Counting the solutions of $\lambda_1 x_1^{k_1}+\cdots +\lambda_t x_t^{k_t}\equiv c\bmod{n}$

TL;DR

This paper derives formulas and algorithms to count solutions of polynomial congruences involving sums of powers modulo integers, with applications to quadratic and linear cases over primes and prime powers.

Contribution

It provides new explicit formulas and an algorithm for counting solutions of polynomial congruences, extending previous results to more general cases and specific polynomial forms.

Findings

Complete solution for linear polynomial solutions over primes.

Explicit counting formulas for quadratic cases when t=2.

General results for polynomial solutions with specific p-adic valuation conditions.

Abstract

Given a polynomial $Q(x_1,\cdots, x_t)=\lambda_1 x_1^{k_1}+\cdots +\lambda_t x_t^{k_t}$, for every $c\in \mathbb{Z}$ and $n\geq 2$, we study the number of solutions $N_J(Q;c,n)$ of the congruence equation $Q(x_1,\cdots, x_t)\equiv c\bmod{n}$ in $(\mathbb{Z}/n\mathbb{Z})^t$ such that $x_i\in (\mathbb{Z}/n\mathbb{Z})^\times$ for $i\in J\subseteq I= \{1,\cdots, t\}$. We deduce formulas and an algorithm to study $N_J(Q; c,p^a)$ for $p$ any prime number and $a\geq 1$ any integer. As consequences of our main results, we completely solve: the counting problem of $Q(x_i)=\sum\limits_{i\in I}\lambda_i x_i$ for any prime $p$ and any subset $J$ of $I$; the counting problem of $Q(x_i)=\sum\limits_{i\in I}\lambda_i x^2_i$ in the case $t=2$ for any $p$ and $J$, and the case $t$ general for any $p$ and $J$ satisfying $\min\{v_p(\lambda_i)\mid i\in I\}=\min\{v_p(\lambda_i)\mid i\in J\}$; the counting problem of $Q(x_i)=\sum\limits_{i\in I}\lambda_i x^k_i$ in the case $t=2$ for any $p\nmid k$ and any $J$, and in the case $t$ general for any $p\nmid k$ and $J$ satisfying $\min\{v_p(\lambda_i)\mid i\in I\}=\min\{v_p(\lambda_i)\mid i\in J\}$.

Tables

Table 1. Table 1. N J ​ ( c , 27 ) subscript 𝑁 𝐽 𝑐 27 N_{J}(c,27) for J 𝐽 J nonempty

$c^{\prime }$	0	1,	3,	9,	2,4,	8,10,	6,12,	else
		26	24	18	23,25	17,19	15,21
$N^{\ast }(c,27)$	0	0	0	0	0	0	0	$3^6$
$N_{\{1,2\}}(c,27)$	0	0	0	0	0	0	$3^6$	$3^6$
$N_{\{1,3\}}(c,27)$	0	0	0	0	0	$3^6$	0	$3^6$
$N_{\{2,3\}}(c,27)$	0	0	0	0	$3^6$	0	0	$3^6$
$N_{\left\{1\right\}}(c,27)$	0	0	0	$3^6$	0	$3^6$	$3^6$	$3^6$
$N_{\left\{2\right\}}(c,27)$	0	0	$3^6$	0	$3^6$	0	$3^6$	$3^6$
$N_{\left\{3\right\}}(c,27)$	0	$3^6$	0	0	$3^6$	$3^6$	0	$3^6$

Equations

$$Q(x_1,\cdots ,x_t)={\lambda }_1{x}_1^{k_1}+\cdots +{\lambda }_t{x}_t^{k_t}\in \mathbb{Z}[x_1,\cdots ,x_t].$$

$$Q(x_1,\cdots ,x_t)\equiv c\mathrm{mod}n$$

$$Q(x_1,\cdots ,x_t)={\lambda }_1{x}_1^{k_1}+\cdots +{\lambda }_t{x}_t^{k_t}\equiv c\mathrm{mod}n,(c\in \mathbb{Z},n\in {\mathbb{Z}}_{\ge 2})$$

$$d_{p,J}=\{\begin{array}{ll}\underset{i\in J}{\min }\{v_p({\lambda }_i{k}_i)+1\},& \text{if}p\text{odd};\\ \underset{i\in J}{\min }\{v_2({\lambda }_i{k}_i)+2\text{if 2∣ki},v_2({\lambda }_i{k}_i)+1\text{if 2∤ki}\},& \text{if}p=2.\end{array}$$

$${\mathbb{N}}_{\varnothing }(0;c,p^a)=N^{\ast }(0;c,p^a)=\{\begin{array}{ll}1,& \text{if}{p}^a\mid c;\\ 0,& \text{if}{p}^a\nmid c.\end{array}$$

$$n=\prod _{p\mid n}{p}^{n_p}.$$

$$N_J(Q;c,n)=\prod _{p\mid n}{N}_J(Q;c,p^{n_p}).$$

$$B_i(J_1,J_2;a)=\{\begin{array}{ll}\{0\},& \text{if}i\notin J_2-J_1;\\ \{0,\cdots ,a\},& \text{if}i\in J_2-J_1,\end{array}B(J_1,J_2;a)=\prod _{i=1}^t{B}_i(J_1,J_2;a).$$

$$J_{\mathbf{b}}=\{i\in I\mid b_i<a\},Q_{\mathbf{b}}=\sum _{j\in J_{\mathbf{b}}}{\lambda }_j{p}^{k_j{b}_j}{x}^{k_j},s(\mathbf{b})=\sum _{j\in J_{\mathbf{b}}}{b}_j.$$

$$N_{J_1}(Q;c,p^a)=\sum _{\mathbf{b}\in B(J_1,J_2;a)}{p}^{-s(\mathbf{b})}{N}_{J_2\cap J_{\mathbf{b}}}(Q_{\mathbf{b}};c,p^a).$$

$$N_J(Q;c,p^{a+1})=p^{t-1}{N}_J(Q;c,p^a)$$

$$N_J(Q;c,p^a)=p^{(t-1)(a-d_{p,J})}{N}_J(Q;c,p^{d_{p,J}}).$$

$${\phi }_a:({\mathbb{F}}_p{)}^t\to \mathbb{Z}/p^a\mathbb{Z},(a_1,\cdots ,a_t)\mapsto Q({\alpha }_1,\cdots ,{\alpha }_t)\mathrm{mod}{p}^a,$$

$$N_J(Q;c,p^a)=p^{(a-1)t}\#{\phi }_{a,J}^{-1}(c\mathrm{mod}{p}^a).$$

$$\#{\phi }_{a,J}^{-1}(c\mathrm{mod}{2}^a)=\#\{T\subseteq \{1,\cdots ,t\}\mid T\supseteq J,v_2(\sum _{i\in T}{\lambda }_i-c)\ge a\}.$$

$$N_J(Q;c(1+p^{f}x),p^a)=N_J(Q;c,p^a).$$

$$N_J(Q;c,p^a)=p^{(t-1)(a-c_p-f_p)}{N}_J(Q;c,p^{c_p+f_p}).$$

$$N_J(Q;c,p^a)=\{\begin{array}{ll}{p}^a{N}_J(Q_{I-\{j\}};c,p^a),& \text{if}j\notin J;\\ p^{a-1}(p-1)N_{J-\{j\}}(Q_{I-\{j\}};c,p^a),& \text{if}j\in J.\end{array}$$

$$N_J(Q;c,p^a)=\{\begin{array}{ll}{p}^{te}{N}_J(Q/p^e;c/p^e,p^{a-e}),& \text{if}e\le \min \{a,c_p\};\\ p^{(at-|J|)}(p-1{)}^{|J|},& \text{if}a\le \min \{e,c_p\};\\ 0,& \text{if}{c}_p<\min \{e,a\}.\end{array}$$

$$J({\chi }_1,\cdots ,{\chi }_t)=\sum _{u_1+\cdots +u_t=1}{\chi }_1(u_1)\cdots {\chi }_t(u_t)$$

$$J_0({\chi }_1,\cdots ,{\chi }_t)=\sum _{u_1+\cdots +u_t=0}{\chi }_1(u_1)\cdots {\chi }_t(u_t).$$

$$Q(x_1,\cdots ,x_t)={\lambda }_1{x}^{k_1}+\cdots {\lambda }_t{x}^{k_t}=c$$

$$N(0,p)=p^{t-1}+\sum _{\begin{array}{c}{\chi }_i^{k_i}=1,{\chi }_i\ne 1\\ {\chi }_1\cdots {\chi }_t=1\end{array}}{\chi }_1({\lambda }_1^{-1})\cdots {\chi }_t({\lambda }_t^{-1})J_0({\chi }_1,\cdots ,{\chi }_t),$$

$$N(c,p)=p^{t-1}+\sum _{\begin{array}{c}{\chi }_i^{k_i}=1\\ {\chi }_i\ne 1\end{array}}{\chi }_1\cdots {\chi }_t(c){\chi }_1({\lambda }_1^{-1})\cdots {\chi }_t({\lambda }_t^{-1})J({\chi }_1,{\chi }_2,\cdots ,{\chi }_t)$$

$$\mathbb{Z}/p^a\mathbb{Z}=\underset{b=0}{\overset{a}{⨆}}(p^b\mathbb{Z}/p^a\mathbb{Z}-p^{b+1}\mathbb{Z}/p^a\mathbb{Z}).$$

$$C_{\mathbf{x}}=\prod _{j\in J_{\mathbf{b}}}{C}_j\subseteq {\Gamma }_{J_2\cap J_{\mathbf{b}}}(Q_{\mathbf{b}};c,p^a).$$

$$N_J(Q;c,p^a)=\sum _{\mathbf{b}\in B(J,I;a)}{p}^{-s(\mathbf{b})}{N}^{\ast }(Q_{\mathbf{b}};c,p^a).$$

$$N_{J_1}(Q;c,p)=\sum _{T\subseteq J_2-J_1}{N}_{J_2-T}(Q_{I-T};c,p).$$

$$N_{J_2}(Q;c,p)=\sum _{T\subseteq J_2-J_1}(-1{)}^{|T|}{N}_{J_1}(Q_{I-T};c,p).$$

$$N_J(Q;c,p)=\sum _{T\subseteq J}(-1{)}^{|T|}N(Q_{I-T};c,p).$$

Full text

1. Introduction and Main results

1.1. Introduction

Given a polynomial

[TABLE]

Let $\mathbf{\lambda}=(\lambda_{1},\cdots,\lambda_{t})\in(\mathbb{Z}-\{0\})^{t}$ and $\mathbf{k}=(k_{1},\cdots,k_{t})\in\mathbb{Z}_{\geq 1}^{t}$. For any $c\in\mathbb{Z}$ and $n\geq 2$, and for a subset $J$ of $I=\{1,\cdots,t\}$, denote by $\Gamma_{J}(c,n)=\Gamma_{J}(Q;c,n)=\Gamma_{J}(\mathbf{\lambda},\mathbf{k};c,n)$ the set of solutions $(x_{1},\cdots,x_{t})$ of the congruence equation

[TABLE]

such that $x_{j}\in(\mathbb{Z}/n\mathbb{Z})^{\times}$ for $j\in J$, and by $N_{J}(Q;c,n)$ the cardinality of $\Gamma_{J}(Q;c,n)$. In particular, write $\Gamma$, $N$, $\Gamma^{*}$ and $N^{*}$ for $\Gamma_{\emptyset}$, $N_{\emptyset}$, $\Gamma_{I}$ and $N_{I}$ respectively. The problem to determine $N_{J}(Q;c,n)$ has been studied by many authors extensively in various special cases:

• (i)

The linear case $\mathbf{k}=(1,\cdots,1)$. For $J=I$, this is a problem proposed by H. Rademacher[7] in 1925 and answered by A. Brauer[2] in 1926, and recovered by many authors later from time to time. For $J=I$ and $\mathbf{\lambda}=(\lambda_{i})$ where the $\lambda_{i}$’s are divisors of $n$, this is the work of Sun and Yang[9] in 2014.

• (ii)

The quadratic case $\mathbf{k}=(2,\cdots,2)$. For $J=\emptyset$, this is studied in the work of Tóth[10] in 2014. For $t=2$ and $\mathbf{\lambda}=(1,1)$, this is the work of Yang and Tang[11] in 2015. For $t=2$, $\mathbf{\lambda}$ arbitrary and $J=I$, this is the work of Sun and Cheng [8] in 2016. For general $t$, $\mathbf{\lambda}=(1,\cdots,1)$ and $J=I$, this is the recent work of Mollahajiaghaei [6]. See also [3] for more development.

• (iii)

The case $t=2$, $\mathbf{\lambda}=(1,1)$ and $\mathbf{k}=(k,k)$. Partial results were obtained by Deaconescu and Du [4].

1.2. Notations.

Before stating our main results, let us fix the following notations.

In this paper, $p$ is always a prime number and $v_{p}$ is the $p$-adic valuation, $a$ is always a positive integer and $I$ is the set $\{1,\cdots,t\}$.

For a set $X$, $\#X$ or $|X|$ means the cardinality of $X$. For two subsets $X$ and $Y$ of the set $U$, the difference set $X-Y$ is the set $\{x\in U\mid x\in X,\ x\notin Y\}$.

For the congruence equation

[TABLE]

we call $t$, $\mathbf{k}$ and $n$ its dimension, degree and level respectively.

For $J$ a nonempty subset of $I$, the depth $d_{p,J}=d_{p,J}(Q)=d_{p,J}(\mathbf{\lambda},\mathbf{k})$ of $Q$ at prime $p$ associated to $J$ is defined by

[TABLE]

Write $d_{p}$ for $d_{p,I}$ and call it the depth of of $Q$ at $p$.

For $J$ a nonempty subset of $I$, let $\mathbf{\lambda}_{J}=(\lambda_{i})_{i\in J}$, $\mathbf{k}_{J}=(k_{i})_{i\in J}$ and $Q_{J}=\sum\limits_{j\in J}\lambda_{j}x^{k_{j}}\in\mathbb{Z}[x_{j}:\ j\in J]$. Set $Q_{\emptyset}=0$ and

[TABLE]

If $Q$ and $(\mathbf{\lambda},\mathbf{k})$ are clear from the context, we may drop them in our notations.

1.3. Main results

Suppose $n$ has the prime decomposition

[TABLE]

By Chinese Remainder Theorem, the set of solutions of $Q(x_{1},\cdots,x_{t})\equiv c\bmod n$ is in one-to-one correspondence with the product set of solutions of the equations $Q(x_{1},\cdots,x_{t})\equiv c\bmod p^{n_{p}}$ for primes $p\mid n$. Moreover, $x\in(\mathbb{Z}/n\mathbb{Z})^{\times}$ if and only if $x\in(\mathbb{Z}/p^{n_{p}})^{\times}$ for all $p\mid n$. Thus for any $J\subseteq I$, we have the product formula

[TABLE]

So to compute $N_{J}(Q;c,n)$, it suffices to study the prime power case $N_{J}(Q;c,p^{a})$. By simple argument (as seen in Proposition 2.1(2)), we may reduce $Q$ to the case $p\nmid\lambda_{i}$ for some $i\in I$, which we call $Q$ is reduced at $p$.

Our first main result is the following theorem (which we call the decomposition formula):

Theorem A**.**

Given the polynomial $Q$. For subsets $J_{1}\subsetneq J_{2}\subseteq I$, let

[TABLE]

For $\mathbf{b}\in B(\emptyset,I;a)$, but $\mathbf{b}\neq(a,\cdots,a)$, let

[TABLE]

If $\mathbf{b}=(a,\cdots,a)$, let $J_{\mathbf{b}}=\emptyset$, $Q_{\mathbf{b}}=0$ and $s(\mathbf{b})=0$. Then we have the decomposition formula

[TABLE]

Our next two results are consequences of the following lifting formula

[TABLE]

for $a$ sufficiently large under various assumptions. We shall establish this formula by simple $p$-adic analysis, not by the more complicated exponential sum argument employed by other authors. More precisely, we have

Theorem B**.**

Given the polynomial $Q$, and assume it is reduced at prime $p$. Then

(1) For $a\geq d_{p,J}$ and $c\in\mathbb{Z}$,

[TABLE]

(2) For $a\leq d_{p}=d_{p,I}$, the map

[TABLE]

where $\alpha_{i}\in\mathbb{Z}$ is any lifting of $a_{i}\in\mathbb{F}_{p}$, is well defined. Let $\varphi_{a,J}$ be the restriction of $\varphi_{a}$ on $\prod\limits_{i\in I-J}\mathbb{F}_{p}\times\prod\limits_{i\in J}\mathbb{F}^{\times}_{p}$, then

[TABLE]

In particular, if $p=2$ and $a\leq d_{2}$,

[TABLE]

Theorem C**.**

Given polynomial $Q$ and prime $p$. Let $f_{p}=\max\{v_{p}(k_{i})+1\}$ (or $3$ if $p=2$ and $\max\{v_{2}(k_{i})\}=1$). For integer $c\neq 0$, let $c_{p}$ be the $p$-adic valuation of $c$. Then for any $a\geq 1$, any $J\subseteq I$ (empty or not), $f\geq f_{p}$ and any $x\in\mathbb{Z}/p^{a}\mathbb{Z}$,

[TABLE]

In particular, for $a\geq c_{p}+f_{p}$,

[TABLE]

Thus $N_{J}(Q;c,p^{a})$ as $a$ varies is completely determined by $N_{J}(c,p^{a})$ for $a\leq c_{p}+f_{p}$.

Remark*.*

For $J=\emptyset$, even if $p\nmid\prod\limits_{i=1}^{t}k_{i}$, the formula for $N(Q;0,p^{a})$ is much more complicated. In general we don’t always have $N(0,p^{a})=p^{t-1}N(0,p^{a-1})$ for $a$ sufficiently large. For example, consider $Q(x_{1},x_{2})=x_{1}^{3}+px_{2}^{3}$. Then $N(0,p^{3a})=p^{4a}$, $N(0,p^{3a+1})=p^{4a+1}$ and $N(0,p^{3a+2})=p^{4a+2}$.

As a consequence of Theorems A, B and C, we will give an algorithm to effectively compute $N_{J}(Q;c,p^{a})$ for all possible $J$, $c$ and $a$ if the prime number $p\nmid\prod\limits_{i=1}^{t}k_{i}$. Moreover, except the case $J=\emptyset$ and $c=0$, the number of steps to compute $N_{J}(Q;c,p^{a})$ is bounded by a constant independent of $a$.

Using the main theorems and the algorithm, we shall work on the example $Q(x_{1},\cdots,x_{t})=\lambda_{1}x_{1}^{k}+\cdots+\lambda_{t}x^{k}_{t}$. We obtain the following results:

1. (1)

In the linear case ($k=1$), we solve the counting problem in full generality (cf. [9]). Namely, for any prime $p$, we completely determine the value of $N_{J}(Q;c,p^{a})$ for arbitrary $J\subseteq I$, $c\in\mathbb{Z}$ and $a\geq 1$. Our result is stated in Theorem 4.1. 2. (2)

In the quadratic case ($k=2$), for any prime $p$, we completely determine the value of $N_{J}(Q;c,p^{a})$ for any $J\subseteq I$ satisfying $\min\{v_{p}(\lambda_{i})\mid i\in I\}=\min\{v_{p}(\lambda_{i})\mid i\in J\}$, and arbitrary $c\in\mathbb{Z}$ and integer $a\geq 1$. In particular, we get the exact formula for $N^{*}(Q;c,p^{a})$ for any $c\in\mathbb{Z}$ and $a\geq 1$. Our result is stated in Theorem 4.4. This is a vast generalization of Yang-Tang [11], Sun-Cheng [8] and Mollahajiaghaei [6]. 3. (3)

In the general case, for prime $p\nmid k$, we give a more detailed version of our algorithm in Theorem 4.2. We obtain formulas so that $N_{J}(Q;c,p^{a})$ can be computed in finite steps independent of $a$ except the case $c=0$ and $J=\emptyset$. 4. (4)

We study the case $p\nmid k$ and the dimension $t=2$ in full generality. When $k=2$, $N_{J}(c,2^{a})$ is also studied in full generality.

Finally we shall work on the example $Q(x_{1},\cdots,x_{t})=9x_{1}+2x_{2}^{3}+x_{3}^{9}$ for $p=3$, which is not covered by our algorithm, but the main theorems are still applicable.

2. Preliminaries

2.1. Reduce $Q$ to the reduced case.

The following fact is obvious:

Proposition 2.1**.**

Consider the number $N_{J}(Q;c,p^{a})$ for $p$ a prime number and $J\subseteq I$.

(1) (Lowering dimension) If there exists $j\in I$ such that $v_{p}(\lambda_{j})\geq a$, then

[TABLE]

(2) (Lowering level) Let $e=\min\{v_{p}(\lambda_{i})\mid i\in I\}$ and $v_{p}(c)=c_{p}$. Then

[TABLE]

(3) (Lowering degree) If one has $v_{p}(k_{i})\geq a$, replace $k_{i}$ by $k_{i}/p^{v_{p}(k_{i})-a+1}$. Then the new $k_{i}$ has $p$-adic valuation $a$ and $N_{J}(Q;c,p^{a})$ is unchanged.

Proof.

The only thing needs to prove is (3), which follows from Euler’s Theorem that for $x\in(\mathbb{Z}/p^{a}\mathbb{Z})^{\times}$, $x^{p^{s}}=x^{p^{a-1}}$ for all $s\geq a$, and for $x\in p\mathbb{Z}/p^{a}\mathbb{Z}$, $x^{p^{s}}=0$ for all $s\geq a-1$ since $p^{a-1}\geq a$ for any prime $p$ and integer $a\geq 1$. ∎

Based on Proposition 2.1, to compute $N_{J}(Q;c,p^{a})$, it suffices to consider the case that $\min\{v_{p}(\lambda_{i})\}=0$, $\max\{v_{p}(\lambda_{i}),\ v_{p}(k_{i})\mid i=1,\cdots,t\}<a$ and the depth $d_{p}\leq a$. In particular, we can always assume $p\nmid\lambda_{i}$ for some $i\in I$.

2.2. Formulas for $N(Q;c,p)$.

We recall the classical formulas for $N(Q;c,p)$. First recall for complex characters $\chi_{1},\ \cdots,\ \chi_{t}$ of the prime field $\mathbb{F}_{p}$, the Jacobi sum $J(\chi_{1},\cdots,\chi_{t})$ is defined by the formula

[TABLE]

and the Jacobi sum $J_{0}(\chi_{1},\cdots,\chi_{t})$ is defined by the formula

[TABLE]

Then the following theorem is well known:

Theorem 2.2**.**

(1) Suppose $p$ is odd and $\lambda_{1}\cdots\lambda_{t}\neq 0\in\mathbb{F}_{p}$. Then $N(c,p)$, the number of solutions of

[TABLE]

over the prime field $\mathbb{F}_{p}$, is given by

[TABLE]

and

[TABLE]

for $c\neq 0$.

(2) If $2\nmid\lambda_{i}$ for some $i\in I$, then $N(0,2)=N(1,2)=2^{t-1}$.

Proof.

Part (1) follows from Theorem 5 in § 8.7 in [5]. Part (2) is clear, since $x^{k}=x$ in $\mathbb{F}_{2}$. ∎

3. Proof of the main theorems and the algorithm

3.1. The decomposition formula and its special cases.

We now prove Theorem A.

Proof of Theorem A.

Note that $\mathbb{Z}/p^{a}\mathbb{Z}$ has a disjoint decomposition (assuming $p^{a+1}\mathbb{Z}/p^{a}\mathbb{Z}$ is the empty set)

[TABLE]

Suppose $\mathbf{x}=(x_{1},\cdots,x_{t})\in\Gamma_{J_{1}}(Q;c,p^{a})$, and if $J_{1}=\emptyset$ and $J_{2}=I$, suppose $\mathbf{x}\neq{\bf 0}$. Then for $j\in J_{2}-J_{1}$, $x_{j}\in p^{b_{j}}\mathbb{Z}/p^{a}\mathbb{Z}-p^{b_{j}+1}\mathbb{Z}/p^{a}\mathbb{Z}$ for some $0\leq b_{j}\leq a$. Set $b_{j}=0$ for $j\notin J_{2}-J_{1}$. Let $\mathbf{b}=\mathbf{b}(\mathbf{x})=(b_{j})_{j=1,\cdots,t}\in B(J_{2},J_{1};a)$ and $J_{\mathbf{b}}\neq\emptyset$.

For $j\in J_{2}\cap J_{\mathbf{b}}$, the element $\tilde{x}_{j}=x_{j}/p^{b_{j}}$ is a well defined element in $(\mathbb{Z}/p^{a-b_{j}})^{\times}$. Let $C_{j}=\{x\in(\mathbb{Z}/p^{a})^{\times}\mid x\equiv\tilde{x}_{j}\bmod{p^{a-b_{j}}}\}$. For $j\in J_{\mathbf{b}}-J_{2}$, let $C_{j}=\{x_{j}\}$. Then

[TABLE]

On the other hand, if $Q_{\mathbf{b}}\neq 0$, then $J_{\mathbf{b}}$ as the set of $j$’s such that $x_{j}$ appears in $Q_{\mathbf{b}}$ is not empty. For $(y_{j})_{j\in J_{\mathbf{b}}}\in\Gamma_{J_{2}\cap J_{\mathbf{b}}}(Q_{\mathbf{b}};c,p^{a})$, let $\tilde{x}_{j}=y_{j}\bmod{p^{a-b_{j}}}$, then $x_{j}=p^{b_{j}}\tilde{x}_{j}$ is a well defined element in $p^{b_{j}}\mathbb{Z}/p^{a}\mathbb{Z}-p^{b_{j}+1}\mathbb{Z}/p^{a}\mathbb{Z}$. Let $x_{j}=0$ for $j\notin J_{\mathbf{b}}$. Then $\mathbf{x}=(x_{j})\in\Gamma_{J_{1}}(Q;c,p^{a})$. In this way, one element $\mathbf{x}$ corresponds exactly to $p^{\sum_{j:b_{j}<a}b_{j}}=p^{s(\mathbf{b})}$ elements in $\Gamma_{J_{2}\cap J_{\mathbf{b}}}(Q_{\mathbf{b}};c,p^{a})$.

If $J_{1}=\emptyset$ and $J_{2}=I$, then ${\bf 0}\in\Gamma_{J_{1}}(Q;c,p^{a})$ if only if $p^{a}\mid c$, which is corresponding to the case $\mathbf{b}=(a,\cdots,a)$ and $Q_{\mathbf{b}}=0$.

In conclusion, (2) is proved. ∎

Special cases of the decomposition formula. We shall use the following special cases in this paper:

(1) The case $J=J_{1}\subsetneq I=J_{2}$. Then

[TABLE]

This means that if we can determine $N^{*}(Q_{\mathbf{b}};c,p^{a})$ for all $\mathbf{b}\in B(J,I;a)$, then we get $N_{J}(Q;c,p^{a})$.

(2) The case $a=1$. Then

[TABLE]

By the Inclusion-Exclusion Principle, (14) has the following inverse formula

[TABLE]

Take $J_{1}=\emptyset$ and $J_{2}=J$ in (15), then we have

[TABLE]

This means that $N_{J}(Q;c,p)$ is determined by $N(Q_{I-T};c,p)$ for all $T\subseteq J$.

Remark*.*

Another interesting question is to count the number $N_{J_{1},J_{2}}(Q;c,n)$ of solutions of $Q(x_{1},\cdots,x_{t})\equiv c\bmod{n}$ such that $x_{i}\in(\mathbb{Z}/n\mathbb{Z})^{\times}$ for $i\in J_{1}$ and $x_{i}\notin(\mathbb{Z}/n\mathbb{Z})^{\times}$ for $i\in J_{2}$. First one must keep in mind that no product formula exists in general for $N_{J_{1},J_{2}}(Q;c,n)$ if $J_{2}\neq\emptyset$. However, by the Inclusion-Exclusion Principle, we have

[TABLE]

As a consequence, the values $N_{J}(Q;c,n)$ for all $J$ determine $N_{J_{1},J_{2}}(Q;c,n)$ for all disjoint pairs $(J_{1},J_{2})$.

3.2. The lifting formula

We need the following lemma whose proof is an easy exercise of Newton’s Binomial Theorem and $p$-adic analysis:

Lemma 3.1**.**

(1) Let $p$ be an odd prime. For integers $x,y$, $k\geq 1$, and $m\geq 1$, we have

[TABLE]

(2) For integers $x$ and integer $y$, $k\geq 1$, and $m\geq 1$, then

[TABLE]

For odd integer $x$,

[TABLE]

(3) Let $U_{p,a}^{(i)}=\{1+p^{i}x\mid x\in\mathbb{Z}/p^{a}\mathbb{Z}\}\subseteq(\mathbb{Z}/p^{a}\mathbb{Z})^{\times}$. Then for $f>0$, $(U_{p,a}^{(i)})^{p^{f}}=U_{p,a}^{(f+i)}$ if $(p,i)\neq(2,1)$ and $(U_{2,a}^{(1)})^{p^{f}}=U_{2,a}^{(f+2)}$.

We are now ready to prove Theorem B and Theorem C.

Proof of Theorem B.

Write $d=d_{p}$. Let $\psi_{a,b}$ be the natural reduction map from $\Gamma_{J}(c,p^{a})$ to $\Gamma_{J}(c,p^{b})$.

(1) First assume $p$ is odd. Suppose that $j$ satisfies $v_{p}(\lambda_{j}k_{j})=e_{j}+f_{j}=d_{j}<a$. By Lemma 3.1(1), if $(x_{1},\cdots,x_{j},\cdots,x_{t})\in\Gamma_{J}(c,p^{a})$, then $(x_{1},\cdots,x_{j}+p^{a-d_{j}}y_{j},\cdots,x_{t})\in\Gamma_{J}(c,p^{a})$ for any $y_{i}\in\mathbb{Z}/p^{a}\mathbb{Z}$.

If $a>d_{p,J}$, then $a>d_{j}+1$ for some $j\in J$. Let $(a_{1},\cdots,a_{t})\in\Gamma_{J}(c,p^{a-1})$. Let $u\in\{0,\cdots,p-1\}$. Let $x_{i}\in\mathbb{Z}/p^{a}\mathbb{Z}$ be any lifting of $a_{i}$. Then

[TABLE]

Thus there exists exactly one $u\in\{0,\cdots,p-1\}$ such that $(x_{1},\cdots,x_{j}+up^{a-d_{j}-1},\cdots,x_{t})\in\Gamma_{J}(c,p^{a})$, and $\psi_{a,a-1}$ is a $p^{t-1}$-to-$1$ map. Thus we have the lifting formula

[TABLE]

for all $a>d_{p,J}$.

Now assume $p=2$. Assume $a>d_{2,J}$. Then the assumption means that $a>d_{j}+2$ for some $j\in J$ with $k_{j}$ even or $a>d_{j}+1$ for some $j\in J$ with $k_{j}$ odd. Let $(a_{1},\cdots,a_{t})\in\Gamma_{J}(c,2^{a-1})$. Let $x_{i}\in\mathbb{Z}/2^{a}\mathbb{Z}$ be any lift of $a_{i}$. Then

[TABLE]

Thus one of $(x_{1},\cdots,x_{t})$ and $(x_{1},\cdots,x_{j}+2^{a-d_{j}-1},\cdots,x_{t})$ is a solution of $Q(x_{1},\cdots,x_{t})\equiv c\bmod{n}$, and $\psi_{a,a-1}$ is a $2^{t-1}$-to-$1$ map. Again we have the lifting formula.

(2) Assume $a\leq d=d_{p,I}$. Suppose $(a_{1},\cdots,a_{t})\in\mathbb{F}_{p}^{t}$, let $\alpha_{i}\in\mathbb{Z}$ be any lifting of $a_{i}$. Then

[TABLE]

for any $y_{i}\in\mathbb{Z}$, and $Q(\alpha_{1},\cdots,\alpha_{t})\bmod p^{a}$ is a fixed element in $\mathbb{Z}/p^{a}\mathbb{Z}$ independent of the lifting, so the map $\varphi_{a}$ is well-defined. Thus for $(a_{1},\cdots,a_{t})\in\Gamma_{J}(c,p)\subseteq\mathbb{F}_{p}^{t}$,

[TABLE]

Assume furthermore that $p=2$. For $T\subseteq I$, let $e_{T}=(e_{T,i})_{i\in I}$ be the element in $\mathbb{F}^{t}_{2}$ that $e_{T,i}=1$ for $i\in T$ and $e_{T,i}=0$ for $i\notin T$. Then $\Gamma_{J}(c,2)$ consists of elements $e_{T}$ satisfying $T\supseteq J$ and $v_{2}(\sum_{i\in T}\lambda_{T}-c)\geq 1$. Let [math] and $1$ in $\mathbb{Z}$ be the liftings of [math] and $1$ in $\mathbb{F}_{2}$ respectively. Then $\varphi_{a}(e_{T})=\sum\limits_{i\in T}\lambda_{i}\bmod{2^{a}}$. This finishes the proof of Theorem B(2). ∎

Corollary 3.2**.**

Given the polynomial $Q(x_{1},\cdots,x_{t})$. If at prime $p$ one has $d_{p}\geq t$. Then there exists $c\in\mathbb{Z}$ such that $N^{*}(Q;c,p^{d_{p}})=0$.

Proof.

This is because there are $p^{d_{p}}$ conjugacy classes modulo $p^{d_{p}}$ but there are only $(p-1)^{t}$ points in $\mathbb{F}_{p}^{\times\,t}$. ∎

Proof of Theorem C.

Write $k_{i}=p^{f_{i}}k^{\prime}$ such that $(p,k^{\prime})=1$. By Lemma 3.1, if $f\geq f_{p}$, then for any $i\in I$, $1+p^{f}x=(1+py_{i})^{p^{f_{i}}}$ for some $f_{i}\in\mathbb{Z}/p^{a}\mathbb{Z}$. If $a\leq c_{p}+f$, the formula is certainly true. For $a>c_{p}+f$, let $u_{i},v_{i}\in\mathbb{Z}$ such that $u_{i}k^{\prime}_{i}+p^{a-k_{i}}v_{i}=1$, then $1+p^{f}x=((1+py_{i})^{u_{i}k_{i}}=\beta_{i}^{k_{i}}$ for some $\beta_{i}\in(\mathbb{Z}/p^{a}\mathbb{Z})^{\times}$. Thus we have a one-to-one correspondence

[TABLE]

and hence $N_{J}(c,p^{a})=N_{J}(c(1+p^{f}x),p^{a})$.

Now consider the natural map $\psi_{a+1,a}:(\mathbb{Z}/p^{a+1}\mathbb{Z})^{t}\rightarrow(\mathbb{Z}/p^{a}\mathbb{Z})^{t}$. For $a>c_{p}+f_{p}$, $\psi^{-1}_{a+1,a}(\Gamma_{J}(c,p^{a}))$ is the disjoint union of $\Gamma_{J}(c+up^{a},p^{a+1})$ for $u\in\{0,\cdots,p-1\}$, but all $\Gamma_{J}(c+up^{a},p^{a+1})$ are of the same cardinality $N_{J}(c,p^{a+1})$, hence the lifting formula $N_{J}(c,p^{a+1})=p^{t-1}N_{J}(c,p^{a})$ holds. This finishes the proof of Theorem C. ∎

3.3. An algorithm to compute $N_{J}(Q;c,p^{a})$ if $p\nmid\prod\limits_{i\in I}k_{i}$

By Theorems A, B and C, we then have the following algorithm to effectively compute $N_{J}(Q;c,p^{a})$.

1. (1)

Reduce $Q$ to the reduced form at $p$ (i.e., $d_{p}(Q)=1$) by Proposition 2.1. We suppose $Q$ is reduced hereafter. 2. (2)

Compute $N(Q;c,p)$ for all $Q$ by using formulas in Theorem 2.2. 3. (3)

For $J$ nonempty, compute $N_{J}(Q;c,p)$ by the Inclusion-Exclusion Principle formula (16). If $d_{p,J}=1$, use the relation $N_{J}(Q;c,p^{a})=p^{(a-1)(t-1)}N_{J}(Q;c,p)$ by Theorem B to get $N_{J}(Q;c,p^{a})$, in particular, get $N^{*}(Q;c,p^{a})$. 4. (4)

For $J$ nonempty and $d_{p,J}=b+1>1$, use the decomposition formula (13) to compute $N_{J}(Q;c,p^{a})$ for all $1<a\leq b+1$, then $N_{J}(Q;c,p^{a})=p^{(a-b-1)(t-1)}N_{J}(Q;c,p^{b+1})$ for $a\geq b+1$ by Theorem B. (Note: the assumption $p\nmid\prod k_{i}$ means the reduced form of $Q_{\mathbf{b}}$ for any $\mathbf{b}$ in the right hand side of (13) is of depth $1$, hence $N^{*}(Q_{\mathbf{b}};c,p^{a})$ can be computed as in the previous step.) 5. (5)

If $c\neq 0$, let $c_{p}=v_{p}(c)$. Compute $N(Q;c,p^{a})=N(Q;0,p^{a})$ for $a\leq c_{p}$ and $N(Q;c,p^{c_{p}+1})$ by the decomposition formula (13). Then for $a>c_{p}+1$, $N(Q;c,p^{a})=p^{(a-c_{p}-1)(t-1)}N(Q;c,p^{c_{p}+1})$ from Theorem C. 6. (6)

Use the decomposition formula (13) to compute $N(Q;0,p^{a})$ for any given $a$.

Remark*.*

We see that except the last step to compute the case $J=\emptyset$ and $c=0$, the number of steps to compute $N_{J}(Q;c,p^{a})$ is bounded by a constant independent of $a$.

In the case $J$ is nonempty, let $|J|=s$. If $c_{p}=v_{p}(c)<b$, by Theorem C, one can furthermore get

[TABLE]

In particular, if $p\nmid c$, i.e., $c_{p}=0$, then we just need formulas for $N(Q_{I-J};c,p)$ in Theorem 2.2 to get $N_{J}(Q;c,p^{a})$.

4. Applications of the main theorems

In this section, we shall apply the general formulas obtained in the previous section to compute $N_{J}(Q;c,p^{a})$ in many special cases. Without loss of generality, we assume $Q$ is reduced, i.e., $p\nmid\lambda_{i}$ for some $i$ because of (10).

4.1. The linear case $Q(x_{1},\cdots,x_{t})=\sum\limits_{i=1}^{t}\lambda_{i}x_{i}$.

Consider the linear congruence equation

[TABLE]

Theorem 4.1**.**

Suppose $p\nmid\lambda_{i}$ for some $i\in I$. For any subset $J$ of $I$ and prime $p$, let $s=\#J$ and $s_{p}=\#J_{p}$ where $J_{p}=\{j\in J\mid p\nmid\lambda_{j}\}$. Then

(1) The lifting formula holds for all $a\geq 1$:

[TABLE]

(2) If there exists $i\notin J$, $p\nmid\lambda_{i}$, then

[TABLE]

if for all $i\notin J$, $p\mid\lambda_{i}$, then

[TABLE]

where $\delta_{c}=1$ if $p\mid c$ and $=0$ if $p\nmid c$.

Proof.

If there exists $i\notin J$, $p\nmid\lambda_{i}$, then one can choose all possible $x_{j}$ for $j\neq i$, and then $x_{i}$ is decided by the $x_{j}$’s, so $N_{J}(Q;c,p^{a})=p^{a(t-s-1)}\cdot\varphi(p^{a})^{s}$. Thus (20) holds, so does (19) in this situation.

If for all $i\notin J$, $p\mid\lambda_{i}$, then there exists $i\in J$ such that $p\nmid\lambda_{i}$, so $d_{p,J}=1$ and (19) holds in this situation by Theorem B. Now one easily has $N_{J}(Q;c,p)=p^{t-s}(p-1)^{s-s_{p}}N^{*}(Q_{J_{p}};c,p)$, and by (15),

[TABLE]

The theorem is proved. ∎

4.2. The case $Q(x_{1},\cdots,x_{t})=\sum\limits_{i=1}^{t}\lambda_{i}x_{i}^{k}$.

In this subsection, we consider the congruence equation

[TABLE]

4.2.1. A general result.

The following Theorem is a more detailed version of our algorithm:

Theorem 4.2**.**

Suppose prime $p\nmid k$ and $Q$ is reduced at $p$. For $c\neq 0$, let $c_{p}$ be the $p$-adic valuation of $c$. Let $I_{p}=\{i\in I\mid p\nmid\lambda_{i}\}$ and $t_{p}=\#I_{p}$. For $J$ a nonempty subset of $I$, let $J_{p}=\{i\in J\mid p\nmid\lambda_{i}\}$, $s=\#J$ and $s_{p}=\#J_{p}$. Then

(1) For $c\neq 0$, $N(Q;c,p^{a})$ for all $a\geq 1$ is completely determined by $N(Q;0,p^{a})$ for $1\leq a\leq c_{p}$ and $N(Q;c,p^{c_{p}+1})$ through the formula

[TABLE]

In particular, if $p\nmid c$, then for $a\geq 1$,

[TABLE]

where $N(Q_{I_{p}};c,p)$ can be computed by the formulas in Theorem 2.2.

(2) If $J_{p}\neq\emptyset$, i.e., $s_{p}\neq 0$ and $d_{p,J}=1$, then for any $a\geq 1$, for any $c\in\mathbb{Z}$,

[TABLE]

[TABLE]

and

[TABLE]

where $N(Q_{T};c,p)$ can be computed by the formula in Theorem 2.2.

In particular, $N^{*}(Q;c,p^{a})$ can be computable by the formulas above, in this case $J=I$ and $J_{p}=I_{p}$.

(3) If $d_{p,J}=b+1>1$, i.e., $s_{p}=0$, then for $c\in\mathbb{Z}$,

[TABLE]

If moreover, $c_{p}<b$, then

[TABLE]

Here $N_{J}(Q;c,p^{a})$ for $a\leq b+1$ and $N(Q_{I-J};c,p^{a})$ for $a\leq c_{p}+1$ can be computed by the decomposition formula (13).

In particular, if $p\nmid c$, then for $a\geq 1$,

[TABLE]

where $N(Q_{I_{p}};c,p)$ can be computed by Theorem 2.2.

4.2.2. The quadratic case.

In this case, we recall the following well-known result:

Proposition 4.3**.**

Suppose $Q(x_{1},\cdots,x_{t})=\lambda_{1}x_{1}^{2}+\cdots+\lambda_{t}x_{t}^{2}$. For odd prime $p$, let \bigl{(}\frac{\cdot}{p}\bigr{)} be the Legendre symbol. If $p\nmid\prod\limits_{i=1}^{t}\lambda_{i}$, then

[TABLE]

Proof.

This follows from §8.6 in [5], and can also be found in [1]. ∎

Remark*.*

The above formula holds for $I=\emptyset$. In this case $t=0$ and $N(0;c,p)=1$ if $p\mid c$ and [math] if not.

Theorem 4.4**.**

Suppose $Q(x_{1},\cdots,x_{t})=\lambda_{1}x_{1}^{2}+\cdots+\lambda_{t}x_{t}^{2}$ and $p\nmid\lambda_{i}$ for some $i\in I$.

(1) For $p$ odd, suppose $p\nmid\lambda_{i}$ for some $i\in I$. Let $I_{p}=\{p\in I\mid p\nmid\lambda_{i}\}$, let $t_{p}=\#I_{p}$ and $r_{p}=\#\{i\in I\mid\lambda_{i}\ \text{is a quadratic non-residue modulo}\ p\}$. Write $p^{*}=p\cdot\left(\frac{-1}{p}\right)$, and for $i\geq j\geq 0$, write

[TABLE]

[TABLE]

Then for $a\geq 1$, we have

[TABLE]

where $N^{*}(Q_{I_{p}};c,p)$ is given by

[TABLE]

(2) Moreover, for $J\subseteq I$ such that $d_{p,J}=1$, i.e., if there exists $i\in J$ such that $p\nmid\lambda_{j}$. Let $J_{p}=\{p\in I\mid p\nmid\lambda_{i}\}$, let $s=\#J$, $s_{p}=\#J_{p}$ and $r_{p,J}=\#\{i\in J\mid\lambda_{i}\ \text{is a quadratic non-residue modulo}\ p\}$. Then for $a\geq 1$, we have

[TABLE]

where

[TABLE]

(3) For $p=2$, for $J\subseteq I$ such that $d_{2,J}=3$, i.e. if there exists $j\in J$ such that $2\nmid\lambda_{j}$, then for $a\geq 3$,

[TABLE]

and for $1\leq a\leq 3$,

[TABLE]

In particular, for $J=I$, let $c^{\prime}_{2}=v_{2}(\sum\limits_{i\in I}\lambda_{i}-c)$. Then

[TABLE]

Remark*.*

For general $Q$ (reduced or not), if we replace the assumption $p\nmid\lambda_{i}$ for some $i\in J$ by the assumption $\min\{v_{p}(\lambda_{i})\mid i\in I\}=\min\{v_{p}(\lambda_{i})\mid i\in J\}$, along with Proposition 2.1(2), we get the formula for $N_{J}(Q;c,p^{a})$ for all $c\in\mathbb{Z}$ and $a\geq 1$.

Proof.

Part (3) follows from Theorem B(2), Part (1) is a special case of (2), and (33) follows from Theorem B(1), we just need to prove (34) in Part (2).

By the Inclusion-Exclusion principle, we know

[TABLE]

We use (30) and the above formula to compute $N_{J_{p}}(Q_{I_{p}};c,p)$. We compute the main term and the error term separately. The main term is

[TABLE]

For the error term, we need the following identities

[TABLE]

[TABLE]

In the case $t_{p}$ is odd and $p\nmid c$, for the subset $T$ of even order, suppose there are $i$ quadratic residues in $\{\lambda_{m}\mid m\in T\}$ and $j$ quadratic non-residues, the contribution of the error term in $N(Q_{I_{p}-T};c,p)$ is

[TABLE]

So the contribution for all $T$ of even order is $(-1)^{r_{p}}\left(\frac{c}{p}\right)(\sqrt{p^{*}})^{t_{p}-1}\times$

[TABLE]

which is

[TABLE]

Similarly for all $T$ of odd order, the error term contribution is

[TABLE]

The other three cases in (34) are obtained by the same method. ∎

4.2.3. The case $t=2$ and $p\nmid k$.

For this case, note that if $p\nmid\lambda_{1}$, let $\lambda_{1}^{-1}$ be the $p$-adic inverse of $\lambda_{1}$, then

[TABLE]

Thus we may assume

[TABLE]

such that $p\nmid\lambda$ and $e\geq 0$. We want to compute $N_{J}(c,p^{a})$ for $J=\emptyset,\ \{1\},\{2\}$ and $I=\{1,2\}$, $c\in\mathbb{Z}$ and $a\geq 1$.

If $p\nmid c$ and $e=0$, by Theorem 2.2 and note that $J_{0}(\chi,\chi^{-1})=(p-1)\chi(-1)$ if $\chi\neq 1$, $=p$ if $\chi=1$, then

[TABLE]

For $J=\{1\}$ or $I$, then $d_{p,J}=1$. By Theorem B, we have $N_{J}(c,p^{a})=p^{a-1}N_{J}(c,p)$. Then by (16), we have

Proposition 4.5**.**

Let $Q(x_{1},x_{2})=x^{k}_{1}+\lambda p^{e}x_{2}^{k}$ such that $p\nmid\lambda k$ and $e\geq 0$. Then

[TABLE]

[TABLE]

Here $N(c,p)$ and $N(0,p)$ are given by (38) and (39) respectively.

Remark*.*

In the quadratic case, Theorem 4.4 gives more precise formulas for the cases $J=\{1\}$ or $I$, or $J=\{2\}$ and $e=0$.

For $J=\emptyset$ and $\{2\}$, the situation for $N_{J}(c,p^{a})$ is much more complicated. We first have

Proposition 4.6**.**

Let $Q(x_{1},x_{2})=x^{k}_{1}+\lambda p^{e}x_{2}^{k}$ such that $p\nmid\lambda k$ and $e\geq 0$. For $c\neq 0$, let $c_{p}$ be the $p$-adic valuation of $c$ and $c^{\prime}=c/p^{c_{p}}$. For $c=0$, let $c_{p}=+\infty$. Let $J=\{2\}$ or $\emptyset$. Then

1. (1)

$N_{J}(Q;c,p^{a})=p^{a-c_{p}-1}N_{J}(Q;c,p^{c_{p}+1})$* for $c\neq 0$.* 2. (2)

If $e\geq a$, then $N_{\{2\}}(Q;c,p^{a})=p^{a-1}(p-1)N(x_{1}^{k};c,p^{a})$ and $N(Q;c,p^{a})=p^{a}N(x_{1}^{k};c,p^{a})$, and

[TABLE]

Here $\lceil x\rceil$ meanings the smallest integer $\geq x$. 3. (3)

If $e<a$, $N_{\{2\}}(Q;c,p^{a})=p^{a-e-1}N_{\{2\}}(Q;c,p^{e+1})$.

Consequently, the study of $N_{J}(Q;c,p^{a})$ for the set $J=\emptyset$ and $\{2\}$ is reduced to the study of $N(Q;up^{a},p^{a+1})$ for $u\in\{0,\cdots,p-1\}$ and $e\leq a$, and $N_{\{2\}}(Q;up^{e},p^{e+1})$ for $u\in\{0,\cdots,p-1\}$.

Proof.

Part (1) follows from Theorem C and Part (3) follows from Theorem B. The first half of (2) follows from Proposition 2.1(1). For the second half of (2), the solutions of $x_{1}^{k}\equiv 0\pmod{p^{a}}$ are of the form $x_{1}=p^{\lceil\frac{a}{k}\rceil}x_{1}^{\prime}$ for $x_{1}^{\prime}$ arbitrary. If $c_{p}<a$, then $x_{1}^{k}\equiv c\pmod{p^{a}}$ is solvable only if $k\mid c_{p}$, in this case

[TABLE]

but $N^{*}(x^{k};c^{\prime},p)=N(x^{k};c^{\prime},p)=\sum\limits_{\chi:\ \chi^{k}=1}\chi(c^{\prime})$. ∎

For the quadratic case, we have

Proposition 4.7**.**

Let $Q(x_{1},x_{2})=x_{1}^{2}+\lambda p^{e}x_{2}^{2}$ such that $p\nmid 2\lambda$. Then

(1) For $u\in\{1,\cdots,p-1\}$,

[TABLE]

For $u=0$,

[TABLE]

(2) For $u\in\{1,\cdots,p-1\}$ and $a\geq e$,

[TABLE]

For $e<a$,

[TABLE]

Proof.

We use the decomposition formula in Theorem A to count the number.

(1) Take $J_{1}=\{2\}$ and $J_{2}=I$ in Theorem A, then the decomposition formula for $N_{\{2\}}(Q;up^{e},p^{e+1})$ is

[TABLE]

If $j<e/2$, $N^{*}(p^{2j}x_{1}^{2}+\lambda p^{e}x_{2}^{2};up^{e},p^{e+1})=0$. If $j>e/2$,

[TABLE]

If $j=e/2$, then

[TABLE]

Combine the results we get the formula for $N_{\{2\}}(Q;up^{e},p^{e+1})$.

The decomposition formula for $N_{\{2\}}(Q;0,p^{e+1})$ is

[TABLE]

If $j\neq e/2$, $N^{*}(p^{2j}x_{1}^{2}+\lambda p^{e}x_{2}^{2};0,p^{a})=0$ and $N^{*}(\lambda p^{e}x_{2}^{2};0,p^{e+1})=0$; for $j=e/2$, N^{*}(p^{2j}x_{1}^{2}+\lambda p^{e}x_{2}^{2};0,p^{e+1})=p^{2e}(p-1)(1+\bigl{(}\frac{-\lambda}{p}\bigr{)}). So we get the formula for $N_{\{2\}}(Q;0,p^{e+1})$.

(2) Take $J_{1}=\emptyset$ and $J_{2}=\{2\}$, then the decomposition formula for $N(Q;up^{a},p^{a+1})$ is

[TABLE]

If $j\geq(a+1-e)/2$, then

[TABLE]

and N(x_{1}^{2};up^{a},p^{a+1})=p^{a/2}(1+\bigl{(}\frac{u}{p}\bigr{)}) if $2\mid a$ and [math] if $2\nmid a$, so

[TABLE]

If $j<(a-e)/2$, then

[TABLE]

If $j=(a-e)/2$, then

[TABLE]

We now can just use results in (1) to obtain the formula for $N(Q;up^{a},p^{a+1})$.

The decomposition formula for $N(Q;0,p^{a})$ is

[TABLE]

If $j\geq(a-e)/2$, then

[TABLE]

If $j<(a-e)/2$, then

[TABLE]

which is given by formulas in (1). Combine these results, we get the formula for $N(Q;0,p^{a})$. ∎

Remark*.*

For completeness, let us study $N_{J}(Q;c,2^{a})$ for $Q(x_{1},x_{2})=x_{1}^{2}+2^{e}\lambda x_{2}^{2}$ and $2\nmid\lambda$. The cases $J=\{1\}$ and $\{1,2\}$ are given in part (3) of Theorem 4.4. Here we give steps to compute $N_{J}(Qc,2^{a})$ for $J=\{2\}$ or $\emptyset$.

(1) We first compute $N(x_{1}^{2};c,2^{a})$. Assume that $c=2^{c_{2}}u$ with $u$ odd for $c\neq 0$. Then

• •

if $c=0$ or $c_{2}\geq a$, $N(x_{1}^{2};0,2^{a})=2^{a-\lceil\frac{a}{2}\rceil}$;

• •

if $a\geq c_{2}+3$, $N(x_{1}^{2};c,2^{a})=N(x_{1}^{2};c,2^{c_{2}+3})$ (by Theorem C);

• •

if $c_{2}+1\leq a\leq c_{2}+3$, $N(x_{1}^{2};c,2^{a})=2^{a-\frac{c_{2}}{2}-1}$ if $2\mid c_{2}$ and $u\equiv 1\pmod{2^{a-c_{2}}}$ or [math] if otherwise.

(2) For $J=\{2\}$, if $a>e+3$, by Theorem B, we have

[TABLE]

If $a\leq e+3$, since $2^{e}x_{2}^{2}\equiv 2^{e}\pmod{2^{a}}$ for any $x_{2}\in(\mathbb{Z}/2^{a}\mathbb{Z})^{\times}$,

[TABLE]

with $N(x_{1}^{2};c-2^{e}\lambda,2^{a})$ be given in part (1).

(3) For $J=\emptyset$, by the decomposition formula in Theorem A, we have

[TABLE]

where $N_{\{2\}}(x_{1}^{2}+\lambda p^{e+2j}x_{2}^{2};c,2^{a})$ is given in part (2) and $N(x_{1}^{2};c,2^{a})$ is given in part (1).

For the general case, we have

Proposition 4.8**.**

Let $Q(x_{1},x_{2})=x^{k}_{1}+\lambda p^{e}x_{2}^{k}$ such that $p\nmid\lambda k$ and $e\geq 0$. Let $C=N(x^{k}_{1}+\lambda x_{2}^{k};u,p)$ and $C_{0}^{*}=N(x^{k}_{1}+\lambda x_{2}^{k};0,p)-1$ given by (38) and (39) respectively. Then

(1) For $u\in\{1,\cdots,p-1\}$,

[TABLE]

For $u=0$,

[TABLE]

(2) For $u\in\{1,\cdots,p-1\}$ and $a\geq e$,

[TABLE]

For $e<a$,

[TABLE]

Here the sum $\sum$ is over all characters $\chi$ such that $\chi^{k}=1$, and $[n]$ means the largest integer $\leq n$.

Proof.

The proof of part (1) is similar to the proof of Proposition 4.7. We just show how to get the formulas of part (2).

Take $J_{1}=\emptyset$ and $J_{2}=\{2\}$, then the decomposition formula for $N(Q;up^{a},p^{a+1})$ is

[TABLE]

If $e+kj>a$, i.e. $j\geq[\frac{a-e}{k}]+1$, then

[TABLE]

and $N(x_{1}^{k};up^{a},p^{a+1})=p^{a-\frac{a}{k}}\sum\chi(u)$ if $k\mid a$ and [math] if $k\nmid a$, so

[TABLE]

If $e+kj<a$, i.e. $j\leq\lceil\frac{a-e}{k}\rceil-1$, then

[TABLE]

By (48), we have

[TABLE]

If $e+kj=a$, i.e $j=\frac{a-e}{k}$, then by (47) we have

[TABLE]

Thus we get the formula for $N(Q;up^{a},p^{a+1})$.

The decomposition formula for $N(Q;0,p^{a})$ is

[TABLE]

If $e+kj\geq a$, i.e. $j\geq\lceil\frac{a-e}{k}\rceil$ then

[TABLE]

so

[TABLE]

If $e+kj<a$, i.e. $j\leq\lceil\frac{a-e}{k}\rceil-1$, then

[TABLE]

and

[TABLE]

thus we get the formula for $N(Q;0,p^{a})$. ∎

Remark*.*

The case $t\geq 3$ can also be computed, but the discussion is a little bit tedious.

4.3. The example $Q(x_{1},x_{2},x_{3})=9x_{1}+3x_{2}^{3}+x_{3}^{9}$ for $p=3$.

At last we consider the congruence equation

[TABLE]

which is not included in the algorithm.

For $c\neq 0$, write $c=3^{c_{3}}c^{\prime}$; for $c=0$ set $c_{3}=+\infty\geq a$ for any $a$. Since for any $J\neq\emptyset$, $d_{3,J}=3$, by Theorem B, we have

[TABLE]

After simple calculation, we then get $N_{J}(Q;c,27)$ in Table 1.

For $J=\emptyset$, the map $\varphi_{3}:(a_{1},a_{2},a_{3})\mapsto Q(\alpha_{1},\alpha_{2},\alpha_{3})\bmod{27}$ from $(\mathbb{Z}/3\mathbb{Z})^{3}$ to $\mathbb{Z}/27\mathbb{Z}$ is found to be one-to-one. Note that any solution $(\beta_{1},\beta_{2},\beta_{3})\in\Gamma(Q;c,27)$ is a lifting of some $(a_{1},a_{2},a_{3})\in\varphi_{3}^{-1}(c)$, but we always have

[TABLE]

Thus for any $c\in\mathbb{Z}$, we have $N(Q;c,27)=3^{6}$. In fact, we have $N(Q;c,3^{a})=3^{2a}$ for $a\leq 3$. For the case $a>3$, we use the notation $N_{J_{1},J_{2}}$ introduced in the remark of §3.1, then

[TABLE]

We compute the right hand side term by term:

• •

if $c_{3}=0$, then $N_{\emptyset,\{2,3\}}=N_{\{2\},\{3\}}=0$, $N_{\{3\},\{2\}}=3^{2a}$ for $c^{\prime}\equiv 1,8,10,17,19,26(\bmod 27)$, and $N_{\{2,3\}}=3^{2a}$ for $c^{\prime}\equiv 2,4,5,7,11,13,\\ 14,16,20,22,23,25(\bmod 27)$ from Table 1;

• •

if $c_{3}=1$, then $N_{\emptyset,\{2,3\}}=N_{\{3\},\{2\}}=N_{\{2,3\}}=0$, and $N_{\{2\},\{3\}}=3^{2a}$;

• •

if $c_{3}\geq 2$, $N_{\{2\},\{3\}}=N_{\{3\},\{2\}}=N_{\{2,3\}}=0$, and $N_{\emptyset,\{2,3\}}=3^{2a}$.

Thus we have

[TABLE]

for any $a>0$.

Acknowledgement

Research is partially supported by National Key Basic Research Program of China (Grant No. 2013CB834202) and National Natural Science Foundation of China (Grant No. 11571328).