estimates for the measures.

Abstract

For an element $r$ of a ring $R$, a Diophantine $D(r)$ $m$-tuple is an $m$-tuple $(a_1,a_2,\ldots,a_m)$ of elements of $R$ such that for all $i,j$ with $i\neq j$, $a_ia_j+r$ is a perfect square in $R$. In this article, we compute and estimate the measures of the sets of $D(r)$ $m$-tuples in the ring $\mathbb{Z}_p$ of $p$-adic integers, as well as its residue field $\mathbb{F}_p$.

Equations

$$a_i{a}_j+r\in \square (R),\text{for all }1\le i<j\le m.$$

$$\frac{1}{2^{\left(\genfrac{}{}{0ex}{}{m}{2}\right)}}{\int }_{\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_p^m}\prod _{1\le i<j\le m}\left(1+\left(\frac{a_i{a}_j+1}{p}\right)\right)d\mu ,$$

$$v_p(a)=\sup \left\{k\in \text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_{\ge 0}:a\in p^k\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_p\right\}.$$

$$v_2(\alpha )\equiv 0(\mathrm{mod}2)\text{and}\beta \equiv 1(\mathrm{mod}8).$$

$${\mathrm{dioph}}_2^1(\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_2)=\frac{1}{3}.$$

$$A_{2k}=\{(a,b)\in {\mathrm{Dioph}}_2^1(\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_2):v_2(ab+1)=2k\}.$$

$$(a,b)(\mathrm{mod}8)\in \{(\pm 2,4),(4,\pm 2),(4,4)\}.$$

$$A_0=\frac{7+7+1+2+2+1}{64}=\frac{5}{16}.$$

$$ab+1=2^{2k}\cdot \alpha ,\alpha \equiv 1(\mathrm{mod}8)$$

$$ab+1\equiv 2^{2k}\mathrm{mod}{2}^{2k+3}.$$

$$A_{2k}=\frac{1}{2}\cdot \frac{1}{2^{2k+3}}=\frac{1}{2^{2k+4}}.$$

$${\mathrm{dioph}}_2^1(\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_2)=\mu \{(a,b)\in \text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_2^2:ab+1\in \square (\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_2)\}=A_0+\sum _{k=1}^{\infty }{A}_{2k}=\frac{5}{16}+\sum _{k=1}^{\infty }\frac{1}{2^{2k+4}}=\frac{5}{16}+\frac{1}{48}=\frac{1}{3}.$$

$${\mathrm{dioph}}_2^r(\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_p)=\{\begin{array}{ll}\frac{1}{2}+\frac{1}{p(p+1)}& \left(\frac{r}{p}\right)=1,\\ \frac{1}{2}-\frac{1}{p+1}& \left(\frac{r}{p}\right)=-1,\\ \frac{1}{2}-\frac{p-1}{2(p+1{)}^2}-\frac{(\alpha +2)}{2(p+1{)}^2{p}^{\alpha -1}}-\frac{1}{2(p+1{)}^2{p}^{\alpha }}+\frac{(\alpha -1)}{2(p+1{)}^2{p}^{\alpha +1}}& \alpha \equiv 1(\mathrm{mod}2),\\ \frac{1}{2}+\frac{(\alpha +1)(p-1{)}^2}{2p^{\alpha +2}}-\frac{2p-1}{2(p+1{)}^2}+\frac{1}{2(p+1{)}^{2}p}-\frac{(\alpha +1)}{2(p+1{)}^2{p}^{\alpha -2}}& \\ +\frac{(\alpha -1)}{2(p+1{)}^2{p}^{\alpha }}-\frac{1}{2(p+1{)}^2{p}^{\alpha +1}}-\frac{1}{2(p+1{)}^2{p}^{\alpha +2}}& \alpha >0,\alpha \equiv 0(\mathrm{mod}2),\left(\frac{s}{p}\right)=1,\\ \frac{1}{2}+\frac{\alpha (p-1{)}^2+p^2+1}{2p^{\alpha +2}}-\frac{2p-1}{2(p+1{)}^2}+\frac{1}{2(p+1{)}^{2}p}-\frac{(\alpha +1)}{2(p+1{)}^2{p}^{\alpha -2}}& \\ +\frac{(\alpha -1)}{2(p+1{)}^2{p}^{\alpha }}-\frac{1}{2(p+1{)}^2{p}^{\alpha +1}}-\frac{1}{2(p+1{)}^2{p}^{\alpha +2}}& \alpha >0,\alpha \equiv 0(\mathrm{mod}2),\left(\frac{s}{p}\right)=-1,\end{array}$$

$${\mathrm{dioph}}_2^1(\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_p)=\frac{1}{2}+\frac{1}{p(p+1)}.$$

$${\mathrm{dioph}}_2^r(\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_p)=\frac{1}{2}-\frac{1}{p+1}.$$

$$A_k=\left\{(a,b)\in {\mathrm{Dioph}}_2^r(\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_p):v_p(ab+r)=2k\right\}.$$

$$ab=\beta -r\not\equiv 0(\mathrm{mod}p),$$

$$\mu (A_0)=\frac{1}{p^2}\cdot (p-1)\cdot \frac{(p-1)}{2}=\frac{(p-1{)}^2}{2p^2}.$$

$$ab=p^{2k}\beta -r\not\equiv 0(\mathrm{mod}p)$$

$$\mu (A_{2k})=\frac{1}{p^{4k+2}}\cdot (p^{2k+1}-p^{2k})\cdot \frac{p-1}{2}=\frac{(p-1{)}^2}{2p^{2k+2}}.$$

$${\mathrm{dioph}}_2^r(\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_p)=\sum _{k=0}^{\infty }\mu (A_k)=\sum _{k=0}^{\infty }\frac{(p-1{)}^2}{2p^{2k+2}}=\frac{(p-1{)}^2}{2p^2}\cdot \frac{p^2}{p^2-1}=\frac{(p-1)}{2(p+1)}=\frac{1}{2}-\frac{1}{p+1}.$$

$${\mathrm{dioph}}_2^r(\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_p)=\frac{1}{2}+\frac{1}{p(p+1)}.$$

$$A_k=\left\{(a,b)\in {\mathrm{Dioph}}_2^r(\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_p):v_p(ab+r)=2k\right\}.$$

$$\mu (\{(a,b)\in \text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_p^2:ab\equiv 0(\mathrm{mod}p)\})=\frac{2p-1}{p^2}.$$

$$\mu (A_0)=\frac{2p-1}{p^2}+\frac{1}{p^2}\cdot (p-1)\cdot \left(\frac{(p-1)}{2}-1\right)=\frac{p^2+1}{2p^2}.$$

$$ab=p^{2k}\beta -r\not\equiv 0(\mathrm{mod}{p}^{2k+1}).$$

$$\mu (A_{2k})=\frac{1}{p^{4k+2}}\cdot (p^{2k+1}-p^{2k})\cdot \frac{p-1}{2}=\frac{(p-1{)}^2}{2p^{2k+2}}.$$

$${\mathrm{dioph}}_2^r(\text{use@mathgroup}\text{M@U}\text{symAMSb}{Z}_p)$$

$$=\frac{p^2+1}{2p^2}+\sum _{k=1}^{\infty }\frac{(p-1{)}^2}{2p^{2k+2}}$$

$$=\frac{p^2+1}{2p^2}+\frac{p-1}{2p^2(p+1)}$$

$$=\frac{p^2+p+2}{2p(p+1)}$$

$$=\frac{1}{2}+\frac{1}{p(p+1)}.$$

Full text

1. Introduction

The study of Diophantine $m$-tuples has a long history, attracting the attention of the some of the most prominent mathematicians. The story begins in the 3rd century with Diophantus, who first observed that the quadruple $(\frac{1}{16},\frac{33}{16},\frac{17}{4},\frac{105}{16})$ has the curious property that the product of any two elements in the set was one less than a rational square. It was not until Fermat that the first quadruple of integers with this property was discovered, namely $(1,3,8,120)$: $1\cdot 3+1=2^{2}$, $3\cdot 8+1=5^{2}$, $\ldots$, $3\cdot 120+1=19^{2}$. Euler found infinitely many Diophantine quadruples in the positive integers and also found a way of extending Fermat’s set to a rational quintuple with this property, namely $(1,3,8,120,\frac{777480}{8288641})$. Such sets of numbers are now known as Diophantine $m$-tuples, precisely defined below:

Definition 1.1**.**

For a commutative ring with unit $R$, an $m$-tuple $(a_{1},\ldots,a_{m})\in R^{m}$ is said to be a Diophantine $m$-tuple over $R$ if for every $i,j$ with $1\leq i<j\leq m$, $a_{i}a_{j}+1\in\square(R)$, where $\square(R)$ denotes the set of square elements in ring $R$.

Here, we will leverage some notation that enables us to include a more general family of sets than just the Diophantine $m$-tuples, as investigated in a number of works, such as [Duj04].

Definition 1.2**.**

For $r\in R$, an $m$-tuple $(a_{1},\ldots,a_{m})\in R^{m}$ is said to be a $\operatorname{D}(r)$* $m$-tuple over $R$* if

[TABLE]

We denote by $\operatorname{Dioph}_{m}^{r}(R)\subset R^{m}$ the set of $\operatorname{D}(r)$ $m$-tuples over $R$. Note that $\operatorname{Dioph}_{m}^{1}(R)$ is the set of Diophantine $m$-tuples over $R$.

In this notation, Diophantus’s original example $(\frac{1}{16},\frac{33}{16},\frac{17}{4},\frac{105}{16})$ is in $\operatorname{Dioph}_{4}^{1}(\use@mathgroup\M@U\symAMSb{Q})$, and after clearing denominators, we see that $(1,33,68,105)\in\operatorname{Dioph}_{4}^{256}(\use@mathgroup\M@U\symAMSb{Z})$. Fermat’s first Diophantine (i.e. $\operatorname{D}(1)$) quadruple in the positive integers is $(1,3,8,120)\in\operatorname{Dioph}_{4}^{1}(\use@mathgroup\M@U\symAMSb{Z})$.

The problem of extending Fermat’s triple $\{1,3,8\}$ to a quadruple was one of the early successes of Baker’s results on linear forms in logarithms; see [BD69]. More recently, Gibbs in [Gib06] found a rational Diophantine sextuple, namely $(\frac{11}{192},\frac{35}{192},\frac{155}{27},\frac{512}{27},\frac{1235}{48},\frac{180873}{16})\in\operatorname{Dioph}_{6}^{1}(\use@mathgroup\M@U\symAMSb{Q})$. Recently, He, Togbé, and Ziegler in [HTZ16] showed that there is no Diophantine quintuple in positive integers, i.e. that $\operatorname{Dioph}_{5}^{1}(\use@mathgroup\M@U\symAMSb{Z})=\varnothing$.

A curious property about Diophantine tuples is that any triple, over any ring (or even any semiring) can be extended to a Diophantine quadruple. This is special to the case of $\operatorname{D}(1)$ (or more generally $\operatorname{D}(k^{2})$), as $\operatorname{D}(r)$ triples do not generally extend to $\operatorname{D}(r)$ quadruples in interesting ways. (Of course, one can always extend any $\operatorname{D}(r)$ $m$-tuple to a $\operatorname{D}(r)$ $(m+1)$-tuple by appending 0, but this is rather trivial.) An open question of considerable interest is whether or not every Diophantine triple $(a,b,c)$ in the positive integers can be uniquely extended to a Diophantine quadruple $(a,b,c,d)$ with $d\in\use@mathgroup\M@U\symAMSb{Z}$ and $d>\max(a,b,c)$.

Most of the past work in Diophantine and $\operatorname{D}(r)$ tuples has been in the case of $R=\use@mathgroup\M@U\symAMSb{Z}$ (or even $\use@mathgroup\M@U\symAMSb{N}$) and $R=\use@mathgroup\M@U\symAMSb{Q}$. There have been a few papers studying the case where $R$ is the ring of integers in a quadratic field (e.g. [Duj97, Fra04]) and a cubic field (e.g. [Fra13]). The question of Diophantine tuples over finite fields was considered in [DK16]. In recent years, there has been a flurry of activity on Diophantine and $\operatorname{D}(r)$ tuples. Instead of summarizing more of it, we instead refer the interested reader to Dujella’s extensive bibliography of the field, which can be found at https://web.math.pmf.unizg.hr/~duje/ref.html.

In the present paper, we investigate some questions similar in spirit to those of [DK16]. In [DK16] and frequently in this literature, it is assumed that elements of a Diophantine tuple must be distinct. However, our main goal is an understanding of the Haar measure of tuples over $\use@mathgroup\M@U\symAMSb{Z}_{p}$. To do so, it is often convenient to reduce modulo $p$, so we do not require our tuples to have distinct elements.

The main goal of this article is to evaluate or approximate the Haar measure of the set $\operatorname{Dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$, for certain values of $m$, $r$, and $p$. Then our goal is to evaluate $\operatorname{dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ defined below:

Definition 1.3**.**

For $r\in R$, let $\operatorname{dioph}_{m}^{r}(R)=\mu(\operatorname{Dioph}_{m}^{r}(R))$ be the measure of the set of $\operatorname{D}(r)$ $m$-tuples in $R^{m}$, where $\mu$ is the product measure induced by a choice of finite measure $\mu$ on $R$.

Throughout this article, we will assume that $R$ is a commutative ring equipped with a probability measure $\mu$, i.e. one in which $\mu(R)=1$. The main case we will be focusing on is the case in which $R$ is the ring $\use@mathgroup\M@U\symAMSb{Z}_{p}$ of $p$-adic integers equipped with its Haar measure $\mu$, normalized so that $\mu(\use@mathgroup\M@U\symAMSb{Z}_{p})=1$. We will also look at the case where $R=\mathcal{O}_{K}$ is the valuation ring of a finite extension $K$ of $\use@mathgroup\M@U\symAMSb{Q}_{p}$, again equipped with its normalized Haar measure. We will also consider the case where $R=\use@mathgroup\M@U\symAMSb{F}_{q}$ is a finite field, equipped with the normalized counting measure, where the normalization is such that $\mu(\use@mathgroup\M@U\symAMSb{F}_{q})=1$.

Unless otherwise specified, we take $a,b,c,d\in\use@mathgroup\M@U\symAMSb{Z}_{p}$. All of our results over $\use@mathgroup\M@U\symAMSb{Z}_{p}$ can be easily modified to obtain bounds over $\use@mathgroup\M@U\symAMSb{F}_{p}$ with respect to the counting measure on $\use@mathgroup\M@U\symAMSb{F}_{p}$ and vice versa, modulo lower order terms.

In the case where $m$ is small, we can usually give an exact value for $\operatorname{dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$, whereas when $m\geq 4$, we can only give an asymptotic as $p\to\infty$, with a lower-order error term. One of our main techniques is to study the reduction to $\operatorname{dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ and then use results about character sums in $\use@mathgroup\M@U\symAMSb{F}_{p}$. More precisely, we note that $\operatorname{dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ is quite close to

[TABLE]

which can be calculated exactly for $m\leq 3$ and approximated up to an error term of the form $O(p^{-1/2})$ as $p\to\infty$ when $m\geq 4$. Here and elsewhere, $\left(\frac{\cdot}{p}\right)$ denotes the Legendre symbol on $\use@mathgroup\M@U\symAMSb{Z}_{p}$.

We proceed as follows in this article. In Section 2 we collect some notation and observe that the Haar measure of Diophantine pairs over $\use@mathgroup\M@U\symAMSb{Z}_{2}$ is $\frac{1}{3}$. In Section 3, we prove Theorem 3.1 and exactly compute the Haar measure of Diophantine $D(r)$ pairs over $\use@mathgroup\M@U\symAMSb{Z}_{p}$ for all odd primes $p$. We observe in Section 4 that when $p=3$, we can compute the Haar measure exactly of $D(r)$ $m$-tuples for all $\left(\frac{r}{3}\right)=1$ and $m\geq 2$. We proceed in Section 5 to give asymptotic bounds in Theorem 5.2 on $\operatorname{dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$, exactly computing $\operatorname{dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ in Theorem 5.3 as an intermediate step. We extend our asymptotic analysis of $\operatorname{Dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ to $m\geq 4$ in Section 6. Finally, we are able to carry over many of our methods when we replace $\use@mathgroup\M@U\symAMSb{Z}_{p}$ by a finite extension thereof as we observe in Section 7, for example exactly computing the Haar measure of Diophantine $D(r)$ pairs in Theorem 7.1 when the residue characteristic is odd.

2. $\operatorname{Dioph}_{2}^{1}(\use@mathgroup\M@U\symAMSb{Z}_{2})$

Before stating any results, let us recall some notation that will be used throughout the article.

Definition 2.1**.**

For $a\in\use@mathgroup\M@U\symAMSb{Z}_{p}$, we let $v_{p}(a)$ denote the $p$-adic valuation of $a$, given by

[TABLE]

Definition 2.2**.**

More generally, if $R=\mathcal{O}_{K}$ is a valuation ring, we let $v_{R}$ denote the normalized valuation so that if $\pi\in\mathcal{O}_{K}$ is a uniformizer, then $v_{R}(\pi)=1$.

Note that $(a,b)\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{2}$ is a Diophantine pair over $\use@mathgroup\M@U\symAMSb{Z}_{2}$ if $ab+1\in\square(\use@mathgroup\M@U\symAMSb{Z}_{2})$. An element $\alpha=2^{v_{2}(\alpha)}\beta\in\square(\use@mathgroup\M@U\symAMSb{Z}_{2})$ if

[TABLE]

Proposition 2.3**.**

The Haar measure of Diophantine pairs over $\use@mathgroup\M@U\symAMSb{Z}_{2}$ is given by

[TABLE]

Proof.

We consider cases based on the residues of $ab$ modulo $8$ and the $2$-adic valuation of $ab+1$ in $\use@mathgroup\M@U\symAMSb{Z}_{2}$. To that end, we will define the measures of disjoint subsets of $\use@mathgroup\M@U\symAMSb{Z}_{2}^{2}$. For $k\geq 0$, let

[TABLE]

If $ab+1\not\equiv 0\pmod{4}$, $ab+1=\square\in\use@mathgroup\M@U\symAMSb{Z}_{2}$ exactly when $ab\equiv 0\pmod{8}$. This occurs when either one of $a,b$ is $0\pmod{8}$ or

[TABLE]

This yields

[TABLE]

If $v_{2}(ab+1)=2$, then $ab+1=\square\in\use@mathgroup\M@U\symAMSb{Z}_{2}$ exactly when $ab+1\equiv 4\pmod{32}$. Considering residue classes of $a,b\pmod{32}$ yields $A_{2}=\frac{1}{2}\cdot\frac{1}{32}$. Similarly we obtain for $v_{2}(ab+1)=2k$ for $k>1$, that $ab+1=\square$ exactly when

[TABLE]

or in other words when

[TABLE]

The above result implies that for $k\geq 1$, we have that

[TABLE]

We therefore obtain the desired result:

[TABLE]

∎

3. $\operatorname{Dioph}_{2}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$

Theorem 3.1**.**

Let $p$ be an odd prime and let $\alpha=v_{p}(r)$ with $r=p^{\alpha}s$. Then, we have the following:

[TABLE]

where $(\frac{a}{b})$ denotes the Legendre symbol. In particular, we have that the measure of the Diophantine pairs over $\use@mathgroup\M@U\symAMSb{Z}_{p}$ is given by

[TABLE]

Remark 3.2*.*

We will repeatedly use the following characterization of $\square(\use@mathgroup\M@U\symAMSb{Z}_{p})$. For an odd prime $p$, $\alpha\in\square(\use@mathgroup\M@U\symAMSb{Z}_{p})$ exactly if $\alpha=p^{2k}\beta$ for some nonnegative integer $k$ and some $\beta\in\use@mathgroup\M@U\symAMSb{Z}_{p}$ with $\left(\frac{\beta}{p}\right)=1$.

Lemma 3.3**.**

Let $p$ be an odd prime and take $r\in\use@mathgroup\M@U\symAMSb{Z}_{p}$ so that $\left(\frac{r}{p}\right)=-1$. Then,

[TABLE]

Proof.

We compute $\operatorname{dioph}_{2}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ by summing the measures of disjoint subsets of $\operatorname{Dioph}_{2}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$. For $k\geq 0$, let

[TABLE]

First, we consider $A_{0}$. If $(a,b)\in A_{0}$, then $ab+r=\beta$ where $\left(\frac{\beta}{p}\right)=1$. Thus

[TABLE]

since $\left(\frac{r}{p}\right)=-1$. This implies that for each choice of nonzero residue class of $a\pmod{p}$ and every nonzero square residue class choice of $\beta\pmod{p}$, there is exactly one choice of residue class $\pmod{p}$ for $b$ that results in $ab+r\in A_{0}$. Since there are $(p-1)/2$ nonzero square residue classes $\pmod{p}$, we have that

[TABLE]

Similarly, for $k\geq 1$, if $(a,b)\in A_{k}$, then $ab+r=p^{2k}\cdot\beta$ where $\left(\frac{\beta}{p}\right)=1$. Here

[TABLE]

since $r\not\equiv 0\pmod{p}$. such that $a\not\equiv 0\pmod{p}$ and the resulting possibilities for $b\pmod{p^{2k+1}}$, we obtain

[TABLE]

This yields the desired result:

[TABLE]

∎

Lemma 3.4**.**

Let $p$ be an odd prime and take $r\in\use@mathgroup\M@U\symAMSb{Z}_{p}$ so that $\left(\frac{r}{p}\right)=1$. Then,

[TABLE]

Proof.

As above, we compute $\operatorname{dioph}_{2}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ by summing the measures of disjoint subsets of $\operatorname{Dioph}_{2}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$. For $k\geq 0$, let

[TABLE]

We first consider $A_{0}$. If $(a,b)\in A_{0}$, then $ab+r=\beta$ where $\left(\frac{\beta}{p}\right)=1$. Since $\left(\frac{r}{p}\right)=1$, if either $a,b\equiv 0\pmod{p}$, then $ab+r\equiv r\pmod{p}$ is nonzero square in $\use@mathgroup\M@U\symAMSb{F}_{p}$ and thus a square in $\use@mathgroup\M@U\symAMSb{Z}_{p}$. We note that

[TABLE]

Else, we write $ab=\beta-r\pmod{p}$ This expression is nonzero as long as $\beta\neq r$. This implies that for each choice of nonzero residue class of $a\pmod{p}$ and every nonzero square residue class choice of $\beta\pmod{p}$ except $r\pmod{p}$, there is exactly one choice of residue class $\pmod{p}$ for $b$ that results in $ab+r\in A_{0}$. Since there are $(p-1)/2$ nonzero square residue classes $\pmod{p}$, we have that

[TABLE]

Similar to Lemma 3.3, for $k\geq 1$, if $(a,b)\in A_{k}$, then $ab+r=p^{2k}\cdot\beta$ where $\left(\frac{\beta}{p}\right)=1$. Here

[TABLE]

since $r\not\equiv 0\pmod{p}$. Considering all residue class of $a$ $\pmod{p}^{3}$ such that $a\not\equiv 0\pmod{p}$ and the resulting possibilities for $b\pmod{p}^{3}$, we obtain

[TABLE]

This yields the desired result:

[TABLE]

∎

Lemma 3.5**.**

Fix some $r\in\use@mathgroup\M@U\symAMSb{Z}_{p}$ such that $p\mid r$, with $v_{p}(r)=\alpha>0$. Then for all $\beta\equiv 0\pmod{2}$ such that $\beta>\alpha$, if

[TABLE]

then

[TABLE]

Proof.

We can write $r=p^{\alpha}s$ with $s\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$, and note that $\alpha\geq 1$. Note that for any $(a,b)\in B_{\beta}$, $ab+r=p^{\beta}q^{2}$ for some $q\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$. Thus, $(a,b)\in B_{\beta}$ iff there exists some $q\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$ such that

[TABLE]

Thus, we can compute $\mu(B_{\beta})$ using a probability over the residue classes of $a,b$ in $\use@mathgroup\M@U\symAMSb{Z}/p^{\beta+1}\use@mathgroup\M@U\symAMSb{Z}$. We can write $a=p^{e_{a}}a^{\prime},b=p^{e_{b}}b^{\prime}$ where $a^{\prime},b^{\prime}\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$, and $e_{a}+e_{b}=\alpha$. Then, there are $p^{\beta+1-e_{a}}-p^{\beta-e_{a}}$ choices for the residue $a\pmod{p}^{\beta+1}$, such that there exists some $q$ and choice(s) of $b\pmod{p}^{\beta+1}$ to satisfy the above equivalence.

For every such choice of $a$, there are $\frac{p-1}{2}$ choices for $b^{\prime}\pmod{p^{\beta-\alpha+1}}$, corresponding to the $\frac{p-1}{2}$ choices for the residue $q^{2}\pmod{p}$. There are $p^{\alpha}$ ways to lift this residue $\pmod{p^{\beta+1-\alpha}}$ to a residue $\pmod{p^{\beta+1}}$, yielding $p^{\alpha}\cdot(p-1)/2$ residue choices of $b\pmod{p^{\beta+1}}$ for each choice of $a$, such that there exists some $q\in\use@mathgroup\M@U\symAMSb{Z}_{p}$ where the above equivalence holds. Since $e_{a}$ can be any integer from [math] to $\alpha$, inclusive, this gives the desired result:

[TABLE]

∎

Lemma 3.6**.**

Fix some $r\in\use@mathgroup\M@U\symAMSb{Z}_{p}$ such that $p\mid r$, with $v_{p}(r)=\alpha$. Let $\beta\equiv 0\pmod{2}$ be such that $\beta<\alpha$, and set

[TABLE]

Then

[TABLE]

Proof.

As above, we write $r=p^{\alpha}s$ with $s\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$, and note that $\alpha\geq 1$. Note that for any $(a,b)\in B_{\beta}$, $ab+r=p^{\beta}q^{2}$ for some $q\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$. Thus, $(a,b)\in B_{\beta}$ iff there exists some $q\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$ such that

[TABLE]

Thus, we can compute $\mu(B_{\beta})$ using a probability over the residue classes of $a,b$ in $\use@mathgroup\M@U\symAMSb{Z}/p^{\beta+1}\use@mathgroup\M@U\symAMSb{Z}$. We can write $a=p^{e_{a}}a^{\prime},b=p^{e_{b}}b^{\prime}$ where $a^{\prime},b^{\prime}\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$, and $e_{a}+e_{b}=\beta$. Then, there are $p^{\alpha+1-e_{a}}-p^{\alpha-e_{a}}$ choices for the residue $a\pmod{p^{\alpha+1}}$, such that there exists some $q$ and choice(s) of $b\pmod{p^{\alpha+1}}$ to satisfy the above equivalence.

For every such choice of $a$, there are $\frac{p-1}{2}$ choices for $b^{\prime}\pmod{p}$, corresponding to the $\frac{p-1}{2}$ choices for the residue $q^{2}\pmod{p}$. There are $p^{\alpha-e_{b}}$ ways to lift this residue $b^{\prime}\pmod{p}$ to a residue of $b\pmod{p^{\alpha+1}}$, yielding $p^{\alpha-e_{b}}\cdot(p-1)/2$ residue choices of $b\pmod{p^{\beta+1}}$ for each choice of $a$, such that there exists some $q\in\use@mathgroup\M@U\symAMSb{Z}_{p}$ where the above equivalence holds. Since $e_{a}$ can be any integer from [math] to $\alpha$, inclusive, this gives the desired result:

[TABLE]

∎

Lemma 3.7**.**

Fix some $r\in\use@mathgroup\M@U\symAMSb{Z}_{p}$ with $v_{p}(r)=\alpha>0$ and let $r=p^{\alpha}s$. If

[TABLE]

then

[TABLE]

Proof.

Note that if $\alpha\equiv 1\pmod{2}$, then no element $x\in\use@mathgroup\M@U\symAMSb{Z}_{p}$ with $v_{p}(x)=\alpha$ is a square. So now assume that $\alpha\equiv 0\pmod{2}$. For any $(a,b)\in B_{\alpha}$, we have $ab+r=p^{\alpha}q^{2}$ for some $q\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$. Rearranging, we find that any $(a,b)\in B_{\alpha}$ satisfies $ab=p^{\alpha}(q^{2}-s)$ for some $q\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$. Note that if $v_{p}(q^{2}-s)=\gamma$, we can check if $(a,b)\in\operatorname{Dioph}_{2}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ by seeing if $ab+r$ is a square $\pmod{p^{\alpha+\gamma+1}}$. Thus, we consider two cases for $\gamma$.

If $\gamma=0$, then $q^{2}-s\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$, and we can check if $ab+r\in\operatorname{Dioph}_{2}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ by computing the residue $\pmod{p^{\alpha+1}}$. As above, we write $a=p^{e_{a}}a^{\prime},b=p^{e_{b}}b^{\prime}$ where $a^{\prime},b^{\prime}\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$ and $e_{a}+e_{b}=\alpha$. There are $p^{\alpha-e_{a}}(p-1)$ choices for $a\pmod{p^{\alpha+1}}$ such that there exists $b\in\use@mathgroup\M@U\symAMSb{Z}_{p}$ with $ab+r\in\operatorname{Dioph}_{2}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$. For each such fixed choice of a residue class for $a$, there are $\frac{p-1}{2}$ choices for $b^{\prime}\pmod{p}$ if $\left(\frac{s}{p}\right)=-1$, and $\frac{p-3}{2}$ choices for $b^{\prime}\pmod{p}$ if $\left(\frac{s}{p}\right)=1$, and $p^{\alpha-e_{b}}$ ways to lift each such choice of $b^{\prime}\pmod{p}$ to a residue $b\pmod{p^{\alpha+1}}$ such that

[TABLE]

for some $q\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$. In particular, this implies that if $\left(\frac{s}{p}\right)=-1$, then

[TABLE]

If $\gamma\geq 1$, then $\left(\frac{s}{p}\right)=1$. We can handle all of these cases together, since the residue of $q^{2}-s$ is restricted $\pmod{p}$, but not modulo higher powers of $p$. Let $q^{2}-s=p^{\gamma}z$ where $z\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{\times}$. Evaluating whether $ab+r=p^{\alpha}q^{2}\in\operatorname{Dioph}_{2}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ in these cases reduces to a check of $ab\pmod{p^{\alpha+\gamma+1}}$. Using the same decomposition of $a,b$ as above, $a=p^{e_{a}}a^{\prime},b=p^{e_{b}}b^{\prime}$ with $e_{a}+e_{b}=\alpha+\gamma$, we have $p^{\alpha+\gamma-e_{a}}(p-1)$ choices for $a\pmod{p^{\alpha+\gamma+1}}$ such that there exists some $b$ making the following equivalence hold:

[TABLE]

For each such choice of $a$, there are $p^{\alpha+\gamma-e_{b}}(p-1)$ choices for the residue $b\pmod{p^{\alpha+\gamma+1}}$ such that there is some $z$ satisfying the above equivalence. This yields a

[TABLE]

fraction of the residue classes $(a,b)\pmod{p}^{\alpha+\gamma+1}$ such that $ab+r\in\operatorname{Dioph}_{2}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ with $ab+r=p^{\alpha}q^{2}$ such that $v_{p}(q^{2}-s)=\gamma$. Combining these disjoint cases gives the desired equality when $\left(\frac{s}{p}\right)=1$:

[TABLE]

∎

Proof of Theorem 3.1.

The result follows immediately by applying Lemmas 3.3 and 3.4 when $\left(\frac{r}{p}\right)\neq 0$. Next, let $v_{p}(r)=\alpha>0$ and let $r=p^{\alpha}s$. We take two cases based on the parity of $\alpha$ and leverage the results of Lemmas 3.5,  3.6, and 3.7. First suppose $\alpha$ is odd. Then,

[TABLE]

Next suppose that $\alpha$ is even. Then

[TABLE]

Thus, when $\left(\frac{s}{p}\right)=1$,

[TABLE]

[TABLE]

and when $\left(\frac{s}{p}\right)=-1$,

[TABLE]

[TABLE]

∎

4. $\operatorname{Dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{3})$

We can exactly compute the Haar measure of $\operatorname{D}(r)$ $m$-tuples over $\use@mathgroup\M@U\symAMSb{Z}_{3}$ for a broad class of values $r$ and all $m\geq 2$. More precisely, we will show the following result:

Theorem 4.1**.**

For any $r\in\use@mathgroup\M@U\symAMSb{Z}_{3}$ with $\left(\frac{r}{3}\right)=1$, the Haar measure of $\operatorname{D}(r)$ $m$-tuples in $\use@mathgroup\M@U\symAMSb{Z}_{3}^{m}$ for $m\geq 2$ is

[TABLE]

Recall that $\alpha\in\square(\use@mathgroup\M@U\symAMSb{Z}_{3})$ iff $v_{3}(\alpha)$ is even and

[TABLE]

This leads us to the following observation:

Lemma 4.2**.**

Fix $r\in\use@mathgroup\M@U\symAMSb{Z}_{3}$ so that $(\frac{r}{3})=1$. Then any $\operatorname{D}(r)$ triple $(a,b,c)$ over $\use@mathgroup\M@U\symAMSb{Z}_{3}$ has at least one of $a,b,c\equiv 0\pmod{3}$.

Proof.

If $(\frac{r}{3})=1$, then $r\equiv 1\pmod{3}$. Suppose to the contrary that $a,b,c\not\equiv 0\pmod{3}$. Then either two of $a,b,c$ are $1\pmod{3}$ or $-1\pmod{3}$. Suppose WLOG that $a\equiv b\pmod{3}$. Then

[TABLE]

and thus

[TABLE]

so $ab+r$ is not a square over $\use@mathgroup\M@U\symAMSb{Z}_{3}$. Thus, $(a,b,c)$ cannot be a $\operatorname{D}(r)$ triple, a contradiction. ∎

Proof of Theorem 4.1.

Fix $r\in\use@mathgroup\M@U\symAMSb{Z}_{3}$ such that $\left(\frac{r}{3}\right)=1$. Lemma 4.2 implies that for a $\operatorname{D}(r)$ $m$-tuple $(a_{1},\ldots a_{m})$ with $(\frac{r}{3})=1$, all but at most two of the $a_{i}$’s are congruent to $0\pmod{3}$. We consider cases:

**Case 1: **

$a_{1},\ldots,a_{m}\equiv 0\pmod{3}$. Then for any $i\neq j$,

[TABLE]

Thus all of the above $m$-tuples are in $\operatorname{Dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{3})$. Therefore,

[TABLE]

**Case 2: **

Exactly one of the $a_{i}$’s is not congruent to $0\pmod{3}$. We have that whenever $i\neq j$

[TABLE]

since for all such pairs $i,j$ at least one of $a_{i},a_{j}\equiv 0\pmod{3}$. Since there are $m$ ways to choose which element of the triple is not $0\pmod{3}$ and two choices of its nonzero residue mod $3$, we have that

[TABLE]

**Case 3a: **

Suppose exactly two of $a_{1},\ldots a_{m}$ are not congruent to $0\pmod{3}$, $a_{i},a_{j}$, and those two are distinct modulo 3 with $a_{i}\equiv 1\pmod{3},a_{j}\equiv 2\pmod{3}$. Then without loss of generality $a_{1}\equiv 1\pmod{3},a_{2}\equiv 2\pmod{3}$. For $k_{1}\neq k_{2}$, $(k_{1},k_{2})\neq(i,j),(j,i)$, we have that

[TABLE]

Now, we compute the measure of pairs $(a_{i},a_{j})$ in $\use@mathgroup\M@U\symAMSb{Z}_{3}^{2}$ such that $a_{i}a_{j}+r\in\square(\use@mathgroup\M@U\symAMSb{Z}_{3})$. For $k=1,2,\ldots$, let

[TABLE]

Following the method of Lemma 3.4, we find that

[TABLE]

The measure inside $\use@mathgroup\M@U\symAMSb{Z}_{3}^{2}$ of the desired set is then

[TABLE]

Since there are $m(m-1)$ ways to pick the pair of elements that are $1,-1\pmod{3}$ in the $m$-tuple,

[TABLE]

**Case 3b: **

If for $(a_{1},\ldots a_{m})$ there are exactly 2 elements $a_{i},a_{j}\not\equiv 0\pmod{3}$, but $a_{i}\equiv a_{j}\pmod{3}$, then

[TABLE]

meaning that $(a_{1},\ldots a_{m})$ cannot be a $D(r)$-tuple over $\use@mathgroup\M@U\symAMSb{Z}_{3}$.

**Case 4: **

If at least $3$ $a_{i}$ (WLOG $a_{1},a_{2},a_{3}$) are not $0\mod 3$, then via Lemma 4.2, we can find $i\neq j$ with $a_{i}a_{j}+1\not\in\square(\use@mathgroup\M@U\symAMSb{Z}_{3})$, so $(a_{1},\ldots a_{m})$ is not a Diophantine $m$-tuple.

The above allows us to compute the Haar measure of $\operatorname{D}(r)$ $m$-tuples in $\use@mathgroup\M@U\symAMSb{Z}_{3}^{m}$ for $r$ such that $(\frac{r}{3})=1$ by summing the Haar measures over disjoint subsets of $\use@mathgroup\M@U\symAMSb{Z}_{3}^{m}$.

[TABLE]

∎

5. $\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$

Throughout this section, we take $p$ to be an odd prime. We give asymptotic bounds on $\operatorname{dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$, thereby showing that over triples $(a,b,c)\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{3}$ the events that $(a,b),(b,c),(a,c)$ are $\operatorname{D}(r)$ pairs are asymptotically independent.

Definition 5.1**.**

For a prime $p$, $m\geq 2,r\in\use@mathgroup\M@U\symAMSb{Z}_{p}$, we let

[TABLE]

and

[TABLE]

Then, let $\widetilde{\operatorname{Dioph}}_{m}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ be the image of $\widetilde{\operatorname{Dioph}}_{m}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ under the reduction map $\use@mathgroup\M@U\symAMSb{Z}_{p}\rightarrow\use@mathgroup\M@U\symAMSb{F}_{p}$ and define $\widetilde{\operatorname{dioph}}_{m}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})=\mu(\widetilde{\operatorname{Dioph}}_{m}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p}))$.

Theorem 5.2**.**

For fixed $r\in\use@mathgroup\M@U\symAMSb{Z}$ viewed in $\use@mathgroup\M@U\symAMSb{Z}_{p}$, we have that as $p\rightarrow\infty$,

[TABLE]

To prove Theorem 5.2, we will first understand the intermediate quantity $\operatorname{dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$.

Theorem 5.3**.**

For fixed $r\in\use@mathgroup\M@U\symAMSb{Z}$ viewed in $\use@mathgroup\M@U\symAMSb{F}_{p}$ with $p\nmid r$ (for odd primes $p$), we have that

[TABLE]

We recall a result about character sums that will be useful in counting Diophantine $D(r)$-tuples over $\use@mathgroup\M@U\symAMSb{F}_{p}$:

Lemma 5.4** (Theorem 5.48 [LN97]).**

Let $f(x)=a_{2}x^{2}+a_{1}x+a_{0}\in\use@mathgroup\M@U\symAMSb{F}_{p}[x]$ and $a_{2}\neq 0$. Let $\Delta=a_{1}^{2}-4a_{0}a_{2}$ be the discriminant of $f$. Then,

[TABLE]

Lemma 5.5**.**

For fixed $r\in\use@mathgroup\M@U\symAMSb{Z}$ viewed in $\use@mathgroup\M@U\symAMSb{F}_{p}$ that satisfies $\left(\frac{r}{p}\right)=1$, we have that

[TABLE]

Proof.

Let $r=s^{2}\neq 0$. Note that if at least $2$ of $a,b,c$ are divisible by $p$, then

[TABLE]

and thus $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$. The size of this subset of $\use@mathgroup\M@U\symAMSb{F}_{p}^{3}$ is $3(p-1)+1=3p-2$. Now suppose that exactly one of $a,b,c$ is divisible by $p$, which we shall assume without loss of generality to be $a$. Then

[TABLE]

Consequently, $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ iff $bc+r\in\square(\use@mathgroup\M@U\symAMSb{F}_{p})$. Since $r=s^{2}$ and $bc\not\equiv 0\pmod{p}$, for each of the $p-1$ choices of $b$, there are $(p-1)/2$ choices of $c$ that yield squares, and thus, we have

[TABLE]

triples $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ with exactly one of $a,b,c\equiv 0\pmod{p}$. This gives the desired result:

[TABLE]

∎

Lemma 5.6**.**

For fixed $r\in\use@mathgroup\M@U\symAMSb{Z}$ viewed in $\use@mathgroup\M@U\symAMSb{F}_{p}$ that satisfies $\left(\frac{r}{p}\right)=-1$, we have that

[TABLE]

Proof.

If $abc\equiv 0\pmod{p}$, then without loss of generality $a\equiv 0\pmod{p}$. Then $ab+r\equiv r\pmod{p}$, so $ab+r\not\in\square(\use@mathgroup\M@U\symAMSb{F}_{p})$, since $\left(\frac{r}{p}\right)=-1$. Therefore, $(a,b,c)\not\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$, giving the desired result. ∎

Lemma 5.7**.**

For fixed $r\in\use@mathgroup\M@U\symAMSb{Z}$ viewed in $\use@mathgroup\M@U\symAMSb{F}_{p}$ such that $p\nmid r$, we have that

[TABLE]

Proof.

We consider cases:

1. (1)

Suppose $ab+r\equiv ac+r\equiv ab+r\equiv 0\pmod{p}$. Then, since $a\not\equiv 0\pmod{p}$,

[TABLE]

By symmetry, this yields, $a\equiv b\equiv c\pmod{p}$, yielding two points in $\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ if $\left(\frac{-r}{p}\right)=1$, else no point. Thus, there are

[TABLE]

points in $\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ with $ab+r\equiv ac+r\equiv ab+r\equiv 0\pmod{p}$. 2. (2)

Now without loss of generality suppose $ab+r\equiv bc+r\equiv 0\pmod{p}$ but $ac+r\not\equiv 0\pmod{p}$. We consider values for $b$ in a triple $(a,b,c)$ and note that $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ exactly if

[TABLE]

Note that as $b$ ranges over all nonzero ($ab+r\equiv 0\pmod{p}$ implies $b\not\equiv 0\pmod{p}$) elements of $\use@mathgroup\M@U\symAMSb{F}_{p}$, $r^{2}/b^{2}$ ranges over all nonzero squares in $\use@mathgroup\M@U\symAMSb{F}_{p}$. Thus let $r/b=X$ and consider the following quadratic function in $\use@mathgroup\M@U\symAMSb{F}_{p}[X]$:

[TABLE]

We observe that for each $X\neq 0$ with $\left(\frac{f(X)}{p}\right)=1$, we have a single choice of $b=r/X$ that gives a point $(a,b,c)\in\operatorname{Dioph}_{3}^{r}$. We proceed to count the number of such $X$ as follows:

[TABLE]

where to obtain ($\ast$), we apply Lemma 5.4. Thus, we have $\frac{1}{2}\left(p-1-\left(\frac{r}{p}\right)+\left(\frac{-r}{p}\right)\right)$ such points $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$. 3. (3)

Finally, suppose that exactly one of $ab+r,bc+r,ac+r\equiv 0\pmod{p}$, WLOG $ab+r\equiv 0\pmod{p}$, so $a=-rb^{-1}$. Then, we wish to count the number of choices of $b,c\in\use@mathgroup\M@U\symAMSb{F}_{p}\backslash\{0\}$ such that

[TABLE]

Since $r\not\equiv 0\pmod{p}$, it suffices for $bc+r\in\square(\use@mathgroup\M@U\symAMSb{F}_{p})\backslash\{0\},\left(\frac{1-cb^{-1}}{p}\right)=\left(\frac{r}{p}\right)$.

Note that for any of the $(p-1)/2$ nonzero values $t^{2}\in\use@mathgroup\M@U\symAMSb{F}_{p}$, $r(1-cb^{-1})=t^{2}$ exactly if we take $c=b(1-r^{-1}t^{2})$. For any such fixed $t^{2}$, $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ with $bc+r\neq 0$ for all choices of $b$ such that

[TABLE]

Since $c\neq 0$, $t^{2}\neq r$. Therefore, we can find all such $b$ by applying Lemma 5.4 to

[TABLE]

finding that the number of such $b$ is given by

[TABLE]

We can sum over the all the possibilities of $t^{2}\in\square(\use@mathgroup\M@U\symAMSb{F}_{p}^{\times})$ to count the number of $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ with $ab+r\equiv 0\pmod{p}$ and $(bc+r)(ac+r)abc\not\equiv 0\pmod{p}$:

[TABLE]

where as above $(\ast)$ follows by Lemma 5.4 applied to $g(t)=1-r^{-1}t^{2}$.

Using the symmetry between $a,b,c$, we can use the above three cases to obtain the desired result:

[TABLE]

∎

Lemma 5.8**.**

For $r\in\use@mathgroup\M@U\symAMSb{Z}$ with $p\nmid r$ viewed as an element of $\use@mathgroup\M@U\symAMSb{F}_{p}$, we have that

[TABLE]

Proof.

We first compute the count of triples in $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ such that $abc(ab+r)(bc+r)(ac+r)\neq 0.$ Let $\alpha=ab,\beta=bc$ and let $\gamma=b^{-1}$, so that $\alpha\beta\gamma^{2}=ac$. This gives the following:

[TABLE]

Recall that $p\nmid r$ and first consider the two outer sums. As $\alpha,\beta$ range over the summation, they range over all values in $\use@mathgroup\M@U\symAMSb{F}_{p}$ except $r$, but contribute a nonzero value only when $\left(\frac{\alpha}{p}\right)=\left(\frac{\beta}{p}\right)=1$. We first consider the inner summation, recalling that $\alpha,\beta\neq r$:

[TABLE]

where as earlier, we apply Lemma 5.4 in $(\ast)$. This simplification gives

[TABLE]

where we use Lemma 5.4 in $(\ast)$. This implies that

[TABLE]

∎

Proof of Theorem 5.3.

We add the results of Lemmas 5.5–5.8 to give a unified expression for all $p\nmid r$. We thus obtain the desired Haar measure

[TABLE]

∎

Proof of Theorem 5.2.

The result follows by noting that

[TABLE]

as if $abc(ab+r)(bc+r)(ac+r)\not\equiv 0\pmod{p}$, then $ab+r,bc+r,ac+r\in\square(\use@mathgroup\M@U\symAMSb{F}_{p})$ if and only if $ab+r,bc+r,ac+r\in\square(\use@mathgroup\M@U\symAMSb{Z}_{p})$. Further,

[TABLE]

Combining this bound with the result of Theorem 5.3 gives the asymptotic result:

[TABLE]

This implies that

[TABLE]

as desired. ∎

6. $\operatorname{Dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ for $m\geq 4$

In this section, we take $p$ to be an odd prime. It appears to be difficult to compute $\operatorname{dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ or $\operatorname{dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ exactly, but it is possible to give asymptotics. In the case of $r=1$, this was done for $\use@mathgroup\M@U\symAMSb{F}_{p}$ by Dujella and Kazalicki in [DK16]; they proved that

[TABLE]

Their technique is similar to ours in the previous section, involving character sums evaluated at the values of polynomials. However, the best analogue of Lemma 5.4 in this case is only an inequality. To do this in the case of $r=1$, Dujella and Kazalicki consider the sum

[TABLE]

They note that the main contribution to this sum is

[TABLE]

obtained by choosing the 1 from each factor of the form $1+\left(\frac{a_{i}a_{j}+1}{p}\right)$. The remaining terms can be bounded using the Weil bound on character sums (or, equivalently, the Hasse–Weil bound for the number of $\use@mathgroup\M@U\symAMSb{F}_{p}$-points on a genus-$g$ curve): for fixed $m$, the remaining terms are $O(p^{-1/2})$ by [LN97, Theorem 5.41].

The arguments in [DK16] do not make any use of $r=1$; they only rely on $p\nmid r$. Thus we have the following:

Theorem 6.1**.**

For all fixed $m\geq 3$ and $r\in\use@mathgroup\M@U\symAMSb{Z}$, we have that

[TABLE]

as $p\to\infty$.

If we instead work with $\use@mathgroup\M@U\symAMSb{Z}_{p}$ instead of $\use@mathgroup\M@U\symAMSb{F}_{p}$, we have

[TABLE]

as in the previous section, and that

[TABLE]

so that

[TABLE]

as well.

We also note that there is a second technique for estimating $\operatorname{dioph}_{4}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ and $\operatorname{dioph}_{4}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$, which we now describe.

Proposition 6.2**.**

For any fixed $r\in\use@mathgroup\M@U\symAMSb{Z}$ viewed in $\use@mathgroup\M@U\symAMSb{Z}_{p}$, we have that as $p\rightarrow\infty$

[TABLE]

It is helpful to begin by showing the relevant result for tuples over $\use@mathgroup\M@U\symAMSb{F}_{p}$:

Proposition 6.3**.**

For any fixed $r\in\use@mathgroup\M@U\symAMSb{Z}$ viewed in $\use@mathgroup\M@U\symAMSb{F}_{p}$, we have that as $p\rightarrow\infty$

[TABLE]

Proof.

Let $(a,b,c,d)\in\use@mathgroup\M@U\symAMSb{F}_{p}^{4}$. Note that $(a,b,c,d)\in\operatorname{Dioph}_{4}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ only if $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ and if $ab+r,bc+r,ac+r\in\square(\use@mathgroup\M@U\symAMSb{F}_{p})$. Assume that $a,b,c$ are all distinct and let

[TABLE]

Fix $(a_{0},b_{0},c_{0})\in S_{0}$ and consider the elliptic curve $E$ defined as

[TABLE]

The Hasse–Weil bound gives that

[TABLE]

Excluding the point at infinity, we find that this gives between $\frac{p-2\sqrt{p}}{2}-2$ and $\frac{p+2\sqrt{p}}{2}$ choices of $d\neq 0\in\use@mathgroup\M@U\symAMSb{F}_{p}$, such that there is a point $(X,Y)\in E(\use@mathgroup\M@U\symAMSb{F}_{p})$ with $X=d$. Of course, not all of these choices $d\in\use@mathgroup\M@U\symAMSb{F}_{p}$ make $a_{0}d+r,b_{0}d+r,c_{0}d+r\in\square(\use@mathgroup\M@U\symAMSb{F}_{p})$; they merely guarantee that the product $(a_{0}d+r)(b_{0}d+r)(c_{0}d+r)\in\square(\use@mathgroup\M@U\symAMSb{F}_{p})$. For our fixed $(a_{0},b_{0},c_{0})\in S_{0}$, the values $d\in\use@mathgroup\M@U\symAMSb{F}_{p}$ that make $(a_{0},b_{0},c_{0},d)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ correspond exactly to the choices of $X$ for which there is a $Y\in\use@mathgroup\M@U\symAMSb{F}_{p}$ with $(X,Y)\in 2E(\use@mathgroup\M@U\symAMSb{F}_{p})$ by [Kna92, Theorem 4.2]. $E(\use@mathgroup\M@U\symAMSb{F}_{p})$ has full 2-torsion as seen via the factored form of $E$ in Equation (6.1). Thus $E(\use@mathgroup\M@U\symAMSb{F}_{p})\cong\use@mathgroup\M@U\symAMSb{Z}/t_{1}\use@mathgroup\M@U\symAMSb{Z}\times\use@mathgroup\M@U\symAMSb{Z}/t_{2}\use@mathgroup\M@U\symAMSb{Z}$, where $t_{1},t_{2}\equiv 0\pmod{2}$. Thus, $\frac{1}{4}$ of the points on $E(\use@mathgroup\M@U\symAMSb{F}_{p})$ are in $2E(\use@mathgroup\M@U\symAMSb{F}_{p})$ and $(X,Y)\in 2E(\use@mathgroup\M@U\symAMSb{F}_{p})$ if and only if $(X,-Y)\in 2E(\use@mathgroup\M@U\symAMSb{F}_{p})$, which since $p>2$, gives the following bound (noting that the point at infinity is guaranteed to be a multiple of $2$):

[TABLE]

Notice that

[TABLE]

Further, observe that

[TABLE]

Consequently,

[TABLE]

Using the bound of Equation (6.2) and recalling that via Theorem 5.3, $\operatorname{dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})=\frac{1}{8}+O\left(\frac{1}{p}\right)$ in conjunction with the above, we find that

[TABLE]

∎

Proof of Proposition 6.2.

Let $(a,b,c,d)\in\use@mathgroup\M@U\symAMSb{Z}_{p}^{4}$. Note that $(a,b,c,d)\in\operatorname{Dioph}_{4}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ if and only if both $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ and $ab+r,bc+r,ac+r\in\square(\use@mathgroup\M@U\symAMSb{Z}_{p})$. For every $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$, $(a,b,c)\in\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ where we view $a,b,c$ under the natural map $\use@mathgroup\M@U\symAMSb{Z}_{p}\rightarrow\use@mathgroup\M@U\symAMSb{F}_{p}$. For $a_{1},a_{2}\in\use@mathgroup\M@U\symAMSb{Z}_{p}$ unless $a_{1}a_{2}+r\equiv 0\pmod{p}$, then $a_{1}a_{2}+r\in\square(\use@mathgroup\M@U\symAMSb{Z}_{p})$ if and only if $a_{1}a_{2}+r\in\square(\use@mathgroup\M@U\symAMSb{F}_{p})$. Thus there are at most $3$ choices of $d\pmod{p}$, such that $(ad+r)(bd+r)(cd+r)\equiv 0\pmod{p}$ for $a,b,c,d\in\use@mathgroup\M@U\symAMSb{F}_{p}$. This computation implies that

[TABLE]

and consequently as desired,

[TABLE]

∎

A similar technique can be used to show that $\operatorname{dioph}_{5}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$ and $\operatorname{dioph}_{5}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ are $\frac{1}{1024}+O\left(\frac{1}{\sqrt{p}}\right)$, since the curve $y^{2}=(ax+r)(bx+r)(cx+r)(dx+r)$ is again an elliptic curve. There is an analogue of [Kna92, Theorem 4.2] for hyperelliptic curves (see [Zar16]). Let $C$ be the hyperelliptic curve $y^{2}=(a_{1}x+r)\cdots(a_{m-1}x+r)$, where $m-1$ is odd and $a_{1},\ldots,a_{m-1}\in\use@mathgroup\M@U\symAMSb{F}_{p}$ are distinct. Let $(x_{0},y_{0})\in C(\use@mathgroup\M@U\symAMSb{F}_{p})$ be a point. Then $a_{i}x_{0}+r$ is a perfect square for all $i$ with $1\leq i\leq m-1$ if and only if the point $[(x_{0},y_{0})-\infty]\in 2\operatorname{Jac}(C)(\use@mathgroup\M@U\symAMSb{F}_{p})$, where $\operatorname{Jac}(C)$ is the Jacobian of $C$. So, in order to estimate $\operatorname{dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{F}_{p})$, it suffices to understand the proportion of points of the form $[(x_{0},y_{0})-\infty]$ in $\operatorname{Jac}(C)(\use@mathgroup\M@U\symAMSb{F}_{p})$ that are multiples of 2 in $\operatorname{Jac}(C)$. While we know that the 2-torsion in $\operatorname{Jac}(C)(\use@mathgroup\M@U\symAMSb{F}_{p})$ is isomorphic to $(\use@mathgroup\M@U\symAMSb{Z}/2\use@mathgroup\M@U\symAMSb{Z})^{2g}$, and so that the proportion of points in $\operatorname{Jac}(C)(\use@mathgroup\M@U\symAMSb{F}_{p})$ which are multiples of 2 is $\frac{1}{2^{2g}}$, what we need to know is that of the points of the more specific form $[(x_{0},y_{0})-\infty]\in\operatorname{Jac}(C)(\use@mathgroup\M@U\symAMSb{F}_{p})$, the proportion which are multiples of 2 is $\frac{1}{2^{2g}}$, up to a small error term. This does not appear to be so easy, since we cannot guarantee that points of this specific form in the Jacobian are “representative” with respect to being multiples of 2.

7. $\operatorname{Dioph}_{m}^{r}(\mathcal{O}_{K})$ for $K/\use@mathgroup\M@U\symAMSb{Q}_{p}$

Many of the methods highlighted above carry over when we replace $\use@mathgroup\M@U\symAMSb{Z}_{p}$ by a finite extension. Let $K/\use@mathgroup\M@U\symAMSb{Q}_{p}$ be a finite extension with valuation ring $\mathcal{O}_{K}$ and residue field $\kappa=\use@mathgroup\M@U\symAMSb{F}_{q}$ for $q=p^{f}$ and $f\geq 1$.

Applying the methods of Section 3 replacing $p$ by $q$ as appropriate and the Legendre symbol with $\chi$, the quadratic multiplicative character over $\use@mathgroup\M@U\symAMSb{F}_{q}$, yields the following result:

Theorem 7.1**.**

Suppose $q$ is a power of an odd prime. With notation as above, let $\alpha=v_{\pi}(r)$ with $r=\pi^{\alpha}s$ where $\pi$ is a uniformizer and $s\in\mathcal{O}_{K}^{\times}$. Then, we have the following:

[TABLE]

where $\chi$ denotes the quadratic multiplicative character on $\use@mathgroup\M@U\symAMSb{F}_{q}$. In particular, we have that the measure of the Diophantine pairs over $\mathcal{O}_{K}$ is given by

[TABLE]

In the case that $K/\use@mathgroup\M@U\symAMSb{Q}_{3}$ is totally ramified, we obtain an analogue of Theorem 4.1 in Section 4:

Theorem 7.2**.**

If $K/\use@mathgroup\M@U\symAMSb{Q}_{3}$ is a finite, totally ramified extension, then for any $r\in\mathcal{O}_{K}$ with $\chi(r)=1$, the Haar measure of $\operatorname{D}(r)$ $m$-tuples in $\mathcal{O}_{K}^{m}$ for $m\geq 2$ is

[TABLE]

In Sections 5 and 6, the computations of $\operatorname{Dioph}_{3}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ and $\operatorname{Dioph}_{m}^{r}(\use@mathgroup\M@U\symAMSb{Z}_{p})$ extend to a bound when we replace $\use@mathgroup\M@U\symAMSb{Z}_{p}$ with $\mathcal{O}_{K}$ for an arbitrary finite $K/\use@mathgroup\M@U\symAMSb{Q}_{p}$. (Note that the result of Lemma 5.4 generalizes to one over $\use@mathgroup\M@U\symAMSb{F}_{q}$, replacing the Legendre symbol with $\chi$ discussed above.)

Theorem 7.3**.**

Let $K_{1},K_{2},\ldots$ be a sequence of characteristic [math] local fields, and let $p_{i}$ be the residue characteristic of $K_{i}$ such that $p_{i}\rightarrow\infty$ as $i\to\infty$. Then, as $i\to\infty$,

[TABLE]

Theorem 7.4**.**

Let $m\geq 4$ be an integer, let $K_{1},K_{2},\ldots$ be a sequence of characteristic [math] local fields, and let $p_{i}$ be the residue characteristic of $K_{i}$ such that $p_{i}\rightarrow\infty$ as $i\to\infty$. Then, as $i\to\infty$,

[TABLE]