Counting the Unit Solutions of Certain Quartic Diagonal Congruence Modulo n

Junyong ZHAO; Shuangnian HU; Qiuyu YIN

doi:10.1051/wujns/2024295391

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Volume 29 / No 5 (October 2024)

Wuhan Univ. J. Nat. Sci., 29 5 (2024) 391-396

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Issue		Wuhan Univ. J. Nat. Sci. Volume 29, Number 5, October 2024


Page(s)		391 - 396
DOI		https://doi.org/10.1051/wujns/2024295391
Published online		20 November 2024

Wuhan University Journal of Natural Sciences, 2024, Vol.29 No.5, 391-396

Mathematics

CLC number: O156

Counting the Unit Solutions of Certain Quartic Diagonal Congruence Modulo n

一类模n四次对角同余方程的单位解的个数

Junyong ZHAO (赵军勇)¹, Shuangnian HU (胡双年)¹^† and Qiuyu YIN (尹秋雨)²

¹ School of Mathematics and Physics, Nanyang Institute of Technology, Nanyang 473004, Henan, China
² School of Science, Chongqing University of Posts and Telecommunications, Chongqing 400065, China

^† Corresponding author. E-mail: hushuangnian@163.com

Received: 6 May 2024

Abstract

Given a positive integer $n$ and the residue class ring $ℤ_{n} ∶ = ℤ / n ℤ$ , we set $ℤ_{n}^{\times}$ to be the group of units in $ℤ_{n}$ , i.e., $ℤ_{n}^{\times} ∶ = {x \in ℤ_{n} : g c d (x, n) = 1}$ . Let $N_{m} (n)$ be the number of solutions of $x_{1}^{4} + \dots + x_{m}^{4} \equiv 0 (m o d n)$ with $x_{1}, \dots, x_{m} \in ℤ_{n}^{\times}$ . In this note, we determine an explicit expression of $N_{m} (n)$ . This extends the results of Sun and Yang in 2014.

摘要

给定正整数 $n$ 和剩余类环 $ℤ_{n} ∶ = ℤ / n ℤ$ , 记 $ℤ_{n}^{\times} ∶ = {x \in ℤ_{n} : g c d (x, n) = 1}$ 为 $ℤ_{n}$ 中所有单位构成的群。设 $N_{m} (n)$ 表示 $x_{1}^{4} + \dots + x_{m}^{4} \equiv 0 (m o d$ $n)$ , $x_{1}, \dots, x_{m} \in ℤ_{n}^{\times}$ 的解的个数, 本文给出了 $N_{m} (n)$ 的一个明确表达式, 这推广了孙和杨于2014年给出的结论。

Key words: quartic diagonal congruence / character sums / units solutions

关键字 : 四次对角同余方程 / 特征和 / 单位解

Cite this article: ZHAO Junyong, HU Shuangnian, YIN Qiuyu. Counting the Unit Solutions of Certain Quartic Diagonal Congruence Modulo n[J]. Wuhan Univ J of Nat Sci, 2024, 29(5): 391-396.

Biography: ZHAO Junyong, male, Ph. D., Lecturer, research direction: number theory. E-mail: jyzhao_math@163.com

Fundation item: Supported by the Natural Science Foundation of Henan Province (232300420123), the National Natural Science Foundation of China (12026224)

© Wuhan University 2024

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

Let $n$ be a positive integer and $ℤ_{n} ∶ = ℤ / n ℤ = {0,1, \dots, n - 1}$ be the residue classes ring modulo $n$ .We set $ℤ_{n}^{\times}$ to be the set of units in $ℤ_{n}$ , i.e., $ℤ_{n}^{\times} ∶ = {x \in ℤ_{n} : g c d (x, n) = 1}$ , and let $ℤ_{n}^{*} = ℤ_{n} \ {0}$ . For $n > 1$ , we set $r a d (n)$ to be the radical of $n$ , i.e., $r a d (n)$ is the product of all distinct prime factors of $n$ . For any prime number $p$ , we set $v_{p} (n)$ to be the $p$ -adic valuation of $n .$ Let $c \in ℤ_{n}$ , in 1926, Brauer^[1] derived an explicit expression of

$x_{1} + \dots + x_{m} \equiv c (m o d n), w h e r e x_{1}, \dots, x_{m} \in ℤ_{n}^{\times}$

The results in Ref. [1] answered a problem in Ref. [2]. Sun and Yang^[3], in 2014, extended Brauer's theorem in Ref. [1] by deriving a more explicit expression for the number of solutions of the following diagonal congruence

$k_{1} x_{1} + \dots + k_{m} x_{m} \equiv c (m o d n) w i t h x_{1}, \dots, x_{m} \in ℤ_{n}^{\times}$

For the diagonal quadratic congruence, Ramy^[4] derived an expression for the number of solutions of

$k_{1} x_{1}^{2} + \dots + k_{m} x_{m}^{2} \equiv c (m o d n) w i t h x_{1}, \dots, x_{m} \in ℤ_{n}^{\times} a n d g c d (k_{1}, \dots, k_{m}, n) = 1$

These results generalized the theorem of Yang and Tang^[5]. For the following cubic congruence

$k_{1} x_{1}^{3} + \dots + k_{m} x_{m}^{3} \equiv c (m o d n)$

Chowla et al^[6] presented an explicit expression for the number of zeros of the above cubic congruence when $n = p$ is prime, $c = 0$ and $k_{1} = \dots = k_{m} = 1$ . In addition, Hu et al^[7] gave a formula for the number of solutions of the cubic diagonal equation over finite field.

Naturally, we ask for the number of solutions of the following quartic congruence:

$k_{1} x_{1}^{4} + \dots + k_{m} x_{m}^{4} \equiv c (m o d n)$

where $k_{1}, \dots, k_{m}, c \in ℤ_{n} a n d g c d (k_{1}, \dots, k_{m}, n) = 1 .$ We will investigate some special cases of the above congruence in this note, mainly focusing on their unit solutions. That is, we study the following congruence:

$x_{1}^{4} + \dots + x_{m}^{4} \equiv 0 (m o d n) w i t h x_{1}, \dots, x_{m} \in ℤ_{n}^{\times} a n d n > 1 b e i n g a n o d d i n t e g e r .$ (1)

We set $N_{m} (n)$ to be the number of solutions of (1) and we will present the formulas of $N_{m} (n)$ . Explicitly, we have the following two main results.

Theorem 1 Let $m$ be a positive integer and $p$ be an odd prime. Then the following statements are true.

1) Let $p \equiv 1 (m o d 4)$ , we have

$\begin{array}{l} N_{1} (p) = 0, \\ N_{2} (p) = {\begin{matrix} 4 (p - 1), i f p \equiv 1 (m o d 8), \\ 0, i f p \equiv 5 (m o d 8), \end{matrix} \\ N_{3} (p) = {\begin{matrix} p^{2} - 6 (r + 2) p + (6 r + 11), i f p \equiv 1 (m o d 8), \\ p^{2} - 6 r p + (6 r - 1), i f p \equiv 5 (m o d 8), \end{matrix} \\ N_{4} (p) = {\begin{matrix} p^{3} + 13 p^{2} + (4 r^{2} + 24 r + 7) p - (4 r^{2} + 24 r + 21), i f p \equiv 1 (m o d 8), \\ p^{3} - 11 p^{2} + (4 r^{2} + 24 r + 7) p - (4 r^{2} + 24 r - 3), i f p \equiv 5 (m o d 8), \end{matrix} \end{array}$

and for every $m \geq 5$ , we have

(i) if $p \equiv 1 (m o d 8)$ , then

$\begin{array}{l} N_{m} (p) + 4 N_{m - 1} (p) + (6 - 6 p) N_{m - 2} (p) + (8 p r - 12 p + 4) N_{m - 3} (p) + (p^{2} + 8 p r - 4 p r^{2} - 6 p + 1) N_{m - 4} (p) \\ = {(p - 1)}^{m - 4} (p^{3} - 6 p^{2} + (8 r + 1) p - 4 r^{2}) \end{array}$

(ii) if $p \equiv 5 (m o d 8)$ , then

$\begin{array}{l} N_{m} (p) + 4 N_{m - 1} (p) + (6 + 2 p) N_{m - 2} (p) + (8 p r + 4 p + 4) N_{m - 3} (p) + (9 p^{2} + 8 p r - 4 p r^{2} + 2 p + 1) N_{m - 4} (p) \\ = {(p - 1)}^{m - 4} (p^{3} + 2 p^{2} + (8 r + 9) p - 4 r^{2}) \end{array}$

where $r$ is uniquely determined by $p = r^{2} + 4 t^{2}, r \equiv 1 (m o d 4)$ .

2) Let $p \equiv 3 (m o d 4)$ , then

$N_{m} (p) = \frac{1}{p} ({(p - 1)}^{m} + \frac{p - 1}{2} ({(\sqrt[]{p} δ - 1)}^{m} + {(- \sqrt[]{p} δ - 1)}^{m})),$

where $δ = \sqrt[]{η (- 1)}$ and $η$ is the quadratic character of $ℤ_{p}$ .

Corollary 1 Let $n$ be an odd integer and $m$ be a positive integer. Then

$N_{m} (n) = \sum_{p | r a d (n)} p^{(m - 1) (v_{p} (n) - 1)} N_{m} (p),$

where $N_{m} (p)$ was obtained in Theorem 1.

The note is organized as follows. In Section 1, we present some preliminary lemmas. Then, we give the proofs of Theorem 1 and Corollary 1 in Section 2.

1 Preliminaries

In this section, we always set $p$ as an odd prime integer. Let $A$ be a finite set, we denote the number of elements of $A$ by $| A |$ . For each integer $a$ , we set ${〈 a 〉}_{p}$ to be the element in the ring $ℤ_{p}$ with $a \equiv {〈 a 〉}_{p} (m o d p) .$ We define $T_{a} ∶ = \sum_{x \in ℤ_{p}^{*}} e x p (\frac{2 π i a x^{4}}{p}) .$ Let $a_{1}, \dots, a_{n}$ be complex numbers. For any positive integer $k$ , we denote the $k$ -elementary symmetrical polynomial of $a_{1}, \dots, a_{n}$ by $σ_{k} (a_{1}, \dots, a_{n})$ , i.e.,

$σ_{k} (a_{1}, \dots, a_{n}) = \sum_{1 \leq j_{1} \leq \dots \leq j_{k} \leq n} a_{j_{1}} \dots a_{j_{k}}$

In the following, we will present some lemmas which are needed in the proofs of our results.

Lemma 1^[8] Let $n_{1}, \dots, n_{r} \in ℤ^{+}$ be pairwise relatively prime and $f (x_{1}, \dots, x_{m}) \in ℤ [x_{1}, \dots, x_{m}] .$ Let $N_{i}$ be the number of solutions of $f (x_{1}, \dots, x_{m}) \equiv 0 (m o d n_{i})$ and let $N$ be the number ofsolutions of $f (x_{1}, \dots, x_{m}) \equiv 0 (m o d n_{1}, \dots, n_{r})$ . Then $N = N_{1}, \dots, N_{r} .$

Lemma 2^[8] Let $a \in ℤ_{p},$ then

$\sum_{b \in ℤ_{p}} e x p (\frac{2 π i a b}{p}) = {\begin{matrix} p, i f p | a, \\ 0, i f p ∤ a . \end{matrix}$

Next, we present the following famous theorem^[9].

Lemma 3^[9] In the notation of above statement, then $T_{1} + 1, T_{g} + 1, T_{g^{2}} + 1, T_{g^{3}} + 1$ are theroots of equation $x^{4} - 6 p x^{2} + 8 p r x + p^{2} - 4 p r^{2} = 0$ if $p \equiv 1 (m o d 8)$ , or the roots of equation $x^{4} + 2 p x^{2} + 8 p r x + 9 p^{2} - 4 p r^{2} = 0$ if $p \equiv 5 (m o d 8)$ ,where $r$ is uniquely determined by $p = r^{2} + 4 t^{2}, r \equiv 1 (m o d 4)$ and $g$ is a primitive root modulo $p$ .

Lemma 4 In the notation of above statement, if $S = {1, g, g^{2}, g^{3}}$ with $g$ a primitive root modulo $p$ ,then there exists a unique $b \in S$ satisfying $T_{a} = T_{b}$ for each $a \in ℤ_{p}^{*} .$

Proof For each $a \in ℤ_{p}^{*}$ ,because $g$ is a primitive root modulo $p$ , $a \equiv g^{c} (m o d p)$ with $c \in ℤ (0 \leq c < p)$ is true. Suppose $c = 4 q + r,$ where $q, r \in ℤ, 0 \leq r < 3$ . One has

$T_{a} = \sum_{x \in ℤ_{p}^{*}} e x p (\frac{2 π i g^{c} x^{4}}{p}) = \sum_{x \in ℤ_{p}^{*}} e x p (\frac{2 π i g^{r} (g^{q} {x)}^{4}}{p}) = T_{g^{r}} .$

This completes the proof of Lemma 4.

Let $F (x_{1}, \dots, x_{m})$ be a polynomial over $ℤ_{p}$ . We denote the number of zeros of $F (x_{1}, \dots, x_{m}) = 0$ by $N (F = 0)$ . That is, $N (F = 0) = | {(x_{1}, \dots, x_{m}) \in ℤ_{p}^{m} : F (x_{1}, \dots, x_{m}) = 0} | .$ Then the following lemma is true.

Lemma 5 Let $p \equiv 3 (m o d 4)$ and $δ = \sqrt[]{η (- 1)}$ with $η$ a quadratic character of $ℤ_{p}$ . One has

$A_{m} (p) ∶ = N (x_{1}^{4} + \dots + x_{m}^{4} = 0) = p^{m - 1} + \frac{1 + {(- 1)}^{m}}{2} (p - 1) p^{\frac{m - 2}{2}} δ^{m} .$

Proof Let $k \in ℤ^{+}, c \in ℤ_{p}$ and let $φ$ be a multiplicative character of $ℤ_{p}$ of order $d = g c d (k, p - 1) .$ It is famous that (for example, Ref. [10]) $N (x^{k} = c) = \sum_{j = 0}^{d - 1} φ^{j} (c) .$ Since $p \equiv 3 (m o d 4),$ one has $N (x^{4} = c) = N (x^{2} = c)$ . Hence,

$N (x_{1}^{4} + \dots + x_{m}^{4} = 0) = \sum_{\begin{array}{l} (x_{1}, \dots, x_{m}) \in ℤ_{p}^{m} \\ x_{1}^{4} + \dots + x_{m}^{4} = 0 \end{array}} 1 = \sum_{\begin{array}{l} (x_{1}, \dots, x_{m}) \in ℤ_{p}^{m} \\ c_{1} + \dots + c_{m} = 0 \end{array}} N (x_{1}^{4} = c_{1}) \dots N (x_{m}^{4} = c_{m})$

$= \sum_{\begin{array}{l} (x_{1}, \dots, x_{m}) \in ℤ_{p}^{m} \\ c_{1} + \dots + c_{m} = 0 \end{array}} N (x_{1}^{2} = c_{1}) \dots N (x_{m}^{2} = c_{m}) = \sum_{\begin{array}{l} (x_{1}, \dots, x_{m}) \in ℤ_{p}^{m} \\ x_{1}^{2} + \dots + x_{m}^{2} = 0 \end{array}} 1 = N (x_{1}^{2} + \dots + x_{m}^{2} = 0) .$

Then the desired results are true by Theorem 6.26 and Theorem 6.27 in Ref. [10].

2 Proofs of Theorem 1 and Corollary 1

In this section, we present the proofs of Theorem 1 and Corollary 1.

Proof of Theorem 1 The proof will be divided into two cases.

Case 1 $p \equiv 1 (m o d 4)$ . From Lemma 4 and Lemma 2, one can get

$N_{m} (p) = \frac{1}{p} \sum_{(x_{1}, \dots, x_{m}) \in (ℤ_{p}^{*})^{m}} \sum_{a = 0}^{p - 1} e x p (\frac{2 π i a (x_{1}^{4} + \dots + x_{m}^{4})}{p}) = \frac{{(p - 1)}^{m}}{p} + \frac{1}{p} \sum_{a = 1}^{p - 1} \sum_{(x_{1}, \dots, x_{m}) \in (ℤ_{p}^{*})^{m}} e x p (\frac{2 π i a (x_{1}^{4} + \dots + x_{m}^{4})}{p})$

$= \frac{{(p - 1)}^{m}}{p} + \frac{1}{p} \sum_{a = 1}^{p - 1} \sum_{x \in ℤ_{p}^{*}} {(e x p (\frac{2 π i a x^{4}}{p}))}^{^{m}} = \frac{{(p - 1)}^{m}}{p} + \frac{p - 1}{4 p} (T_{1}^{m} + T_{g}^{m} + T_{g^{2}}^{m} + T_{g^{3}}^{m})$ (2)

If $p \equiv 1 (m o d 8)$ , from Lemma 3, it follows that $T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}$ are roots of

$x^{4} + 4 x^{3} + (6 - 6 p) x^{2} + (8 p r - 12 p + 4) x + (p^{2} + 8 p r - 4 p r^{2} - 6 p + 1) = 0$ (3)

where $r$ satisfies the condition in Lemma 3. Therefore,

${\begin{array}{l} σ_{1} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) = - 4, \\ σ_{2} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) = 6 - 6 p, \\ σ_{3} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) = 12 p - 8 p r - 4, \\ σ_{4} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) = p^{2} + 8 p r - 4 p r^{2} - 6 p + 1 . \end{array}$ (4)

Similarly, if $p \equiv 1 (m o d 8)$ , from Lemma 3, it follows that $T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}$ are roots of

$x^{4} + 4 x^{3} + (6 + 2 p) x^{2} + (8 p r + 4 p + 4) x + (9 p^{2} + 8 p r - 4 p r^{2} + 2 p + 1) = 0$ (5)

where $r$ satisfies the condition in Lemma 3, which implies that

${\begin{array}{l} σ_{1} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) = - 4, \\ σ_{2} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) = 6 + 2 p, \\ σ_{3} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) = - 4 p - 8 p r - 4, \\ σ_{4} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) = 9 p^{2} + 8 p r - 4 p r^{2} + 2 p + 1 . \end{array}$ (6)

Now, using (2), (4) and (6), we can get the explicit formula of $N_{m} (p)$ . First, it is easy to see that $N_{1} (p) = 0$ . Next, by (2), one has

$\begin{array}{l} N_{2} (p) = \frac{{(p - 1)}^{2}}{p} + \frac{p - 1}{4 p} (T_{1}^{2} + T_{g}^{2} + T_{g^{2}}^{2} + T_{g^{3}}^{2}) \\ = \frac{{(p - 1)}^{2}}{p} + \frac{p - 1}{4 p} (σ_{1} {(T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}})}^{2} - 2 σ_{2} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}})) . \end{array}$

By (4), (6) and direct calculation, one has

$N_{2} (p) = {\begin{matrix} 4 (p - 1), i f p \equiv 1 (m o d 8), \\ 0, i f p \equiv 5 (m o d 8) . \end{matrix}$

Similarly, we can get that

$\begin{array}{l} N_{3} (p) = \frac{{(p - 1)}^{3}}{p} + \frac{p - 1}{4 p} (T_{1}^{3} + T_{g}^{3} + T_{g^{2}}^{3} + T_{g^{3}}^{3}) \\ = \frac{{(p - 1)}^{3}}{p} + \frac{p - 1}{4 p} (σ_{1} {(T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}})}^{3} + 3 σ_{3} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) - 3 σ_{1} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) σ_{2} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}})) \\ = {\begin{matrix} p^{2} - 6 (r + 2) p + (6 r + 11), i f p \equiv 1 (m o d 8), \\ p^{2} - 6 r p + (6 r - 1), i f p \equiv 5 (m o d 8), \end{matrix} \end{array}$

and

$\begin{array}{l} N_{4} (p) = \frac{{(p - 1)}^{4}}{p} + \frac{p - 1}{4 p} (T_{1}^{4} + T_{g}^{4} + T_{g^{2}}^{4} + T_{g^{3}}^{4}) = \frac{{(p - 1)}^{4}}{p} + \frac{p - 1}{4 p} (σ_{1} {(T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}})}^{4} - 4 σ_{1} {(T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}})}^{2} σ_{2} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) + 4 σ_{1} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) σ_{3} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}}) + 2 σ_{2} {(T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}})}^{2} - 4 σ_{4} (T_{1}, T_{g}, T_{g^{2}}, T_{g^{3}})) \\ = {\begin{matrix} p^{3} + 13 p^{2} + (4 r^{2} + 24 r + 7) p - (4 r^{2} + 24 r + 21), i f p \equiv 1 (m o d 8), \\ p^{3} - 11 p^{2} + (4 r^{2} + 24 r + 7) p - (4 r^{2} + 24 r - 3), i f p \equiv 5 (m o d 8) . \end{matrix} \end{array}$

Now, let $a \in {1, g, g^{2}, g^{3}}$ . For each integer $m$ with $m \geq 5$ , using (3) and (5), one has

$T_{a}^{m} + 4 T_{a}^{m - 1} + (6 - 6 p) T_{a}^{m - 2} + (8 p r - 12 p + 4) T_{a}^{m - 3} + (p^{2} + 8 p r - 4 p r^{2} - 6 p + 1) T_{a}^{m - 4} = 0$ (7)

if $p \equiv 1 (m o d 8)$ , and

$T_{a}^{m} + 4 T_{a}^{m - 1} + (6 + 2 p) T_{a}^{m - 2} + (8 p r + 4 p + 4) T_{a}^{m - 3} + (9 p^{2} + 8 p r - 4 p r^{2} + 2 p + 1) T_{a}^{m - 4} = 0$ (8)

if $p \equiv 5 (m o d 8)$ , where $r$ satisfies the condition in Lemma 3.

If $p \equiv 1 (m o d 8)$ , then it follows from (2) and (7) that

$\begin{array}{l} (N_{m} (p) - \frac{{(p - 1)}^{m}}{p}) + 4 (N_{m - 1} (p) - \frac{{(p - 1)}^{m - 1}}{p}) + (6 - 6 p) (N_{m - 2} (p) - \frac{{(p - 1)}^{m - 2}}{p}) \\ + (8 p r - 12 p + 4) (N_{m - 3} (p) - \frac{{(p - 1)}^{m - 3}}{p}) + (p^{2} + 8 p r - 4 p r^{2} - 6 p + 1) (N_{m - 4} (p) - \frac{{(p - 1)}^{m - 4}}{p}) = 0 . \end{array}$

Therefore, we have

$\begin{array}{l} N_{m} (p) + 4 N_{m - 1} (p) + (6 - 6 p) N_{m - 2} (p) + (8 p r - 12 p + 4) N_{m - 3} (p) + (p^{2} + 8 p r - 4 p r^{2} - 6 p + 1) N_{m - 4} (p) \\ = {(p - 1)}^{m - 4} (p^{3} - 6 p^{2} + (8 r + 1) p - 4 r^{2}) \end{array}$

as expected.

If $p \equiv 5 (m o d 8)$ , one can use the similar argument to get the desired result by (2) and (8), and we omit it here.

Theorem 1 is proved in this case.

Case 2 $p \equiv 3 (m o d 4)$ . For each integer $i$ with $1 \leq i \leq m$ , define

$A_{m, i} (p) ∶ = {(x_{1}, \dots, x_{i - 1}, 0, x_{i + 1}, \dots, x_{m}) \in {(ℤ_{p})}^{m} : x_{1}^{4} + \dots + x_{m}^{4} \equiv 0 (m o d p)} .$

By principle of cross-classification, one can get

$N_{m} (p) = | A_{m} (p) ∖ ⋃_{i = 1}^{m} A_{m, i} (p) | = | A_{m} (p) | + \sum_{t = 1}^{m} {(- 1)}^{t} \sum_{1 \leq i_{1} < \dots < i_{t} \leq m} | ⋂_{j = 1}^{t} A_{m, i_{j}} (p) |$ (9)

For any integer $i$ with $1 \leq i \leq m - 1$ , and for each $t$ -tuple integer $(i_{1}, \dots, i_{t})$ with $1 \leq i_{1} < \dots < i_{t} \leq m$ , one can deduce that

$\sum_{1 \leq i_{1} < \dots < i_{t} \leq m} | ⋂_{j = 1}^{t} A_{m, i_{j}} (p) | = (\binom{m}{t}) | A_{m - t} (p) |$ (10)

Thus by Lemma 5, (9) and (10) one has

$\begin{array}{l} N_{m} (p) = | A_{m} (p) | + \sum_{t = 1}^{m} {(- 1)}^{t} (\begin{matrix} m \\ t \end{matrix}) | A_{m - t} (p) | = | A_{m} (p) | + \sum_{t = 1}^{m - 1} {(- 1)}^{t} (\binom{m}{t}) | A_{m - t} (p) | + {(- 1)}^{m} \\ = {(- 1)}^{m} + \frac{1}{p} ({(p - 1)}^{m} - {(- 1)}^{m} + \frac{p - 1}{2} ({(\sqrt[]{p} δ - 1)}^{m} + {(- \sqrt[]{p} δ - 1)}^{m} - 2 {(- 1)}^{m})) \\ = \frac{1}{p} ({(p - 1)}^{m} + \frac{p - 1}{2} ({(\sqrt[]{p} δ - 1)}^{m} + {(- \sqrt[]{p} δ - 1)}^{m})) . \end{array}$

The proof of Theorem 1 is completed.

Finally, we present the proof of Corollary 1.

Proof of Corollary 1 Let $n = \prod_{p | r a d (n)} p^{v_{p} (n)}$ be the prime decomposition of $n$ . It follows that $N_{m} (n) = \prod_{p | r a d (n)} N_{m} (p^{v_{p} (n)})$ from Lemma 1. It is easy to obtain $N_{m} (n)$ from $N_{m} (p^{v_{p} (n)})$ .

Let $(y_{1}, \dots, y_{m})$ be a solution of $x_{1}^{4} + \dots + x_{m}^{4} \equiv 0 (m o d p^{k})$ with $g c d (y_{i}, p^{k}) = 1,1 \leq i \leq m .$ It is easy to get that

${(y_{1} + k_{1} p^{k})}^{4} + \dots + {(y_{m} + k_{m} p^{k})}^{4} \equiv 0 (m o d p^{k + 1}), k_{1}, \dots, k_{m} \in ℤ_{p},$

if and only if

$y_{1}^{4} + \dots + y_{m}^{4} + 4 (y_{1}^{3} k_{1} p^{k} + \dots + y_{m}^{3} k_{m} p^{k}) \equiv 0 (m o d p^{k + 1})$ (11)

Since $g c d (p, 4) = g c d (y_{i}, p) = 1$ with $1 \leq i \leq m$ , for any $(m - 1)$ -tuple $(k_{1}, \dots, k_{m - 1})$ , there is a unique $k_{m} \in ℤ_{p}$ such that (11) holds. Therefore, we have

$N_{m} (p^{k + 1}) = p^{m - 1} N_{m} (p^{k}) .$

It then follows that

$N_{m} (p^{v_{p} (n)}) = p^{(m - 1) (v_{p} (n) - 1)} N_{m} (p) .$

The proof of Corollary 1 is completed.

References

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[1] Brauer A. Losung der Aufgabe 30[J]. Jahresber Dtsch Math-Ver, 1926, 35: 92-94. [Google Scholar]

[2] Rademacher H. Aufgabe 30[J]. Jahresbe Dtsch Math-Ver, 1925, 34:158. [Google Scholar]

[3] Sun C F, Yang Q H. On the sumset of atoms in cyclic groups[J]. International Journal of Number Theory, 2014, 10(6): 1355-1363. [CrossRef] [MathSciNet] [Google Scholar]

[4] Ramy T E. On the number of incongruent solutions to a quadratic congruence over algebraic integers[J]. International Journal of Number Theory, 2019, 15(1):105-130. [CrossRef] [MathSciNet] [Google Scholar]

[5] Yang Q H, Tang M. On the addition of squares of units and nonunits modulo [J]. Journal of Number Theory, 2015, 155: 1-12. [CrossRef] [MathSciNet] [Google Scholar]

[6] Chowla S, Cowles J, Cowles M. On the number of zeros of diagonal cubic forms[J]. Journal of Number Theory, 1977, 9(4): 502-506. [CrossRef] [MathSciNet] [Google Scholar]

[7] Hu S N, Wang S H, Li Y Y, et al. Note on the number of solutions of cubic diagonal equations over finite fields [J]. Wuhan University Journal of Natural Sciences, 2023, 28(5): 369-372. [CrossRef] [EDP Sciences] [Google Scholar]

[8] Apostol T M. Introduction to Analytic Number Theory[M]. New York: Springer-Verlag, 1976. [Google Scholar]

[9] Berndt B C, Evans R J, Williams K S. Gauss and Jacobi Sums[M]. New York: John Wiley & Sons, Inc, 1999. [Google Scholar]

[10] Lidl R, Niederreiter H. Finite Fields: Encyclopedia of Mathematics and Its Applications[M]. Cambridge: Cambridge University Press, 1997. [Google Scholar]