On a problem of Perron

TL;DR

This paper characterizes the set of squares in finite fields using additive properties, providing a new purely additive perspective on Perron's problem.

Contribution

It offers a novel additive characterization of squares in finite fields, linking set partitions to algebraic properties.

Findings

Partition into squares and non-squares is characterized by additive properties.

Provides a purely additive criterion for identifying squares in finite fields.

Connects set partitions with algebraic structure in finite fields.

Abstract

It is shown that a partition $\mathfrak A\cup \mathfrak B$ of the set $\mathbb F_{p^m}^*=\mathbb F_{p^m}-\{0 \}$, with $|\mathfrak A|=|\mathfrak B|$, is the separation into squares and non squares, if and only if the elements of $\mathfrak A$ and $\mathfrak B$ satisfy certain additive properties, thus providing a purely additive characterization of the set of squares in $\mathbb F_{p^m}$.

Equations

$$d_{p^{m}}=\left\{\begin{array}[]{ll}\displaystyle\frac{p^{m}-1}{4}~{}~{},&~{}~{}\mbox{if}~{}~{}p=1~{}\bmod 4\\ &\\ \displaystyle\frac{p^{m}-(-1)^{m}}{4}~{}~{},&~{}~{}\mbox{if}~{}~{}p=3~{}\bmod 4~{}~{},\end{array}\right.$$

$$\beta \in {\mathbb{F}}_{p^m}\iff [b_1,b_2,\dots b_m]\in {\mathbb{F}}_p^m,$$

$${\mathfrak{Q}}_{p^m}\iff q_r(\mathbf{x})=\sum _{\beta \in {\mathfrak{Q}}_{p^m}}\prod _{j=1}^m{x}_j^{b_j},{\mathfrak{N}}_{p^m}\iff q_n(\mathbf{x})=\sum _{\beta \in {\mathfrak{N}}_{p^m}}\prod _{j=1}^m{x}_j^{b_j}.$$

$$1+q_r(\mathbf{x})+q_n(\mathbf{x})=\prod _{j=1}^m\frac{x_j^p-1}{x_j-1},$$

$$\begin{array}[]{lclcl}q_{r}(\mathbf{x})^{2}&=&\sum_{\beta\in\mathfrak{Q}_{r}}A_{b_{1},\ldots,b_{m}}\prod_{j=1}^{m}x_{j}^{b_{j}}&{}{}\hfil&\bmod\langle(x_{1}^{p}-1),(x_{2}^{p}-1),\cdots,(x_{m}^{p}-1)\rangle\\ q_{n}(\mathbf{x})^{2}&=&\sum_{\beta\in\mathfrak{Q}_{n}}B_{b_{1},\ldots,b_{m}}\prod_{j=1}^{m}x_{j}^{b_{j}}&{}{}\hfil&\bmod\langle(x_{1}^{p}-1),(x_{2}^{p}-1),\cdots,(x_{m}^{p}-1)\rangle\\ \end{array}$$

$$\begin{array}[]{l}A_{b_{1},\ldots,b_{m}}=A_{a_{1},\ldots,a_{m}}+(\mathcal{N}(\alpha)|p)\\ B_{b_{1},\ldots,b_{m}}=B_{a_{1},\ldots,a_{m}}-(\mathcal{N}(\alpha)|p)\\ \end{array}$$

$$q_r(\mathbf{x})-q_n(\mathbf{x})=Q(\mathbf{x})\prod _{j=1}^m(x_j-1)$$

$$q_r(\mathbf{x})+q_n(\mathbf{x})=-1+\prod _{j=1}^m\frac{x_j^p-1}{x_j-1},$$

$$q_r(\mathbf{x}{)}^2-q_n(\mathbf{x}{)}^2=-Q(\mathbf{x})\prod _{j=1}^m(x_j-1)+Q(\mathbf{x})\prod _{j=1}^m(x_j^p-1)=-Q(\mathbf{x})\prod _{j=1}^m(x_j-1)\mathrm{mod}\prod _{j=1}^m(x_j^p-1).$$

$$q_n(\mathbf{x}{)}^2-q_r(\mathbf{x}{)}^2=q_r(\mathbf{x})-q_n(\mathbf{x})\Rightarrow q_r(\mathbf{x}{)}^2+q_r(\mathbf{x})=q_n(\mathbf{x}{)}^2+q_n(\mathbf{x})(\mathrm{mod}\prod _{j=1}^m(x_j^p-1)),$$

$$\eta =\beta \alpha =\beta x+\beta y$$

$$\frac{p^m-1}{2}{A}_1+\frac{p^m-1}{2}{A}_{-1}+A_0={\left(\frac{p^m-1}{2}\right)}^2,$$

$$A_{1}+A_{-1}=\left\{\begin{array}[]{lcl}\displaystyle\frac{p^{m}-3}{2}&{}{}\hfil&\mbox{if $p\equiv 1\pmod{4}$}\\ &&\\ \displaystyle\frac{p^{m}-2-(-1)^{m}}{2}&{}{}\hfil&\mbox{if $p\equiv 3\pmod{4}$}\\ \end{array}\right.~{}~{},$$

$$d_{p^{m}}=\left\{\begin{array}[]{lcl}\displaystyle\frac{p^{m}-1}{4}&{}{}\hfil&\mbox{if $p\equiv 1\pmod{4}$}\\ &&\\ \displaystyle\frac{p^{m}-(-1)^{m}}{4}&{}{}\hfil&\mbox{if $p\equiv 3\pmod{4}$}~{}.\\ \end{array}\right.$$

$$a(\mathbf{x})=\sum _{\beta \in \mathfrak{A}}\prod _{j=1}^m{x}_j^{b_j},b(\mathbf{x})=\sum _{\beta \in \mathfrak{B}}\prod _{j=1}^m{x}_j^{b_j}.$$

$$\begin{array}[]{lcl}a(\mathbf{x})^{2}&=&(d_{p^{m}}-1)a(\mathbf{x})+d_{p^{m}}b(\mathbf{x})+c_{p^{m}}\\ &=&d_{p^{m}}(-1+\prod_{j=1}^{m}\frac{x_{j}^{p}-1}{x_{j}-1})-a(\mathbf{x})+c_{p^{m}}\\ &=&d_{p^{m}}(-1+\prod_{j=1}^{m}(x_{j}-1)^{p-1})-a(\mathbf{x})+c_{p^{m}}~{}~{}\bmod\prod_{j=1}^{m}(x_{j}^{p}-1),\pmod{p}~{}~{}.\end{array}$$

$$a(\mathbf{x}{)}^2+a(\mathbf{x})=d_{p^m}(\prod _{j=1}^m(x_j-1{)}^{p-1}-1)+c_{p^m}(\mathrm{mod}\prod _{j=1}^m(x_j^p-1)),(\mathrm{mod}p),$$

$$b(\mathbf{x}{)}^2+b(\mathbf{x})=d_{p^m}(\prod _{j=1}^m(x_j-1{)}^{p-1}-1)+c_{p^m}(\mathrm{mod}\prod _{j=1}^m(x_j^p-1)).$$

$$a(\mathbf{x}{)}^2+⟨\prod _{j=1}^m(x_j-1{)}^{N+1}⟩=b(\mathbf{x}{)}^2+⟨\prod _{j=1}^m(x_j-1{)}^{N+1}⟩=c(\mathbf{x}{)}^2+⟨\prod _{j=1}^m(x_j-1{)}^{N+1}⟩=g(\mathbf{x}{)}^2+⟨\prod _{j=1}^m(x_j-1{)}^{N+1}⟩.$$

$$a(\mathbf{x}{)}^2+⟨\prod _{j=1}^m(x_j-1{)}^N⟩=b(\mathbf{x}{)}^2+⟨\prod _{j=1}^m(x_j-1{)}^N⟩=c(\mathbf{x}{)}^2+⟨\prod _{j=1}^m(x_j-1{)}^N⟩=g(\mathbf{x}{)}^2+⟨\prod _{j=1}^m(x_j-1{)}^N⟩,$$

$$a(\mathbf{x})+⟨\prod _{j=1}^m(x_j-1{)}^N⟩=b(\mathbf{x})+⟨\prod _{j=1}^m(x_j-1{)}^N⟩.$$

$$\begin{array}[]{lcl}b(\mathbf{x})^{2}+\langle\prod_{j=1}^{m}(x_{j}-1)^{N+1}\rangle&=&a(\mathbf{x})^{2}+\langle\prod_{j=1}^{m}(x_{j}-1)^{N+1}\rangle\\ &=&(b(\mathbf{x})+\prod_{j=1}^{m}(x_{j}-1)^{N}f(\mathbf{x}))^{2}+\langle\prod_{j=1}^{m}(x_{j}-1)^{N+1}\rangle\\ &=&b(\mathbf{x})^{2}+2b(\mathbf{x})f(\mathbf{x})\prod_{j=1}^{m}(x_{j}-1)^{N}+\\ &{}\hfil&\hskip 56.9055ptf(\mathbf{x})^{2}\prod_{j=1}^{m}(x_{j}-1)^{2N}+\langle\prod_{j=1}^{m}(x_{j}-1)^{N+1}\rangle\\ &=&b(\mathbf{x})^{2}+2b(\mathbf{x})f(\mathbf{x})\prod_{j=1}^{m}(x_{j}-1)^{N}+\langle\prod_{j=1}^{m}(x_{j}-1)^{N+1}\rangle.\\ \end{array}$$

$$a(\mathbf{x}{)}^2+a(\mathbf{x})=d_{p^m}(\prod _{j=1}^m(x_j-1{)}^{p-1}-1)+c_{p^m}(\mathrm{mod}\prod _{j=1}^m(x_j^p-1)),$$

$$c_{p^{m}}=\left\{\begin{array}[]{lcl}\frac{p^{m}-1}{2}~{},&{}{}\hfil&\\ 0~{},&{}{}\hfil&\mbox{if}~{}~{}p\equiv 3\bmod 4\bigwedge m\equiv 1\bmod 2~{}~{}.\end{array}\right.$$

$$y^2+y-d_{p^m}(\prod _{j=1}^m(x_j-1{)}^{p-1}-1)-c_{p^m}=0$$

Full text

Mathematics Subject Classification (2010): 11A15, 11N69, 11R32

Keywords: finite field, multivariate polynomial, even partition.

1 Introduction

In 1952, Oskar Perron gave some additive properties of the fibers of the quadratic character on $\mathbb{F}^{∗}_{p}$ [5]. Specifically, he showed that if $\mathfrak{A},\mathfrak{B}\subset\mathbb{F}_{p}$ are the subsets of quadratic residues and non-residues respectively, then, letting $d_{p}=\frac{p-1}{4}$ if $p\equiv 1\pmod{4}$, and $d_{p}=\frac{p+1}{4}$ if $p\equiv 3\pmod{4}$,

Every element of $\mathfrak{A}$ [respectively $\mathfrak{B}$] can be written as a sum of two elements of $\mathfrak{A}$ [respectively $\mathfrak{B}$] in exactly $d_{p}-1$ ways. 2. 2.

Every element of $\mathfrak{A}$ [respectively $\mathfrak{B}$] can be written as a sum of two elements of $\mathfrak{B}$ [respectively $\mathfrak{A}$] in exactly $d_{p}$ ways.

It was natural to inquire about just how strong this result is, and in [3] it is shown that these additive properties uniquely characterize the even partition of $\mathbb{F}^{∗}_{p}$ into quadratic residues and non-residues. Further, in [4] this result has been generalized (and the ”even” restriction removed) to fibers of arbitrary characters on $\mathbb{F}^{∗}_{p}$, with suitable cyclotomic numbers in place of the constants above. In the present paper, it is considered the generalization of the even partition (i.e. by the quadratic character $\chi_{2}$) to every finite field of odd characteristic, that is, the partition of $\mathbb{F}^{∗}_{p^{m}}$ into squares and non-squares. Specifically, the following theorem is proved.

Theorem 1**.**

Let $p$ be an odd prime, $m$ any positive integer, and set

[TABLE]

then

1.

Every square $\beta\in\mathbb{F}^{∗}_{p^{m}}$ [non-square $\eta$], i.e. $\chi_{2}(\beta)=1$ [$\chi_{2}(\eta)=-1$*], can be written as a sum of two squares [non-squares] in exactly $d_{p^{m}}-1$ ways. *

*Every square [non-square] can be written as a sum of two non-squares [squares] in exactly $d_{p^{m}}$ ways. *

Every non-zero square can be written as a sum of a square and a non-square in exactly $p^{m}-1-2d_{p^{m}}$ ways.

2.

Conversely, every partition of $\mathbb{F}^{*}_{p^{m}}$ into two sets of the same cardinality and satisfying the above properties, is necessarily constituted by the partition in squares and non-squares as defined by $\chi_{2}$.

Let $\gamma$ denote a root of an irreducible (possibly primitive) polynomial $p(x)=x^{m}+a_{m-1}x^{m-1}+\cdots+a_{0}$ over $\mathbb{F}_{p}$. The elements of $\mathbb{F}_{p^{m}}$ are represented, in the basis $\mathcal{B}=\{1,\gamma,\ldots,\gamma^{m-1}\}$, as $m$-dimensional vectors in $\mathbb{F}_{p}^{m}$, that is

[TABLE]

and every element of $\mathbb{F}_{p^{m}}$ will be interchangeably denoted with $\beta$ or $[b_{1},b_{2},\ldots b_{m}]$.

Let $\mathfrak{Q}_{p^{m}}$ and $\mathfrak{N}_{p^{m}}$ be the sets of squares and non-squares of $\mathbb{F}_{p^{m}}^{*}$, respectively; recalling that $\beta\in\mathfrak{Q}_{p^{m}}$ if and only if the field norm $\mathcal{N}(\beta)$ is a quadratic residue in $\mathbb{F}_{p}$, the quadratic character $\chi_{2}$ over $\mathbb{F}_{p^{m}}$ can be defined, for $\beta\neq 0$, as $\chi_{2}(\beta)=(\mathcal{N}(\beta)|p)$, where $(.|p)$ indicates the Legendre symbol. To prove Theorem 1, we introduce a description of $\mathfrak{Q}_{p^{m}}$ and $\mathfrak{N}_{p^{m}}$ using multivariate polynomials in $m$ variables

[TABLE]

1.1 Proof of Claim 1.

Since the fibers $\mathfrak{Q}_{p^{m}}$ and $\mathfrak{N}_{p^{m}}$ form a partition of $\mathbb{F}_{p^{m}}$, the $0\in\mathbb{F}_{p^{m}}$ standing alone, we have

[TABLE]

as well as the following proposition

Proposition 1**.**

The set $\{1,q_{r}(\mathbf{x}),q_{n}(\mathbf{x})\}$ is a basis of a subspace $\mathcal{V}_{3}$ of dimension $3$ in the $p^{m}$-dimensional vector space $\mathbb{Q}[\mathbf{x}]/\langle(x_{1}^{p}-1),(x_{2}^{p}-1),\cdots,(x_{m}^{p}-1)\rangle$ of multivariate polynomials of degree at most $p-1$ in each variable.

The representatives of $q_{r}(\mathbf{x})^{2}$ and $q_{n}(\mathbf{x})^{2}$ modulo $\langle(x_{1}^{p}-1),(x_{2}^{p}-1),\cdots,(x_{m}^{p}-1)\rangle$ in $\mathbb{Q}[x]$ are denoted by

[TABLE]

where $A_{b_{1},\ldots,b_{m}}$ and $B_{b_{1},\ldots,b_{m}}$ are non-negative integers smaller than $p^{m}$. It is observed that $A_{b_{1},\ldots,b_{m}}$ [or $B_{b_{1},\ldots,b_{m}}$] is precisely the number of ways in which every $\beta\in\mathbb{F}_{p^{m}}$ can be written as a sum of two squares [or non-squares]. The numbers $A_{b_{1},\ldots,b_{m}}$ and $B_{b_{1},\ldots,b_{m}}$ are elements of the set $\mathcal{R}=\{0,1,2,\ldots,p^{m}-1\}$.

Lemma 1**.**

Let $p$ be an odd prime, $m$ be a positive integer, and $A_{b_{1},\ldots,b_{m}},B_{b_{1},\ldots,b_{m}}$ as defined above. Then for every $\alpha,\beta\in\mathbb{F}_{p^{m}}$, the following hold:

$B_{b_{1},\ldots,b_{m}}-A_{b_{1},\ldots,b_{m}}=(\mathcal{N}(\beta)|p)$. 2. 2.

If $(\mathcal{N}(\beta)|p)=(\mathcal{N}(\alpha)|p)$, then $A_{b_{1},\ldots,b_{m}}=A_{a_{1},\ldots,a_{m}}$ and $B_{b_{1},\ldots,b_{m}}=B_{a_{1},\ldots,a_{m}}$. 3. 3.

If $(\mathcal{N}(\beta)|p)\neq(\mathcal{N}(\alpha)|p)$, then

[TABLE]

Proof.

Let $\mathbf{e}$ be the all-one $m$-dimensional vector, then $q_{r}(\mathbf{e})=q_{n}(\mathbf{e})=\frac{p^{m}-1}{2}$, thus

[TABLE]

which, multiplied by

[TABLE]

gives

[TABLE]

That is

[TABLE]

where the left equality proves item 1.

Suppose now that $\chi_{2}(\alpha)=\chi_{2}(\beta)=1$ in $\mathbb{F}_{p^{m}}$. Then there exist a square $\delta\in\mathbb{F}_{p^{m}}$ so that $\beta=\delta\alpha$. If $\chi_{2}(x)=\chi_{2}(y)=1$, with $\alpha=x+y$, it follows that $\beta=\delta x+\delta y$ and $\delta x,\delta y$ are also squares. Thus $A_{b_{1},\ldots,b_{m}}=A_{a_{1},\ldots,a_{m}}$, and with a similar argument $B_{b_{1},\ldots,b_{m}}=B_{a_{1},\ldots,a_{m}}$.

Suppose $\chi_{2}(\alpha)=1$, and that $\alpha=x+y$ is a sum of two non-squares, let $\beta$ be any non-square, then

[TABLE]

says that a non-square is the sum of two squares, it follows that $A_{\eta}=B_{\alpha}$ with $\eta$ a non-square, and $\alpha$ a square, the same equality holds by exchanging square and non-square. ∎

Let $A_{1}$ and $A_{-1}$ denote the common value of the $A_{\alpha}$ with $(\mathcal{N}(\alpha)|p)=1$ and $-1$, respectively. Similarly, define $B_{1}$ and $B_{-1}$ to be the common values of the $B_{\alpha}$ for $(\mathcal{N}(\alpha)|p)=1$ and $-1$, respectively. Further, from Lemma 1, we have $A_{1}=B_{-1}$, and $B_{1}=A_{-1}$. Let $A_{0}$ denote the number of sums of two squares giving [math], then $A_{0}=0$ if $p\equiv 3\pmod{4}$ and $m$ odd because $\chi_{2}(-1)=-1$, otherwise $A_{0}=\frac{p^{m}-1}{2}$ because $\chi_{2}(-1)=1$, i.e. $-1$ is a square. A direct counting of the number of sums of two squares gives

[TABLE]

therefore, in view of the above observations, we have

[TABLE]

furthermore $A_{1}+A_{-1}=B_{1}+B_{-1}$.

Theorem 2**.**

Let $p^{m}$ be a power of an odd prime and set

[TABLE]

Then every square [non-square] can be written as a sum of two squares [non-squares] in exactly $d_{p^{m}}-1$ ways. Every square [non-square] can be written as a sum of two non-squares in exactly $d_{p^{m}}$ ways. Moreover, every non-zero element can be written as a sum of a square and a non-square in exactly $p^{m}-1-2d_{p^{m}}$ ways.

Proof.

As above, let $A_{1}$, $B_{1}$ denote respectively the number of ways in which a square can be written as a sum of two squares or two non-squares. Let $A_{-1}$, $B_{-1}$ denote respectively the number of ways in which a non-square can be written as a sum of two squares or two non-squares. To show that $d_{p^{m}}=A_{-1}=B_{1}$ and $d_{p^{m}}-1=A_{1}=B_{-1}$, we consider the equation (1) and the equation $A_{-1}-A_{1}=1$ from Lemma 1, which is obtained noting that $A_{-1}=B_{1}$.

Solving the trivial linear system we obtain the claimed values for $A_{1}$, $A_{-1}$, and in turn the values for $B_{1}$, $B_{-1}$.

The number of ways, that every non-zero element is written as a sum of a square and a non-square, is obtained by observing that the equation $x_{1}+x_{2}=\beta\neq 0$ in $\mathbb{F}_{p^{m}}$ has $p^{m}-2$ solutions with neither $x_{1}$ nor $x_{2}$ equal [math]. Therefore, the number of solutions with $x_{1}$ a square, and $x_{2}$ a non-square, or vice-versa, is $p^{m}-2-(2d_{p^{m}}-1)$. ∎

1.2 Proof of Claim 2.

The goal of this section is to show that the additive properties given in Section 1.1 completely characterize the squares in $\mathbb{F}_{p^{m}}$. Let $d_{p^{m}}$ be defined as in Equation (2), and, for the remainder of this section, suppose $\mathfrak{A}$ and $\mathfrak{B}$ form an even partition of $\mathbb{F}_{p^{m}}-\{0\}$ such that

Every element of $\mathfrak{A}$ [$\mathfrak{B}$] can be written as a sum of two elements from $\mathfrak{A}$ [$\mathfrak{B}$] in exactly $d_{p^{m}}-1$ ways. 2. 2.

Every element of $\mathfrak{A}$ [ $\mathfrak{B}$ ] can be written as a sum of two elements from $\mathfrak{B}$ [$\mathfrak{A}$ ] in exactly $d_{p^{m}}$ ways.

Define two polynomials in $\mathbb{F}_{p^{m}}[\mathbf{x}]$,

[TABLE]

 It follows from the assumptions on the sets $\mathfrak{A}$ and $\mathfrak{B}$ that

[TABLE]

where the identity $\frac{x_{j}^{p}-1}{x_{j}-1}=(x_{j}-1)^{p-1}\bmod p$ has been used. Thus, we can write the equation

[TABLE]

where

1. $c_{p^{m}}$ is the number of ways in which zero can be written as a sum of two elements of $\mathfrak{A}$, and can be explicitly computed by evaluating (3) at $\mathbf{x}=\mathbf{e}$, obtaining $c_{p^{m}}=\frac{p^{m}-1}{2}$ if $p=1\pmod{4}$, and $c_{p^{m}}=0$ if $p=3\pmod{4}$ and $m$ is odd.

Similarly, we find

[TABLE]

To show that $\{a(\mathbf{x}),b(\mathbf{x})\}=\{q_{r}(\mathbf{x}),q_{n}(\mathbf{x})\}$, it is sufficient to show the coincidence modulo $p$, since the coefficients of every polynomial are [math] or $1$, we proceed as follows.

Lemma 2**.**

Let $p$ be an odd prime, and $\mathbb{R}_{k}=\mathbb{F}_{p^{m}}[x]/\langle{\prod_{j=1}^{m}(x_{j}-1)^{k}}\rangle$ for $k\geq 1$. Then each invertible element of $\mathbb{R}_{k}$ has at most two distinct square roots.

Proof.

We proceed by induction on $k$. The initial case is obvious since $\mathbb{R}_{1}=\mathbb{F}_{p^{m}}$. Suppose now that the result holds for all $1\leq k\leq N$. Further suppose that the polynomials $a(\mathbf{x}),b(\mathbf{x}),c(\mathbf{x}),g(\mathbf{x})\in\mathbb{F}_{p^{m}}[\mathbf{x}]$ are invertible modulo $\langle\prod_{j=1}^{m}(x_{j}-1)^{N+1}\rangle$ and

[TABLE]

By canonical projection onto $\mathbb{R}_{N}$, it follows that

[TABLE]

so that two of these must be equal by the induction hypothesis, say

[TABLE]

It follows that $a(\mathbf{x})=b(\mathbf{x})+\prod_{j=1}^{m}(x_{j}-1)^{N}f(\mathbf{x})$ for some $f(\mathbf{x})\in\mathbb{F}_{p^{m}}[x]$. Thus

[TABLE]

So $2(\prod_{j=1}^{m}(x_{j}-1)^{N})b(\mathbf{x})f(\mathbf{x})\in\langle\prod_{j=1}^{m}(x_{j}-1)^{N+1}\rangle$, but since $2b(\mathbf{x})$ is invertible modulo $\langle\prod_{j=1}^{m}(x_{j}-1)^{N}\rangle$, it follows that $\prod_{j=1}^{m}(x_{j}-1)|f(\mathbf{x})$, so that $a(\mathbf{x})+\langle\prod_{j=1}^{m}(x_{j}-1)^{N+1}\rangle=b(\mathbf{x})+\langle\prod_{j=1}^{m}(x_{j}-1)^{N+1}\rangle$. ∎

Theorem 3**.**

Let $p$ be an odd prime and let $d_{p^{m}}$ be defined as in Equation (2). Suppose $\mathfrak{A}\in\mathbb{F}_{p^{m}}^{*}$ and $\mathfrak{B}=\mathbb{F}_{p^{m}}^{*}\backslash\mathfrak{A}$. Then $\mathfrak{A}$ is precisely the set of squares of $\mathbb{F}_{p^{m}}^{*}$ if and only if

$|\mathfrak{A}|=\frac{p^{m}-1}{2}$, 2. 2.

$1\in\mathfrak{A}$, 3. 3.

Every element of $\mathfrak{A}$ can be written as a sum of two elements from $\mathfrak{A}$ in exactly $d_{p^{m}}-1$ ways. 4. 4.

Every element of $\mathfrak{B}$ can be written as a sum of two elements from $\mathfrak{A}$ in exactly $d_{p^{m}}$ ways.

Proof.

As in Equation (3), it follows from the hypotheses that

[TABLE]

where

[TABLE]

It is an immediate corollary of Lemma 2 that a quadratic equation in $\mathbb{R}_{k}[y]$ with invertible coefficients has at most two solutions (this follows from a completing-the-square argument). In particular, the equation

[TABLE]

has coefficients invertible in $\mathbb{R}_{p}=\mathbb{F}_{p^{m}}[\mathbf{x}]/\langle{\prod_{j=1}^{m}(x_{j}-1)^{p}}\rangle$ so that it has at most two distinct solutions in $\mathbb{R}_{p}$.   From the proof of Lemma 1, we have that $q_{r}(\mathbf{x})$ and $q_{n}(\mathbf{x})$ are two distinct solutions, so that $a(\mathbf{x})=q_{r}(\mathbf{x})$ or $a(\mathbf{x})=q_{n}(\mathbf{x})$. But since $1\in\mathfrak{A}$ and $\mathfrak{A},\mathfrak{B}$ are disjoint by assumption, it must be the case that   $a(\mathbf{x})=q_{r}(\mathbf{x})$.

∎