Distinct coordinate solutions of linear equations over finite fields

TL;DR

This paper derives explicit formulas for counting solutions with distinct variables to linear equations over finite fields and rings, extending previous results with new combinatorial and graph-theoretic methods.

Contribution

It provides new explicit formulas for the number of distinct solutions to linear equations over finite fields and rings, including special cases and a novel proof of existing theorems.

Findings

Explicit formulas for solution counts over finite fields.

Closed-form solutions for special coefficient cases.

A new proof of a generalized theorem using graph theory.

Abstract

Let $\mathbb{F}_q$ be the finite field of $q$ elements and $a_1,a_2, \ldots, a_k, b\in \mathbb{F}_q$. We investigate $N_{\mathbb{F}_q}(a_1, a_2, \ldots,a_k;b)$, the number of ordered solutions $(x_1, x_2, \ldots,x_k)\in\mathbb{F}_q^k$ of the linear equation $$ a_1x_1+a_2x_2+\cdots+a_kx_k=b$$ with all $x_i$ distinct. We obtain an explicit formula for $N_{\mathbb{F}_q}(a_1,a_2, \ldots, a_k;b)$ involving combinatorial numbers depending on $a_i$'s. In particular, we obtain closed formulas for two special cases. One is that $a_i, 1\leq i\leq k$ take at most three distinct values and the other is that $\sum_{i=1}^ka_i=0$ and $\sum_{i\in I}a_i\neq 0$ for any $I\subsetneq [k]$. The same technique works when $\mathbb{F}_q$ is replaced by $\mathbb{Z}_n$, the ring of integers modulo $n$. In particular, we give a new proof for the main result given by Bibak, Kapron and Srinivasan, which generalizes a theorem of Sch\"{o}nemann via a graph theoretic method.

Equations

$$a_1{x}_1+a_2{x}_2+\cdots +a_k{x}_k=b$$

$$a_1{x}_1+a_2{x}_2+\cdots +a_k{x}_k=b,$$

$$N_D(a_1,a_2,\dots ,a_k;b)=\#\{(x_1,x_2,\dots ,x_k)\in D^k:a_1{x}_1+a_2{x}_2+\cdots +a_k{x}_k=b:x_i\ne x_j,\forall i\ne j\}.$$

$$N_{{\mathbb{F}}_q}(\underset{k}{\underbrace{1,\dots ,1}};b)=\frac{1}{q}(q{)}_k,$$

$$N_{{\mathbb{F}}_q}(\underset{k}{\underbrace{1,\dots ,1}};b)=\frac{1}{q}(q{)}_k+(-1{)}^{k+k/p}\frac{v(b)}{q}k!\left(\genfrac{}{}{0ex}{}{q/p}{k/p}\right).$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=\frac{1}{q}(q{)}_k.$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=\frac{1}{q}(q{)}_k-\frac{v(b)}{q}\left(d(a_1,a_2,\dots ,a_k)+\sum _{i=1}^{k}d(a_1,a_2,\dots ,a_{i-1},\widehat{a_i},a_{i+1},\dots ,a_k)\right),$$

$$d(a_1,a_2,\dots ,a_k)=\{\begin{array}{ll}-v(a_1),& \text{if}k=1;\\ -{\sum }_{i=1}^{k}d(a_1,a_2,\dots ,a_{i-1},\widehat{a_i},a_{i+1},\dots ,a_k),& \text{if}k\ge 2\text{and}{\sum }_{i=1}^k{a}_i\ne 0;\\ (q-k)d(a_1,a_2,\dots ,a_{i-1},\widehat{a_i},a_{i+1},\dots ,a_k),& \text{if}k\ge 2\text{and}{\sum }_{i=1}^k{a}_i=0.\end{array}$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=\frac{1}{q}(q{)}_k+\frac{v(b)}{q}\sum _{i=1}^n(-1{)}^{k-i}p(a_1,a_2,\dots ,a_k;k,i)q^i.$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=\frac{1}{q}(q{)}_k+v(b)(-1{)}^{k-1}(k-1)!.$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\underset{k-2}{\underbrace{1,\dots ,1}};b)=\frac{1}{q}(q{)}_k.$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\underset{k-2}{\underbrace{1,\dots ,1}};b)=\frac{1}{q}(q{)}_k+v(b)(-1{)}^{k-1+\lfloor \frac{k-2}{p}\rfloor }(k-2)!\left(\frac{q\{k-2{\}}_p-p(k-2)}{q-p}+1\right)(\genfrac{}{}{0ex}{}{q/p-1}{\lfloor (k-2)/p\rfloor }).$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\underset{k-2}{\underbrace{1,\dots ,1}};b)=\frac{1}{q}(q{)}_k+v(b)(-1{)}^{k-1+\lfloor \frac{k-1}{p}\rfloor }(k-2)!(k-1-q{1}_A(a_1,a_2))(\genfrac{}{}{0ex}{}{q/p-1}{\lfloor (k-1)/p\rfloor }),$$

$$N_{{\mathbb{F}}_p}(a_1,a_2,\dots ,a_k;0)=\frac{1}{p}(p{)}_k+(-1{)}^{k-1}(k-1)!(p-1).$$

$$N_{\mathbb{Z}/n\mathbb{Z}}(a_1,a_2,\dots ,a_k;b)=\#\{(x_1,x_2,\dots ,x_k)\in (\mathbb{Z}/n\mathbb{Z}{)}^k:a_1{x}_1+a_2{x}_2+\cdots +a_k{x}_k=b:x_i\ne x_j,\forall i\ne j\}.$$

$$N_{\mathbb{Z}/n\mathbb{Z}}(a_{1},a_{2},\ldots,a_{k};b)=\begin{cases}\frac{1}{n}(n)_{k}+(-1)^{k}(k-1)!,&\text{if}\ \gcd(\sum_{i=1}^{k}a_{i},n)\nmid b;\\ \frac{1}{n}(n)_{k}+(-1)^{k-1}(k-1)!\big{(}\gcd(\sum_{i=1}^{k}a_{i},n)-1\big{)},&\text{if}\ \gcd(\sum_{i=1}^{k}a_{i},n)\mid b.\end{cases}$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b+Ac)$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;1)$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=\frac{1}{q}(q{)}_k.$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=\frac{1}{q}(q{)}_k-\frac{v(b)}{q}(N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;1)-N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;0)),$$

$$\sum _{b\in {\mathbb{F}}_q}{N}_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=(q{)}_k.$$

$$a_1{x}_1+\cdots +a_i{x}_i+\cdots +a_j{x}_j+\cdots +a_k{x}_k=0$$

$$a_1{x}_1+\cdots +a_i{x}_j+\cdots +a_j{x}_i+\cdots +a_k{x}_k=0$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=N_{{\mathbb{F}}_q^{\ast }}(a_1,a_2,\dots ,a_k;b-Ac)+\sum _{i=1}^k{N}_{{\mathbb{F}}_q^{\ast }}(a_1,a_2,\dots ,a_{i-1},\widehat{a_i},a_{i+1},\dots ,a_k;b-Ac),$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;1)$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;0)$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=q{N}_{{\mathbb{F}}_q^{\ast }}(a_1,a_2,\dots ,a_{i-1},\widehat{a_i},a_{i+1},\dots ,a_k;b)$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=q{N}_{{\mathbb{F}}_q^{\ast }}(a_2,\dots ,a_k;b)$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;b)=q{N}_{{\mathbb{F}}_q^{\ast }}(a_1,a_2,\dots ,a_{i-1},\widehat{a_i},a_{i+1},\dots ,a_k;b)$$

$$d(a_1,a_2,\dots ,a_k):=N_{{\mathbb{F}}_q^{\ast }}(a_1,a_2,\dots ,a_k;1)-N_{{\mathbb{F}}_q^{\ast }}(a_1,a_2,\dots ,a_k;0).$$

$$N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;1)-N_{{\mathbb{F}}_q}(a_1,a_2,\dots ,a_k;0)=d(a_1,a_2,\dots ,a_k)+\sum _{i=1}^{k}d(a_1,a_2,\dots ,a_{i-1},\widehat{a_i},a_{i+1},\dots ,a_k).$$

Full text

[TABLE]

with all $x_{i}$ distinct. We obtain an explicit formula for $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ involving combinatorial numbers depending on $a_{i}$’s. In particular, we obtain closed formulas for two special cases. One is that $a_{i},1\leq i\leq k$ take at most three distinct values and the other is that $\sum_{i=1}^{k}a_{i}=0$ and $\sum_{i\in I}a_{i}\neq 0$ for any $I\subsetneq\{1,2,\dots,k\}$.

The same technique works when $\mathbb{F}_{q}$ is replaced by $\mathbb{Z}_{n}$, the ring of integers modulo $n$. In particular, we give a new proof for the main result given by Bibak, Kapron and Srinivasan ([2]), which generalizes a theorem of Schönemann via a graph theoretic method.

This work is supported by the National Science Foundation of China (11771280) and the National Science Foundation of Shanghai Municipal (17ZR1415400, 19ZR1424100).

1. Introduction

Let $\mathbb{F}_{q}$ be the finite field of $q$ elements of characteristic $p$ and $D$ be a subset in $\mathbb{F}_{q}$. Given $a_{1},a_{2},\ldots,a_{k},b\in\mathbb{F}_{q}$, we are interested in the number of solutions of the linear equation over $\mathbb{F}_{q}$

[TABLE]

with the restriction that all $x_{i}$ in $D$ are distinct, that is, the cardinality of the set

[TABLE]

This number is closely related to the reduced degree of a map over $\mathbb{F}_{q}$. Any map from $\mathbb{F}_{q}$ to $\mathbb{F}_{q}$ can be uniquely represented by a polynomial of degree at most $q-1$. The degree of such a polynomial is called the reduced degree of the map. Suppose that the range of $f$ is $\{a_{1},a_{2},\ldots,a_{q}\}$(with multiplicity counted). By the Lagrange interpolation formula, it is direct to check that $f$ is a polynomial of degree at most $q-2$ if and only if $\sum_{i=1}^{q}a_{i}=0$, and $f$ is a polynomial of degree at most $q-3$ if and only if $\sum_{i=1}^{q}a_{i}=0$ and $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{q};0)>0$. In particular, it is well-known (see for example, [8]) that $N_{\mathbb{F}_{q}^{*}}(1,\omega,\omega^{2},\dots,\omega^{q-2};0)$ counts the number of permutation polynomials of $\text{degree}\leq q-3$ over $\mathbb{F}_{q}$, where $\omega$ is a primitive element of $\mathbb{F}_{q}$. For more related work we refer to [12, 13, 20].

Furthermore, $N_{D}(a_{1},a_{2},\dots,a_{k};b)$ can be naturally regarded as a counting version of knapsack type problem over finite rings. In particular, when $a_{i}=1$, $1\leq i\leq k$, this is the counting version of subset sum problem, a well-known ${\bf\#P}$ problem in theoretical computer science. It thus has many applications in coding theory and number theory. For details we refer to [4, 5, 6, 7, 9]. Note that if all $a_{i}$’s lie in the prime field $\mathbb{F}_{p}$, then this problem is a restricted composition problem over $\mathbb{F}_{q}$; see [10] for a broad generalization.

Few results are known for arbitrary $a_{i}$’s, even for special cases such as $D=\mathbb{F}_{q}$ or $D=\mathbb{F}_{q}^{*}$. In [9], Gács et al. proved that $N_{\mathbb{F}_{q}}(a_{1},a_{2},\dots,a_{k};0)>0$ always holds except some obvious degenerate cases by using the polynomial method; later, Nagy [21] extended the result to cyclic groups. Another result, proved by Li and Wan [15], gives a first explicit formula for $N_{D}(a_{1},a_{2},\dots,a_{k};b)$ when $a_{1}=a_{2}=\cdots=a_{k}=1$ and $|D|\geq q-2$.

Theorem 1.1** ([15], Theorem 1.2).**

Define $v(b)=-1$ if $b\neq 0$, and $v(b)=q-1$ if $b=0$. If $p\nmid k$, then

[TABLE]

and if $p\mid k$, then

[TABLE]

Along this way, Li and Wan gave a series of asymptotic estimates on $N_{D}({1,1,\ldots,1};b)$ for many different kinds of $D\subseteq\mathbb{F}_{q}$; see for example [17, 18, 19].

In this paper, we first prove that apart from some exceptions that can be classified the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}=1$ always has a solution with distinct coordinates.

Theorem 1.2**.**

Suppose $q\geq 3$ and $k\leq q$. Then the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}=1$ has a solution $(x_{1},x_{2},\ldots,x_{k})\in\mathbb{F}_{q}^{k}$ with all $x_{i}$ distinct, unless one of the following holds: (i) $k<q$ and $a_{1}=a_{2}=\cdots=a_{k}=0$;* (ii) $k=q$ and $a_{1}=a_{2}=\cdots=a_{k}=a$ for an element $a\in\mathbb{F}_{q}$.*

We remark that this theorem together with Theorem 1.2 in [9] allows us to characterize when a linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}=b$ has a solution with distinct coordinates, i.e., when $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)>0$.

Next we obtain a recurrence formula for $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$.

Theorem 1.3**.**

If $\sum_{i=1}^{k}a_{i}\neq 0$, then

[TABLE]

If $\sum_{i=1}^{k}a_{i}=0$, then

[TABLE]

where $v(b)=-1$ if $b\neq 0$, and $v(b)=q-1$ if $b=0$; the hat denotes the omission of an element, and $d(a_{1},a_{2},\ldots,a_{k})$ satisfies

[TABLE]

This immediately gives an $O(k!)$ algorithm for computing $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ explicitly.

Corollary 1.4**.**

The number $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ can be computed in $O(k!)$ field operations.

We also obtain an explicit formula for $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ involving combinatorial numbers depending on $a_{i}$’s. In particular, we obtain a closed formula when $a_{i},1\leq i\leq k$ take at most three distinct values. This generalizes the main result of Theorem 1.1.

Theorem 1.5**.**

Let $p(a_{1},a_{2},\ldots,a_{k};k,i)$ be the number of permutations in $S_{k}$ of $i$ cycles with the sum of $a_{i}$’s over its each cycle vanishing. Then

[TABLE]

In particular, if $\sum_{i=1}^{k}a_{i}=0$ but $\sum_{i\in I}a_{i}\neq 0$ for all $I\subsetneq\{1,2,\ldots,k\}$, then

[TABLE]

Theorem 1.6**.**

Denote by $\{x\}_{p}=x-\lfloor x/p\rfloor p$ the least non-negative residue of $x$ modulo $p$. For a special case $[a_{1},a_{2},a_{3},\dots,a_{k}]=[a_{1},a_{2},1,\dots,1]$, we have

1. (i)

If $a_{1}+a_{2}+k-2\neq 0$, then

[TABLE] 2. (ii)

If $a_{1}+a_{2}+k-2=0$ and $a_{1},a_{2}\notin\mathbb{F}_{p}$, then

[TABLE] 3. (iii)

If $a_{1}+a_{2}+k-2=0$ and $a_{1},a_{2}\in\mathbb{F}_{p}$, then

[TABLE]

where $A=\{(a_{1},a_{2})\in\mathbb{F}_{p}^{2}:a_{1}\neq 1,\ a_{2}\neq 1\ \text{and}\ \{1-a_{1}\}_{p}+\{1-a_{2}\}_{p}\leq p\}$.

When the field is prime and the $a_{i}$’s satisfy some strong conditions, this problem was first considered by Schönemann [22] 180 years ago, and he proved the following result:

Theorem 1.7** (Schönemann).**

Let $p$ be a prime, $a_{1},a_{2},\ldots,a_{k}$ be arbitrary integers. If $\sum_{i=1}^{k}a_{i}\equiv 0\pmod{p}$ but $\sum_{i\in I}a_{i}\not\equiv 0\pmod{p}$ for all $I\subsetneq\{1,\dots,k\}$, then

[TABLE]

One may generalize this problem from $\mathbb{F}_{p}$ to $\mathbb{Z}/n\mathbb{Z}$, i.e., the residue ring modulo $n$. Similarly, given $a_{1},a_{2},\ldots,a_{k},b\in\mathbb{Z}/n\mathbb{Z}$, we define

[TABLE]

By using tools from additive combinatorics and group theory, Grynkiweicz et al. [11] gave necessary and sufficient conditions to characterize when $N_{\mathbb{Z}/n\mathbb{Z}}(a_{1},a_{2},\ldots,a_{k};b)>0$; see also [1, 11] for connections to zero-sum theory and [3] for applications to coding theory.

Bibak et al. generalize Schönemann’s theorem from $\mathbb{F}_{p}$ to $\mathbb{Z}/n\mathbb{Z}$ ([2]). They proved the following result:

Theorem 1.8** ([2], Theorem 2.3).**

Let $a_{1},a_{2},\ldots,a_{k},b,n\in\mathbb{Z}$, $n\geq 1$, and $\gcd(\sum_{i\in I}a_{i},n)=1$ for all $I\subsetneq\{1,\dots,k\}$. The number $N_{\mathbb{Z}/n\mathbb{Z}}(a_{1},a_{2},\ldots,a_{k};b)$ of solutions $(x_{1},x_{2},\ldots,x_{k})\in(\mathbb{Z}/n\mathbb{Z})^{k}$ of linear congruence $a_{1}x_{1}+a_{2}x_{2}+\dots+a_{k}x_{k}\equiv b\pmod{n}$ with all $x_{i}$ distinct modulo $n$, is

[TABLE]

The main technique for counting $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ is a sieve method for distinct coordinate counting developed by Li and Wan in [16] and it works well for the $\mathbb{Z}/n\mathbb{Z}$ case and thus we give another proof of Theorem 1.8.

This paper is organized as follows. Some preliminary results and the proofs of Theorem 1.2 and Theorem 1.3 are given in Section 2. The Li-Wan sieve technique and the proof of Theorem 1.5 are introduced in Section 3. The proof for Theorem 1.6 is given in Section 4 and the proof for Theorem 1.8 is given in Section 5.

Notations. We use $(q)_{k}:=q(q-1)\dots(q-k+1)$ to denote the falling factorial of $q$ and $\lfloor x\rfloor$ to denote the greatest integer less than or equal to $x$. If $A$ is a set, we use $1_{A}(x)$ to denote the indicator function, thus $1_{A}(x)=1$ when $x\in A$ and $1_{A}(x)=0$ otherwise.

2. Preliminary Results and the Proof of Theorem 1.3

The number of ordered $k$-tuples $(x_{1},x_{2},\ldots,x_{k})\in\mathbb{F}_{q}^{k}$ with all $x_{i}$ distinct is $(q)_{k}$, and the sum $\sum_{i=1}^{k}a_{i}x_{i}$ could be any element $b$ of the finite field $\mathbb{F}_{q}$. One expects that in favorable cases that the sums are equally distributed and thus $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ should be roughly $\frac{1}{q}(q)_{k}$. It is indeed the case when the $a_{i}$’s do not sum to zero. A simple observation gives the following result.

Lemma 2.1**.**

If $\sum_{i=1}^{k}a_{i}\neq 0$, then $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ are equal for all $b\in\mathbb{F}_{q}$. If $\sum_{i=1}^{k}a_{i}=0$, then $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k})$ are equal for all $b\in\mathbb{F}_{q}$ expect $b=0$.

Proof.

Pick an element $c\in\mathbb{F}_{q}$. Note that the bijective map $(x_{1},x_{2},\ldots,x_{k})\to(x_{1}+c,x_{2}+c,\dots,x_{k}+c)$ sends the distinct coordinate solutions of the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}=b$ to those of the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}=b+Ac$, where $A=\sum_{i=1}^{k}a_{i}$. Thus

[TABLE]

for any $c\in\mathbb{F}_{q}$. Then observe that, for $A\neq 0$, $b+Ac$ runs over all elements of $\mathbb{F}_{q}$ when $c$ does. Thus $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ are equal for all $b\in\mathbb{F}_{q}$ if $\sum_{i=1}^{k}a_{i}\neq 0$.

For the case $\sum_{i=1}^{k}a_{i}=0$, consider the bijective map $(x_{1},x_{2},\ldots,x_{k})\mapsto(x_{1}/b,x_{2}/b,\dots,x_{k}/b)$, where $b\neq 0$. It sends the distinct coordinate solution of the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{1}x_{k}=b$ to those of the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}=1$. Thus

[TABLE]

for all $b\neq 0$. Therefore $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ are equal for all $b\in\mathbb{F}_{q}$ except $b=0$. ∎

As an immediate consequence of Lemma 2.1, we have

Corollary 2.2**.**

If $\sum_{i=1}^{k}a_{i}\neq 0$, then

[TABLE]

If $\sum_{i=1}^{k}a_{i}=0$, then

[TABLE]

where $v(b)=-1$ if $b\neq 0$, and $v(b)=q-1$ if $b=0$.

Proof.

Note that

[TABLE]

The claim then follows from this equality and Lemma 2.1. ∎

Now we prove Theorem 1.2.

Proof the Theorem 1.2.

If $k<q$, extend the set of $a_{i}$’s to a set of size $q$ with $a_{k+1}=a_{k+2}=\cdots=a_{q}=0$. Then notice that $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};1)=0$ if and only if $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k},a_{k+1},a_{k+2},\dots,a_{q};1)=0$, where $a_{k+1}=a_{k+2}=\cdots=a_{q}=0$, so we only need to consider the case $k=q$. Now assume $k=q$. We shall show that $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{q};1)=0$ if and only if $a_{i}$, $1\leq i\leq q$ are equal. Suppose that the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{q}x_{q}=1$ does not have a solution $(x_{1},x_{2},\ldots,,x_{q})\in\mathbb{F}_{q}^{q}$ with all $x_{i}$ distinct, and thus neither does the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{q}x_{q}=b$ with $b\neq 0$ by Lemma 2.1. This implies that $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{q}x_{q}=0$ for all ordered $q$-tuples $(x_{1},x_{2},\ldots,x_{q})\in\mathbb{F}_{q}^{q}$ with $x_{i}$ distinct. Let $(x_{1},x_{2},\ldots,x_{q})\in\mathbb{F}_{q}^{q}$ be an ordered $q$-tuple with all $x_{i}$ distinct (there exists such an ordered $q$-tuple since $|\mathbb{F}_{q}|=q$), we then have

[TABLE]

Swapping the $i$-th and the $j$-th coordinates of $(x_{1},x_{2},\ldots,x_{q})$, we obtain another ordered $q$-tuple with distinct coordinates and thus

[TABLE]

Subtracting (2.2) from (2.1), we get $(a_{i}-a_{j})(x_{i}-x_{j})=0$, which implies $a_{i}=a_{j}$ since $x_{i}\neq x_{j}$. Since $i,j$ are arbitrary, we conclude that all of the $a_{i}$ are equal if there does not exist distinct $x_{i}\in\mathbb{F}_{q}$ such that $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{q}x_{q}=1$. On the other hand, if $q\geq 3$ and all of the $a_{i}$ are equal, then $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{q}x_{q}=0$ for all $(x_{1},x_{2},\ldots,,x_{q})\in\mathbb{F}_{q}^{q}$ with $x_{i}$ distinct since the sum of all elements of $\mathbb{F}_{q}$ is zero except $\mathbb{F}_{q}$ being $\mathbb{F}_{2}$. Thus there does not exist distinct $x_{i}\in\mathbb{F}_{q}$ satisfying $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{q}x_{q}=1$ when $q\geq 3$ and all of the $a_{i}$ are equal. The proof is completed. ∎

Next we turn to the proof of Theorem 1.3, the recurrence relation of $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$. The main idea is to introduce $N_{\mathbb{F}_{q}^{*}}(a_{1},a_{2},\ldots,a_{k};b)$, the number of solutions $(x_{1},x_{2},\ldots,x_{k})\in(\mathbb{F}_{q}^{*})^{k}$ of the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}=b$ with all $x_{i}$ distinct, which is related to $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ by the Lemma given below.

Lemma 2.3**.**

Let $A=\sum_{i=1}^{k}a_{i}$ and $c\in\mathbb{F}_{q}$. Then we have

[TABLE]

where the hat denotes the omission of an element. In particular, letting $(b,c)$ be $(1,0)$ and $(0,0)$, we obtain

[TABLE]

Proof.

Let $c$ be an element of $\mathbb{F}_{q}$. Then the solutions $(x_{1},x_{2},\ldots,x_{k})\in\mathbb{F}_{q}^{k}$ of the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}=b$ with all $x_{i}$ distinct can be divided into two parts depending on whether $c$ appears. By the linear substitution $y_{i}=x_{i}-c$, $1\leq i\leq k$, the number of solutions in which $c$ does not appear is $N_{\mathbb{F}_{q}^{*}}(a_{1},a_{2},\ldots,a_{k};b-Ac)$, and the number of solutions in which $c$ appears is $\sum_{i=1}^{k}N_{\mathbb{F}_{q}^{*}}(a_{1},a_{2},\ldots,a_{i-1},\widehat{a_{i}},a_{i+1},\dots,a_{k};b-Ac)$. Thus (2.3) follows. ∎

There is an additional relation between $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ and $N_{\mathbb{F}_{q}^{*}}(a_{1},a_{2},\ldots,a_{k};b)$ when $\sum_{i=1}^{k}a_{i}=0$.

Lemma 2.4**.**

Suppose that $\sum_{i=1}^{k}a_{i}=0$. Then we have

[TABLE]

for $1\leq i\leq k$ and $b\in\mathbb{F}_{q}$. In particular, $N_{\mathbb{F}_{q}^{*}}(a_{1},a_{2},\ldots,a_{i-1},\widehat{a_{i}},a_{i+1}\dots,a_{k};b)$ are equal for all $1\leq i\leq k$.

Proof.

By the linear substitution $y_{1}=x_{1}$ and $y_{i}=x_{i}-x_{1}$, $2\leq i\leq k$ and the assumption that $\sum_{i}^{k}a_{i}=0$, the number of solutions $(x_{1},x_{2},\ldots,x_{k})\in\mathbb{F}_{q}^{k}$ of $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}=b$ with all $x_{i}$ distinct is equal to the number of solutions $(y_{1},y_{2}\dots,y_{k})\in\mathbb{F}_{q}\times(\mathbb{F}_{q}^{*})^{k-1}$ of $a_{2}y_{2}+\dots+a_{n}y_{n}=b$ with $y_{i}$ distinct for $2\leq i\leq k$. Since $y_{1}\in\mathbb{F}_{q}$ can be arbitrarily chosen, we have

[TABLE]

The same argument gives

[TABLE]

for all $1\leq i\leq k$. Hence the proof is completed. ∎

Proof of Theorem 1.3.

For an ordered $k$-tuple $(a_{1},a_{2},\ldots,a_{k})\in\mathbb{F}_{q}^{k}$, define

[TABLE]

Subtracting (2.5) from (2.4), we obtain

[TABLE]

It is direct to verify that $d(a_{1})=N_{\mathbb{F}_{q}^{*}}(a_{1};1)-N_{\mathbb{F}_{q}^{*}}(a_{1};0)$ equals $-(q-1)$ if $a_{1}=0$, and equals $1$ if $a_{1}\neq 0$. Thus $d(a_{1})=-v(a_{1})$.

Now suppose $k\geq 2$. If $\sum_{i=1}^{k}a_{i}\neq 0$, then by Lemma 2.1 we have $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};1)=N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};0)$. Thus the left-hand side of (2.7) is zero, which implies

[TABLE]

Now let us consider the case $\sum_{i=1}^{k}a_{i}=0$. In this case, Lemma 2.4 implies that the left-hand side of (2.7) is equal to $qd(a_{1},a_{2},\ldots,a_{i-1},\widehat{a_{i}},a_{i+1},\dots,a_{k})$. Again, by Lemma 2.4, $d(a_{1},a_{2},\ldots,a_{i-1},\widehat{a_{i}},a_{i+1},\dots,a_{k})$ are equal for $1\leq i\leq k$ when $\sum_{i=1}^{k}a_{i}=0$. Thus the right-hand side (2.7) can be simplified into $d(a_{1},a_{2},\ldots,a_{k})+kd(a_{1},a_{2},\ldots,a_{i-1},\widehat{a_{i}},a_{i+1},\dots,a_{k})$. Therefore equality (2.7) yields

[TABLE]

∎

3. Li-Wan’s new sieve and the summation expression

In [16], J. Li and D. Wan proposed a new sieve method for distinct coordinate counting problems. We introduce it here briefly.

Let $D$ be a finite set. For a positive integer $k$, let $D^{k}=D\times D\times\cdots\times D$ be the $k$-fold Cartesian product of $D$ with itself. Let $X$ be a subset of $D^{k}$. Then every element $x\in X$ can written in an ordered $k$-tuple form $x=(x_{1},x_{2},\ldots,x_{k})$ with $x_{i}\in D$, $1\leq i\leq k$. We are interested in the number of the elements in $X$ whose coordinates are distinct, that is, the cardinality of the set

[TABLE]

Let $S_{k}$ denote the symmetric group on the set $\{1,\dots,k\}$. Given a permutation $\tau\in S_{k}$, we can write it as a product of disjoint cycles $\tau=C_{1}C_{2}\cdots C_{\ell(\tau)}$ uniquely apart from the order of the cycles, where $\ell(\tau)$ denotes the number of disjoint cycles of $\tau$. We define the signature of $\tau$ to be $\operatorname{sign}(\tau):=(-1)^{k-\ell(\tau)}$. We also define the set $X_{\tau}$ to be

[TABLE]

We have the following theorem which will be used in the proof of the summation expression of $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$.

Theorem 3.1** ( [16], Theorem 1.1).**

We have

[TABLE]

The unsigned Stirling number of the first kind $c(k,i)$ is defined to be the number of permutations in $S_{k}$ with exactly $i$ cycles. It can also be defined via the following classic identity [23]:

[TABLE]

With these preparations, we are ready to prove Theorem 1.5.

Proof of Theorem 1.5.

Let $X$ be the set of all solutions (not necessarily to have distinct coordinates) of the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}=b$ in $\mathbb{F}_{q}$, i.e.,

[TABLE]

Since $N_{\mathbb{F}_{q}}(a_{1},a_{2},\ldots,a_{k};b)$ counts the number of elements in $\overline{X}$, by Theorem 3.1 we have

[TABLE]

where $\overline{X}$ and $X_{\tau}$ are defined as in (3.1) and (3.2). Let $\tau=C_{1}C_{2}\cdots C_{\ell}$ be the disjoint cycle product of $\tau$. Let $A_{j}=\sum_{i\in C_{j}}a_{i}$, $1\leq j\leq\ell$. By the definition of $X_{\tau}$, we have

[TABLE]

Hence $|X_{\tau}|=q^{\ell-1}=q^{\ell(\tau)-1}$ if $A_{1},A_{2},\ldots,A_{\ell}$ are not all zero, and $|X_{\tau}|=q^{\ell}1_{b=0}=q^{\ell(\tau)-1}(v(b)+1)$ otherwise. Denote by $p(a_{1},a_{2},\ldots,a_{k};k,i)$ the number of permutations in $S_{k}$ of $i$ cycles with the sum of $a_{i}$’s over its each cycle vanishing. We deduce from (3.4) that

[TABLE]

In particular, if $\sum_{i=1}^{k}a_{i}=0$ but $\sum_{i\in I}a_{i}\neq 0$ for all $I\subsetneq\{1,\dots,k\}$, then $p(a_{1},a_{2},\ldots,a_{k},k,i)=0$ for $i\geq 2$ and $p(a_{1},a_{2},\ldots,a_{k},k,i)=(k-1)!$ for $i=1$. Thus we conclude that

[TABLE]

in this special case. ∎

4. Proof of Theorem 1.6

In this section, we prove Theorem 1.6. We first need some combinatorial formulas and equalities.

Lemma 4.1**.**

Let $k,n$ be integers. Then we have

[TABLE]

and

[TABLE]

Proof.

Comparing the coefficients $x^{k}$ on both sides of the identity $(1-x)^{-1}(1-x)^{n}=(1-x)^{n-1}$, we obtain (4.1). Similarly, comparing the coefficients of $x^{k-1}$ on both sides of the identity $(1-x)^{-1}((1-x)^{n})^{\prime}=-n(1-x)^{n-2}$, we obtain (4.2). ∎

Lemma 4.2** ([16], Lemma 3.1).**

Assume $p\mid k$. Let $p(k,i)$ be the number of permutations in $S_{k}$ of $i$ cycles with the length of its each cycle divisible by $p$. Then

[TABLE]

Lemma 4.3**.**

Assume $p\mid(k-j)$. Let $p(k,i,j)$ be the number of permutations in $S_{k}$ of $i$ cycles with a cycle of length $j$ containing $\{1,2\}$ and the length of each remaining $(i-1)$ cycles divisible by $p$. Then

[TABLE]

Proof.

Let $\tau\in S_{k}$ be a cycle described in the Lemma. We can write $\tau$ as a product of two permutations $\tau=\tau_{1}\tau_{2}$, where $\tau_{1}$ denotes the cycle of $\tau$ of length $j$ containing $\{1,2\}$, and $\tau_{2}$ denotes the product of the other $(i-1)$ cycles of $\tau$.

Since the $j$-cycle $\tau_{1}$ contains $\{1,2\}$ already, the remaining $(j-2)$ elements of $\tau_{1}$ must come from the set $\{3,\dots,k\}$ and thus there are $\binom{k-2}{j-2}$ choices of them. The number of $j$-cycles on a $j$-element set is $(j-1)!$, so there are $(j-1)!\binom{k-2}{j-2}$ ways to determine $\tau_{1}$ by the multiplication principle. But $\tau_{2}$ can be viewed as a permutation in $S_{k-j}$ of $(i-1)$ cycles such that the length of each its cycle is divisible by $p$, so there are $p(k-j,i-1)$ choices of $\tau_{2}$ by Lemma 4.2. Since every permutation can be expressed by a product of disjoint cycles uniquely up to the order of the cycles, we see that every ordered pair $(\tau_{1},\tau_{2})$ uniquely corresponds to a $\tau$. Therefore there are

[TABLE]

such $\tau$’s in total. The claim then follows. ∎

Lemma 4.4**.**

Assume $p\mid(k-j_{1}-j_{2})$. Let $p(k,i,j_{1},j_{2})$ be the number of permutations in $S_{k}$ of $i$ cycles with a cycle of length $j_{1}$ containing $\{1\}$ but not containing $\{2\}$, a cycle of length $j_{2}$ containing $\{2\}$ but not containing $\{1\}$, and the length of each remaining $(i-2)$ cycles divisible by $p$. Then we have

[TABLE]

Proof.

Let $\tau$ be a cycle described in the Lemma. We can write $\tau$ as a product of three permutations $\tau=\tau_{1}\tau_{2}\tau_{3}$, where $\tau_{1}$ denotes the cycle of $\tau$ of length $j_{1}$ containing $\{1\}$ but not containing $\{2\}$, $\tau_{2}$ denotes the cycle of $\tau$ of length $j_{2}$ containing $\{2\}$ but not containing $\{1\}$, and $\tau_{3}$ denotes the product of the other $(i-2)$ cycles of $\tau$. By a similar argument used in the proof of Lemma 4.3, we conclude that there are

[TABLE]

such $\tau$’s in total. The claim then follows ∎

Proof of Theorem 1.6.

By Corollary 2.2, it suffices to consider the case $a_{1}+a_{2}+k-2=0$. So assume $a_{1}+a_{2}+k-2=0$. In particular, both $a_{1}$ and $a_{2}$ lie in $\mathbb{F}_{p}$, or neither $a_{1}$ nor $a_{2}$ lies in $\mathbb{F}_{p}$. Let $X$ denote the set of all solutions of the linear equation $a_{1}x_{1}+a_{2}x_{2}+x_{3}+\cdots+x_{k}=b$ in $\mathbb{F}_{q}$, i.e.,

[TABLE]

Then $N_{\mathbb{F}_{q}}(a_{1},a_{2},\underbrace{1,\dots,1}_{k-2};b)$ counts the number of elements in $\overline{X}$. We have from Theorem 3.1 that

[TABLE]

For a permutation $\tau\in S_{k}$, write it as a disjoint cycle product $\tau=C_{1}C_{2}\cdots C_{\ell}$. Then we have two cases: $\{1\}$ and $\{2\}$ are contained in one cycle of $\tau$, or they are contained in two separate cycles of $\tau$, respectively. For the former case, we may assume that $\{1,2\}$ is contained in the cycle $C_{1}$ after rearranging the cycles. Then we have

[TABLE]

where $c_{i}$ denotes the length of the cycle $C_{i}$, $1\leq i\leq\ell$. Thus $|X_{\tau}|=q^{\ell}1_{b=0}=q^{\ell(\tau)-1}(v(b)+1)$ if $c_{1}\equiv 2-a_{1}-a_{2}\equiv k\pmod{p}$ and $p\mid c_{i}$ for $2\leq i\leq\ell$, and $|X_{\tau}|=q^{\ell-1}$ otherwise. For the latter case, we may assume that $\{1\}$ is contained in the cycle $C_{1}$ and $\{2\}$ is contained in the cycle $C_{2}$ after rearranging the cycles. Similarly, we have

[TABLE]

Thus $|X_{\tau}|=q^{\ell-1}(v(b)+1)$ if $c_{1}\equiv 1-a_{1}\pmod{p}$, $c_{2}\equiv 1-a_{2}\pmod{p}$ and $p\mid c_{i}$ for $3\leq i\leq\ell$, and $|X_{\tau}|=q^{\ell-1}$ otherwise. From this classification of $|X_{\tau}|$, we see that (4.3) can be simplified into

[TABLE]

where

[TABLE]

We first consider the case $a_{1},a_{2}\notin\mathbb{F}_{p}$ in which it suffices to evaluate the sum $S_{1}$ since the sum $S_{2}$ vanishes. Applying Lemma 4.3 and Lemma 4.2, we see that

[TABLE]

We can evaluate the above sum by using Lemma 4.1 and thus obtain

[TABLE]

For the case $a_{1},a_{2}\in\mathbb{F}_{p}$, we have to consider the sum $S_{2}$. Similarly, using Lemma 4.4 and Lemma 4.2, we see that

[TABLE]

where $N_{j}$ is defined as

[TABLE]

For $j\equiv k\pmod{p}$, it is direct to check that

[TABLE]

Inserting $N_{j}$ into the sum of $S_{2}$, by a routine computation and Lemma 4.1, we obtain

[TABLE]

where $\varepsilon$ equals $-1$, [math], $1$ in the three cases, respectively. Again, a direct computation shows that $S_{1}+S_{2}$ takes the form

[TABLE]

where $A=\{(a_{1},a_{2})\in\mathbb{F}_{p}^{2}:a_{1}\neq 1,\ a_{2}\neq 1\ \text{and}\ \{1-a_{1}\}_{p}+\{1-a_{2}\}_{p}\leq p\}$. The proof is then completed ∎

5. Proof of Theorem 1.8

For the purpose of our proof, we need the following result of D. N. Lehmer [14] which gives the number of solutions of linear congruence.

Proposition 5.1** ([14]).**

Let $a_{1},a_{2},\ldots,a_{k},b,n\in\mathbb{Z}$, $n\geq 1$. The linear congruence $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}\equiv b\pmod{n}$ has a solution $(x_{1},x_{2},\ldots,x_{k})\in(\mathbb{Z}/n\mathbb{Z})^{k}$ if and only if $d\mid b$, where $d=\gcd(a_{1},a_{2},\ldots,a_{k},n)$. Furthermore, if this conditions is satisfied, then there are $dn^{k-1}$ solutions.

Now we give another proof of Theorem 1.8 via a sieve method.

Proof of Theorem 1.8.

Let $X$ be of set of all solutions of the linear congruence $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{k}x_{k}\equiv b\pmod{n}$ in $\mathbb{Z}/n\mathbb{Z}$, i.e.,

[TABLE]

Then $N_{\mathbb{Z}/n\mathbb{Z}}(a_{1},a_{2},\ldots,a_{k};b)$ counts the number of elements in $\overline{X}$. Thus Theorem 3.1 yields

[TABLE]

Next we compute $|X_{\tau}|$. Let $\tau=C_{1}C_{2}\cdots C_{\ell}$ be a disjoint cycle product of $\tau$, and let $A_{j}=\sum_{i\in C_{j}}a_{i},1\leq j\leq\ell$. From the definition of $X_{\tau}$, we see that

[TABLE]

Since $\gcd(\sum_{i\in I}a_{i},n)=1$ for all $I\subsetneq\{1,\dots,k\}$, by Proposition 5.1, we have $|X_{\tau}|=n^{\ell-1}=n^{\ell(\tau)-1}$ for all $\tau\in S_{k}$ with $\ell(\tau)\geq 2$ . Note that Proposition 5.1 also shows that

[TABLE]

for $\tau\in S_{k}$ with $\ell(\tau)=1$. Substituting this into (5.1), when $\gcd(\sum_{i=1}^{k}a_{i},n)\nmid b$, we obtain

[TABLE]

And when $\gcd(\sum_{i=1}^{k}a_{i},n)\mid b$, we obtain

[TABLE]

This ends the proof ∎

Acknowledgements

Part of the work was done when the second author was a graduate student at Shanghai Jiao Tong University.