On the number of unit solutions of cubic congruence modulo $n$

1.   Introduction

For any positive integer $n$, let $\mathbb Z_n: = \mathbb Z/n\mathbb Z = \{0, \ldots, n-1\}$ be the ring of residue classes modulo $n$, and let $\mathbb{Z}_n^{\times}: = \{x\in \mathbb Z_n|\gcd(x, n) = 1\}$, $\mathbb{Z}_n^*: = \{x\in \mathbb Z_n|x\ne 0\}$ respectively. In 1926, for any fixed $c\in\mathbb Z_n$, A. Brauer ^[2] studied the linear congruence

and gave a formula for the number of incongruent solutions. This answered a problem of H. Rademacher ^[7]. In 2014, Sun and Yang ^[9] generalized A. Brauer's result by giving an explicit formula for the number of incongruent solutions of general linear congruence

where $k_1, \ldots, k_m, c\in \mathbb{Z}_n$.

Recently, Taki Eldin ^[8] studied the quadratic case and provided an explicit formula for the number of incongruent solutions of

where $k_1, \ldots, k_m, c\in \mathbb{Z}_n$ with $\gcd(k_1\cdots k_m, n) = 1$, which extended the result of Yang and Tang ^[10].

Therefore, it is natural to consider the following cubic congruence:

with $k_1, \ldots, k_m, c\in \mathbb{Z}_n$ such that $\gcd(k_1\cdots k_m, n) = 1$.

When $n = p$ is a prime number, $k_1 = \cdots = k_m = 1$, S. Chowla, J. Cowles and M. Cowles ^[3], Hong and Zhu ^[4] gave a formula of the number of incongruent solutions of the above congruence with $c = 0$ and $c\neq0$ respectively. In this note, we investigate the following cubic congruence

Denote by $N_m(n)$ the number of incongruent solutions of (1.1). Li and Ouyang ^[6], in 2018, presented a relation between $N_{m}(p^{a})$ and $N_{m}(p^{b})$ for some certain integers $a$ and $b$ with $a\geq b$. In this paper, we will give an explicit formula of $N_m(n)$, which couldn't be obtained by the results in ^[6]. Particularly, we have the first main theorem as follows.

Theorem 1.1. Let $p$ be a prime number and $m$ be a positive integer. Then each of the following holds.

(1). If $p\equiv 1\pmod 3$, then

and

for all $m\geq 4$, where $c$ is uniquely determined by

(2). If $p\not\equiv 1\pmod 3$, then

For every nonzero integer $n$, let $rad(n)$ be the radical of $n$, i.e., the product of distinct prime divisors of $n$. As usual, for any prime number $p$, let $v_p(n)$ be the $p$-adic valuation of $n$, i.e., $p^{v_p(n)}\mid n$ and $p^{v_p(n)+1}\nmid n$. For any $a\in\mathbb{Z}$, let $\langle a\rangle_n$ be the unique element in $\mathbb{Z}_n$ such that $a\equiv \langle a\rangle_n\pmod n$. Now, we can state our second main Theorem.

Theorem 1.2. Let $n$ and $m$ be positive integers and let $\xi: = \exp(\frac{2\pi i}{9})$. Then

where

$N_m(p)$ was obtained in Theorem 1.1.

The paper is organized as follows: Section 2 provides some notations and lemmas which will be used in the sequel. In Section 3, we give the proofs of Theorem 1.1 and Theorem 1.2.

2.   Preliminaries

Throughout, $p$ denotes a prime number. For any finite set $A$, denote by $|A|$ the cardinality of $A$. For any $a\in\mathbb{Z}$, let

Next, we give some lemmas which are needed in the proofs of Theorem 1.1 and Theorem 1.2. We begin with the following famous result.

Lemma 2.1. (^[1]) (Chinese Remainder Theorem) Let $f(x_1, \dots, x_m)\in\mathbb{Z}[x]$. If $n_1, \dots, n_r$ are pairwise relatively prime positive integers, let $N_i$ be the number of zeros of

and $N$ be the number of zeros of

then $N = N_1\cdots N_r$.

The following classical result was obtained by Gauss.

Lemma 2.2. (^[3]) Let $g$ be a primitive root modulo $p$. Then $T_1+1$, $T_g+1$, $T_{g^2}+1$ are the roots of equation

where $c$ is uniquely determined by

Lemma 2.3. Let $g$ be a primitive root modulo $p$, and let $S = \{1, g, g^2\}$. Then for any $a\in \mathbb{Z}_p^*$, there exists a unique $b\in S$ such that

Proof. Since $g$ is a primitive root modulo $p$, for any $a\in \mathbb{Z}_p^*$, one has $a\equiv g^c\pmod p$ for some integer $c$ with $1\le c\le p-1$. Let $c = 3q+r$ with $q, r\in\mathbb Z$ and $0\le r\le 2$. It then follows that

as desired. So Lemma 2.3 is proved.

Lemma 2.4. (^[1]) For any $a\in \mathbb{Z}_p$, we have

Lemma 2.5. (^[5]) Let $a\in\mathbb{Z}_p^*$. Then $x^3\equiv a\pmod p$ is solvable if and only if $a^{\frac{p-1}{d}}\equiv 1\pmod p$, where $d = \gcd(3, p-1)$.

For any positive integer $m$, let

We have the following lemma.

Lemma 2.6. Suppose $p\not\equiv 1\pmod 3$. Then for any positive integer $m$, we have

Proof. Let $f: \mathbb{Z}_p\mapsto \mathbb{Z}_p$ be a map satisfying that $f(a) = \langle a^3\rangle_p$ for any $a\in\mathbb{Z}_p$. We claim that $f$ is bijective. In fact, by Lemma 2.5, it is easy to see that $f$ is subjective. Moreover, since $|\mathbb Z_p|$ is finite, one has that $f$ is also injective. So the claim is true.

Therefore, we deduce that

as expected.

Lemma 2.7. Let $\xi: = \exp(\frac{2\pi i}{9})$ and let $N'_m(9)$ be the number of solutions of congruence

Then

Proof. Let $S_1$ and $S_{-1}$ be the number of terms of Eq (2.1) for which $x_i = 1$ and $x_i = -1$ $(1\le i\le m)$ respectively. Then the Eq (2.1) has a solution if and only if $S_1\equiv S_{-1}\pmod 9$. We distinguish two cases as follows.

Case 1. Let $\langle m\rangle_9$ be even.

The congruence (2.1) has a solution if and only if

Therefore, one gets

Since $\xi = \exp(\frac{2\pi i}{9})$, we easily obtain the well-known fact:

By (2.3) and computing directly, one gets the following identity:

By (2.2) and (2.4), we have

Case 2. Let $\langle m\rangle_9$ be odd.

Obviously, the congruence (2.1) has no solution if $m = 1, 3, 5, 7$. So we suppose that $m\ge 10$.

The Eq (2.1) has a solution if and only if

From an argument which is similar to that in Case 1, we get

From the above discussion, we can conclude that

This finishes the proof of Lemma 2.7.

3.   Proof of Theorem 1.1 and Theorem 1.2

In this section, we give the proofs of Theorem 1.1 and Theorem 1.2.

Proof of Theorem 1.1. We divide the proof into two cases.

Case 1. Let $p \equiv 1 \pmod 3$. By Lemma 2.3 and Lemma 2.4, we deduce that

By Lemma 2.2, $T_1$, $T_g$, $T_{g^2}$ are roots of equation

where $c$ is uniquely determined by

It then follows from (3.2) that

Clearly, $N_1(p) = 0$. Moreover, using (3.1) and (3.3), we get that

and

Now, let $m$ be any integer with $m\ge 4$. Then for any $a\in \{1, g, g^2\}$, we have

by (3.2). It then follows from (3.1) and (3.4) that

which is equivalent to

So Theorem 1.1 is proved in this case.

Case 2. Let $p\not\equiv 1\pmod 3$. For any integer $i$ with $1\le i\le m$, define

Then using principle of cross-classification, we derive that

Let $t$ be an integer with $1\le t\le m-1$. Then for any integer $t$-tuple $(i_1, \ldots, i_t)$ with $1\le i_1 < \cdots < i_t\le m$, it is obvious that

Thus by Lemma 2.5, (3.5) and (3.6), one gets that

This finishes the proof of Theorem 1.1.

Now, we begin the proof of Theorem 1.2.

Proof of Theorem 1.2. Let $n$ have the prime decomposition

By Lemma 2.1, one has the product formla

So to compute $N_m(n)$, it is enough to study the prime power case $N_m(p^{v_p(n)})$ with $p|n$.

Now, we consider the following two cases with $p|n$.

Case 1. For any $p|n$ it holds either $p\ne3$ or $p = 3, \ v_3(n) = 1$.

If $p\ne 3$, by Theorem B(1) of ^[6], we have

where $N_m(p)$ has been studied in Theorem 1.1.

If $p = 3$ and $v_3(n) = 1$, one has

Hence, for $p\ne3$ or $p = 3, \ v_3(n) = 1$, we get

It then follows from (3.7) and (3.8) that

as expected.

Case 2. $p = 3$ and $v_3(n)\ge 2$.

By Theorem B(1) in ^[6], one has

Since $x^3\equiv 1\pmod 9$ for $x\in\{1, 4, 7\}$ and $x^3\equiv -1\pmod 9$ for $x\in\{2, 5, 8\}$, one gets that

Therefore, by (3.7)–(3.10), we have

By Theorem 1.1, we have

Hence, we get

From the above discussion of Case 1 and Case 2, we can conclude that

where

$N'_m(9)$ and $N_m(p)$ were obtained in Lemma 2.7 and Theorem 1.1, respectively.

This finishes the proof of Theorem 1.2.

4.   Conclusions

In this paper, we present an explicit formula of the number of unit solutions of diagonal cubic form over $\mathbb Z_n$, by using the method of exponential sums. As future directions, one can find the formula of the number of unit solutions of $x_1^3+\cdots +x_n^3\equiv c\pmod n$ over $\mathbb Z_n$ with $c\not\equiv 0\pmod n$.

Conflict of interest

We declare that we have no conflict of interest.