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AN EXACT ROOT COUNT IN A PROBLEM ON OPTIMAL TERNARY CYCLIC CODES

Part of: Finite fields and commutative rings (number-theoretic aspects) Theory of error-correcting codes and error-detecting codes

Published online by Cambridge University Press:  21 September 2026

FRANCISCO-JAVIER SOTO Open the ORCID record for FRANCISCO-JAVIER SOTO [Opens in a new window]
FRANCISCO-JAVIER SOTO*
Affiliation:
Escuela Técnica Superior de Ingeniería Informática, Universidad Rey Juan Carlos (URJC), 28933 Móstoles, Spain Current address: Departamento de Matemáticas, Estadística y Computación, Universidad de Cantabria , 39005 Santander, Spain
*
e-mail: fjavier.soto@unican.es
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Abstract

Let $q=3^m$ with $m$ odd and let $s\geq 1$. We determine exactly the number of roots in $\mathbb {F}_q$ of $ x^{3^s+1}-x^2+1=0. $ This equation appears in the study of open problems of Ding and Helleseth [‘Optimal ternary cyclic codes from monomials’, IEEE Trans. Inform. Theory 59(9) (2013), 5898–5904]. A Cayley transform sends the equation to the norm-one subgroup of $\mathbb {F}_{q^2}^{\times }$, where the question becomes a calculation in a cyclic group. We prove that $ \#\{x\in \mathbb {F}_q:x^{3^s+1}-x^2+1=0\} = \gcd \, (3^m+1,\,3^s+2(-1)^s)-1. $ Consequently, the corresponding no-root condition is equivalent to $ \gcd \, (3^m+1,\,3^s+2(-1)^s)=1. $ As an application, we settle in the negative a conjectural no-root statement from the cyclic-code literature, by giving infinitely many pairs $(m,s)$ satisfying the proposed hypotheses for which the equation has roots.

Keywords

finite fieldsternary cyclic codesCayley transformnorm-one subgrouproot count

MSC classification

Primary: 11T06: Polynomials
Secondary: 94B15: Cyclic codes

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Type
Research Article
Information
Bulletin of the Australian Mathematical Society , First View , pp. 1 - 7
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Creative Common License - CCCreative Common License - BY
This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence (https://creativecommons.org/licenses/by/4.0/), which permits unrestricted re-use, distribution, and reproduction in any medium, provided the original work is properly cited.
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© The Author(s), 2026. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc.

1 Introduction

Let $q=3^m$ q equals 3 Superscript m with $m\ge 1$ m greater than or equals 1 odd and let $s\ge 1$ s greater than or equals 1 . We study the equation

(1.1) $$ \begin{align} x^{3^s+1}-x^2+1=0 \end{align} $$ x Superscript 3 Super Superscript s Superscript plus 1 Baseline minus x squared plus 1 equals 0

over $\mathbb {F}_q$ double struck upper F Subscript q .

The motivation comes from the theory of optimal ternary cyclic codes. Ding and Helleseth [Reference Ding and Helleseth1] formulated several open problems concerning ternary cyclic codes with minimum distance four. In this line of work, one often has to check that certain explicit equations over finite fields have no solutions. This idea has been used in, for example, [Reference Han and Yan2, Reference Wang and Wu4–Reference Zhao, Luo and Sun7].

In particular, Zhao et al. [Reference Zhao, Luo and Sun7] reduce one part of their analysis to the solvability of (1.1) (see [Reference Zhao, Luo and Sun7, Proposition 9 and Theorem 5]). In their conclusion, they formulate the following conjectural no-root statement:

$$ \begin{align*} \text{if }m\text{ is odd},\ \gcd\, (m,3)=1 \text{ and } \gcd\, (m,s)=1, \ \text{then } (1.1) \text{ has no solution in }\mathbb{F}_{3^m}. \end{align*} $$ if m is odd comma gcd left parenthesis m comma 3 right parenthesis equals 1 and gcd left parenthesis m comma s right parenthesis equals 1 comma then left parenthesis 1.1 right parenthesis has no solution in double struck upper F Subscript 3 Sub Superscript m Subscript Baseline period

Thus, (1.1) is a concrete finite-field question arising from the cyclic-code problem.

The purpose of this note is to determine the root count exactly. We prove that the number of roots of (1.1) in $\mathbb {F}_{3^m}$ double struck upper F Subscript 3 Sub Superscript m is

$$ \begin{align*} \gcd\, (3^m+1,\,3^s+2(-1)^s)-1. \end{align*} $$ gcd left parenthesis 3 Superscript m Baseline plus 1 comma 3 Superscript s Baseline plus 2 left parenthesis negative 1 right parenthesis Superscript s Baseline right parenthesis minus 1 period

This gives a simple greatest-common-divisor test for the no-root condition. In particular, the hypotheses $\gcd \, (m,3)=1$ gcd left parenthesis m comma 3 right parenthesis equals 1 and $\gcd \, (m,s)=1$ gcd left parenthesis m comma s right parenthesis equals 1 do not decide the question. We also give an infinite family of pairs $(m,s)$ left parenthesis m comma s right parenthesis satisfying these two hypotheses for which (1.1) has many roots. In this sense, the conjectural statement quoted above is settled in the negative.

The proof uses a Cayley transform. Since $m$ m is odd, $-1$ negative 1 is not a square in $\mathbb {F}_q$ double struck upper F Subscript q . Choosing $\beta \in \mathbb {F}_{q^2}$ beta element of double struck upper F Subscript q squared with $\beta ^2=-1$ beta squared equals negative 1 , the map

$$ \begin{align*} x\longmapsto \frac{x-\beta}{x+\beta} \end{align*} $$ x long right arrow from bar StartFraction x minus beta Over x plus beta EndFraction

identifies $\mathbb {F}_q$ double struck upper F Subscript q with the norm-one subgroup of $\mathbb {F}_{q^2}^{\times }$ double struck upper F Subscript q squared Superscript times after removing one point. Under this change of variable, (1.1) becomes a power equation in a cyclic group.

The main result is as follows.

Theorem 1.1. Let $q=3^m$ q equals 3 Superscript m with $m\ge 1$ m greater than or equals 1 odd and let $s\ge 1$ s greater than or equals 1 . Denote by $N_{m,s}$ upper N Subscript m comma s the number of roots of (1.1) in $\mathbb {F}_q$ double struck upper F Subscript q . Then,

$$ \begin{align*} N_{m,s} = \gcd\, (q+1,\,3^s+2(-1)^s)-1. \end{align*} $$ upper N Subscript m comma s Baseline equals gcd left parenthesis q plus 1 comma 3 Superscript s Baseline plus 2 left parenthesis negative 1 right parenthesis Superscript s Baseline right parenthesis minus 1 period

Equivalently,

$$ \begin{align*} N_{m,s}= \begin{cases} \gcd\, (q+1,3^s-2)-1 & \text{if }s\text{ is odd},\\ \gcd\, (q+1,3^s+2)-1 & \text{if }s\text{ is even}. \end{cases} \end{align*} $$ upper N Subscript m comma s Baseline equals StartLayout Enlarged left brace 1st Row 1st Column gcd left parenthesis q plus 1 comma 3 Superscript s Baseline minus 2 right parenthesis minus 1 2nd Column if s is odd comma 2nd Row 1st Column gcd left parenthesis q plus 1 comma 3 Superscript s Baseline plus 2 right parenthesis minus 1 2nd Column if s is even period EndLayout

In particular, (1.1) has a root in $\mathbb {F}_q$ double struck upper F Subscript q if and only if $ \gcd \, (q+1,\,3^s+2(-1)^s)>1. $ gcd left parenthesis q plus 1 comma 3 Superscript s Baseline plus 2 left parenthesis negative 1 right parenthesis Superscript s Baseline right parenthesis greater than 1 period

2 The Cayley transform

Throughout this section, let $q=3^m$ q equals 3 Superscript m with $m\ge 1$ m greater than or equals 1 odd. Since $q\equiv 3 \pmod 4$ q identical to 3 left parenthesis mod 4 right parenthesis , the polynomial $T^2+1$ upper T squared plus 1 is irreducible over $\mathbb {F}_q$ double struck upper F Subscript q . Choose $\beta \in \mathbb {F}_{q^2}$ beta element of double struck upper F Subscript q squared such that

$$ \begin{align*} \beta^2=-1. \end{align*} $$ beta squared equals negative 1 period

Then, $\beta \notin \mathbb {F}_q$ beta not an element of double struck upper F Subscript q and the Frobenius automorphism satisfies

$$ \begin{align*} \beta^q=-\beta. \end{align*} $$ beta Superscript q Baseline equals negative beta period

Let

$$ \begin{align*} \mu_{q+1}=\{u\in \mathbb{F}_{q^2}^{\times}:u^{q+1}=1\} \end{align*} $$ mu Subscript q plus 1 Baseline equals StartSet u element of double struck upper F Subscript q squared Superscript times Baseline colon u Superscript q plus 1 Baseline equals 1 EndSet

be the norm-one subgroup of $\mathbb {F}_{q^2}^{\times }$ double struck upper F Subscript q squared Superscript times . This group is cyclic of order $q+1$ q plus 1 [Reference Lidl and Niederreiter3, Ch. 2].

Proposition 2.1. The map

$$ \begin{align*} \phi\colon \mathbb{F}_q\to \mu_{q+1}\setminus\{1\}, \quad \phi(x)=\frac{x-\beta}{x+\beta}, \end{align*} $$ StartLayout 1st Row phi colon double struck upper F Subscript q Baseline right arrow mu Subscript q plus 1 Baseline minus StartSet 1 EndSet comma phi left parenthesis x right parenthesis equals StartFraction x minus beta Over x plus beta EndFraction comma EndLayout

is a bijection. Its inverse is

$$ \begin{align*} \phi^{-1}(u)=\beta\,\frac{1+u}{1-u}. \end{align*} $$ phi Superscript negative 1 Baseline left parenthesis u right parenthesis equals beta StartFraction 1 plus u Over 1 minus u EndFraction period

Proof. Let $x\in \mathbb {F}_q$ x element of double struck upper F Subscript q . Since $\beta \notin \mathbb {F}_q$ beta not an element of double struck upper F Subscript q , the denominator $x+\beta $ x plus beta is nonzero. Moreover,

$$ \begin{align*} \phi(x)^q = \bigg(\frac{x-\beta}{x+\beta}\bigg)^q = \frac{x+\beta}{x-\beta} = \phi(x)^{-1}. \end{align*} $$ phi left parenthesis x right parenthesis Superscript q Baseline equals left parenthesis StartFraction x minus beta Over x plus beta EndFraction right parenthesis Superscript q Baseline equals StartFraction x plus beta Over x minus beta EndFraction equals phi left parenthesis x right parenthesis Superscript negative 1 Baseline period

Hence, $\phi (x)^{q+1}=1$ phi left parenthesis x right parenthesis Superscript q plus 1 Baseline equals 1 , so $\phi (x)\in \mu _{q+1}$ phi left parenthesis x right parenthesis element of mu Subscript q plus 1 . In addition, $\phi (x)=1$ phi left parenthesis x right parenthesis equals 1 would imply $\beta =0$ beta equals 0 , which is impossible. Therefore,

$$ \begin{align*} \phi(x)\in\mu_{q+1}\setminus\{1\}. \end{align*} $$ phi left parenthesis x right parenthesis element of mu Subscript q plus 1 Baseline minus StartSet 1 EndSet period

Conversely, let $u\in \mu _{q+1}\setminus \{1\}$ u element of mu Subscript q plus 1 minus StartSet 1 EndSet . Since $u^q=u^{-1}$ u Superscript q Baseline equals u Superscript negative 1 and $\beta ^q=-\beta $ beta Superscript q Baseline equals negative beta ,

$$ \begin{align*} \bigg(\beta\,\frac{1+u}{1-u}\bigg)^q = -\beta\,\frac{1+u^{-1}}{1-u^{-1}} = \beta\,\frac{1+u}{1-u}. \end{align*} $$ left parenthesis beta StartFraction 1 plus u Over 1 minus u EndFraction right parenthesis Superscript q Baseline equals minus beta StartFraction 1 plus u Superscript negative 1 Baseline Over 1 minus u Superscript negative 1 Baseline EndFraction equals beta StartFraction 1 plus u Over 1 minus u EndFraction period

Thus, $\phi ^{-1}(u)\in \mathbb {F}_q$ phi Superscript negative 1 Baseline left parenthesis u right parenthesis element of double struck upper F Subscript q . A direct computation gives

$$ \begin{align*} \phi(\phi^{-1}(u))=u \quad\text{and}\quad \phi^{-1}(\phi(x))=x. \end{align*} $$ StartLayout 1st Row phi left parenthesis phi Superscript negative 1 Baseline left parenthesis u right parenthesis right parenthesis equals u and phi Superscript negative 1 Baseline left parenthesis phi left parenthesis x right parenthesis right parenthesis equals x period EndLayout

This proves the proposition.

3 The root count

The next proposition is the key step.

Proposition 3.1. Let $q=3^m$ q equals 3 Superscript m with $m\ge 1$ m greater than or equals 1 odd, let $s\ge 1$ s greater than or equals 1 and let $x\in \mathbb {F}_q^\times $ x element of double struck upper F Subscript q Superscript times . Put

$$ \begin{align*} u=\phi(x)=\frac{x-\beta}{x+\beta}. \end{align*} $$ u equals phi left parenthesis x right parenthesis equals StartFraction x minus beta Over x plus beta EndFraction period

Then, (1.1) is equivalent to

$$ \begin{align*} u^{\,3^s+2(-1)^s}=1. \end{align*} $$ u Superscript 3 Super Superscript s Superscript plus 2 left parenthesis negative 1 right parenthesis Super Superscript s Baseline equals 1 period

Equivalently,

$$ \begin{align*} \begin{cases} u^{3^s}\!=u^2 & \text{if }s\text{ is odd},\\ u^{3^s}\!=u^{-2} & \text{if }s\text{ is even}. \end{cases} \end{align*} $$ StartLayout Enlarged left brace 1st Row 1st Column u Superscript 3 Super Superscript s Baseline equals u squared 2nd Column if s is odd comma 2nd Row 1st Column u Superscript 3 Super Superscript s Baseline equals u Superscript negative 2 Baseline 2nd Column if s is even period EndLayout

Proof. Since $x\neq 0$ x not equals 0 , (1.1) is equivalent to

(3.1) $$ \begin{align} x^{3^s}=x-x^{-1}. \end{align} $$ x Superscript 3 Super Superscript s Superscript Baseline equals x minus x Superscript negative 1 Baseline period

By Proposition 2.1,

$$ \begin{align*} x=\beta\,\frac{1+u}{1-u}. \end{align*} $$ x equals beta StartFraction 1 plus u Over 1 minus u EndFraction period

Since $\phi (0)=-1$ phi left parenthesis 0 right parenthesis equals negative 1 , the condition $x\neq 0$ x not equals 0 implies $u\neq -1$ u not equals negative 1 . We also have $u\neq 1$ u not equals 1 . Hence, ${1-u^2\neq 0}$ 1 minus u squared not equals 0 .

Now,

$$ \begin{align*} x^{-1} = -\beta\,\frac{1-u}{1+u} \end{align*} $$ x Superscript negative 1 Baseline equals minus beta StartFraction 1 minus u Over 1 plus u EndFraction

and, therefore,

$$ \begin{align*} x-x^{-1} = \beta\bigg(\frac{1+u}{1-u}+\frac{1-u}{1+u}\bigg) = \beta\,\frac{(1+u)^2+(1-u)^2}{1-u^2}. \end{align*} $$ x minus x Superscript negative 1 Baseline equals beta left parenthesis StartFraction 1 plus u Over 1 minus u EndFraction plus StartFraction 1 minus u Over 1 plus u EndFraction right parenthesis equals beta StartFraction left parenthesis 1 plus u right parenthesis squared plus left parenthesis 1 minus u right parenthesis squared Over 1 minus u squared EndFraction period

In characteristic $3$ 3 , we have $2=-1$ 2 equals negative 1 and so

$$ \begin{align*} (1+u)^2+(1-u)^2=2(1+u^2)=-(1+u^2). \end{align*} $$ left parenthesis 1 plus u right parenthesis squared plus left parenthesis 1 minus u right parenthesis squared equals 2 left parenthesis 1 plus u squared right parenthesis equals minus left parenthesis 1 plus u squared right parenthesis period

Thus,

(3.2) $$ \begin{align} x-x^{-1} = -\beta\,\frac{1+u^2}{1-u^2}. \end{align} $$ x minus x Superscript negative 1 Baseline equals minus beta StartFraction 1 plus u squared Over 1 minus u squared EndFraction period

Since $\beta ^3=-\beta $ beta cubed equals negative beta , iteration gives

$$ \begin{align*} \beta^{3^s}=(-1)^s\beta. \end{align*} $$ beta Superscript 3 Super Superscript s Superscript Baseline equals left parenthesis negative 1 right parenthesis Superscript s Baseline beta period

Hence,

(3.3) $$ \begin{align} x^{3^s} = \beta^{3^s}\frac{1+u^{3^s}}{1-u^{3^s}} = (-1)^s\beta\,\frac{1+u^{3^s}}{1-u^{3^s}}. \end{align} $$ x Superscript 3 Super Superscript s Superscript Baseline equals beta Superscript 3 Super Superscript s Superscript Baseline StartFraction 1 plus u Superscript 3 Super Superscript s Superscript Baseline Over 1 minus u Superscript 3 Super Superscript s Superscript Baseline EndFraction equals left parenthesis negative 1 right parenthesis Superscript s Baseline beta StartFraction 1 plus u Superscript 3 Super Superscript s Superscript Baseline Over 1 minus u Superscript 3 Super Superscript s Superscript Baseline EndFraction period

The denominator in (3.3) is nonzero. Indeed, if $u^{3^s}=1$ u Superscript 3 Super Superscript s Baseline equals 1 , then the order of $u$ u divides both $3^s$ 3 Superscript s and $q+1$ q plus 1 . Since $\gcd \, (3^s,q+1)=1$ gcd left parenthesis 3 Superscript s Baseline comma q plus 1 right parenthesis equals 1 , this forces $u=1$ u equals 1 , contrary to $u\in \mu _{q+1}\setminus \{1\}$ u element of mu Subscript q plus 1 minus StartSet 1 EndSet .

Substituting (3.2) and (3.3) into (3.1) yields

$$ \begin{align*} (-1)^s\beta\,\frac{1+u^{3^s}}{1-u^{3^s}} = -\beta\,\frac{1+u^2}{1-u^2}. \end{align*} $$ left parenthesis negative 1 right parenthesis Superscript s Baseline beta StartFraction 1 plus u Superscript 3 Super Superscript s Superscript Baseline Over 1 minus u Superscript 3 Super Superscript s Superscript Baseline EndFraction equals minus beta StartFraction 1 plus u squared Over 1 minus u squared EndFraction period

All denominators in the preceding identity are nonzero, so the following cross-multiplications are reversible.

If $s$ s is odd, the previous equation becomes

$$ \begin{align*} \frac{1+u^{3^s}}{1-u^{3^s}} = \frac{1+u^2}{1-u^2}, \end{align*} $$ StartFraction 1 plus u Superscript 3 Super Superscript s Superscript Baseline Over 1 minus u Superscript 3 Super Superscript s Superscript Baseline EndFraction equals StartFraction 1 plus u squared Over 1 minus u squared EndFraction comma

and cross-multiplication yields $u^{3^s}=u^2$ u Superscript 3 Super Superscript s Baseline equals u squared .

If $s$ s is even, it becomes

$$ \begin{align*} \frac{1+u^{3^s}}{1-u^{3^s}} = -\frac{1+u^2}{1-u^2}. \end{align*} $$ StartFraction 1 plus u Superscript 3 Super Superscript s Superscript Baseline Over 1 minus u Superscript 3 Super Superscript s Superscript Baseline EndFraction equals minus StartFraction 1 plus u squared Over 1 minus u squared EndFraction period

Cross-multiplication gives

$$ \begin{align*} (1+u^{3^s})(1-u^2)=-(1+u^2)(1-u^{3^s}). \end{align*} $$ left parenthesis 1 plus u Superscript 3 Super Superscript s Superscript Baseline right parenthesis left parenthesis 1 minus u squared right parenthesis equals minus left parenthesis 1 plus u squared right parenthesis left parenthesis 1 minus u Superscript 3 Super Superscript s Superscript Baseline right parenthesis period

Equivalently, in characteristic $3$ 3 ,

$$ \begin{align*} u^{3^s+2}=1, \end{align*} $$ u Superscript 3 Super Superscript s Superscript plus 2 Baseline equals 1 comma

or $u^{3^s}=u^{-2}$ u Superscript 3 Super Superscript s Baseline equals u Superscript negative 2 . This gives the asserted equivalence.

Proof of Theorem 1.1.

Since the constant term in (1.1) is $1$ 1 , every root of (1.1) lies in $\mathbb {F}_q^\times $ double struck upper F Subscript q Superscript times . By Propositions 3.1 and 2.1, the roots of (1.1) in $\mathbb {F}_q$ double struck upper F Subscript q are in bijection with the elements

$$ \begin{align*} u\in\mu_{q+1}\setminus\{1,-1\} \end{align*} $$ u element of mu Subscript q plus 1 minus StartSet 1 comma negative 1 EndSet

such that

$$ \begin{align*} u^{\,3^s+2(-1)^s}=1. \end{align*} $$ u Superscript 3 Super Superscript s Superscript plus 2 left parenthesis negative 1 right parenthesis Super Superscript s Baseline equals 1 period

Put

$$ \begin{align*} M=3^s+2(-1)^s. \end{align*} $$ upper M equals 3 Superscript s Baseline plus 2 left parenthesis negative 1 right parenthesis Superscript s Baseline period

Since $\mu _{q+1}$ mu Subscript q plus 1 is cyclic of order $q+1$ q plus 1 , the equation $u^M=1$ u Superscript upper M Baseline equals 1 has exactly

$$ \begin{align*} \gcd\, (q+1,M) \end{align*} $$ gcd left parenthesis q plus 1 comma upper M right parenthesis

solutions in $\mu _{q+1}$ mu Subscript q plus 1 .

The integer $M$ upper M is odd. Hence, $u=-1$ u equals negative 1 does not satisfy $u^M=1$ u Superscript upper M Baseline equals 1 . Thus, among the solutions in $\mu _{q+1}$ mu Subscript q plus 1 , the only element excluded by passing to $\mu _{q+1}\setminus \{1,-1\}$ mu Subscript q plus 1 Baseline minus StartSet 1 comma negative 1 EndSet is $u=1$ u equals 1 . It follows that

$$ \begin{align*} N_{m,s} = \gcd\, (q+1,M)-1 = \gcd\, (q+1,3^s+2(-1)^s)-1. \end{align*} $$ upper N Subscript m comma s Baseline equals gcd left parenthesis q plus 1 comma upper M right parenthesis minus 1 equals gcd left parenthesis q plus 1 comma 3 Superscript s Baseline plus 2 left parenthesis negative 1 right parenthesis Superscript s Baseline right parenthesis minus 1 period

This proves the theorem.

4 Application to the conjectural no-root condition

We now give the consequence in the form in which (1.1) is used in the cyclic-code setting. Whenever the nonexistence of roots of (1.1) over $\mathbb {F}_{3^m}$ double struck upper F Subscript 3 Sub Superscript m is required, the obstruction is completely determined by the following greatest common divisor.

Corollary 4.1. Let $m\ge 1$ m greater than or equals 1 be odd and let $s\ge 1$ s greater than or equals 1 . Then, (1.1) has no root in $\mathbb {F}_{3^m}$ double struck upper F Subscript 3 Sub Superscript m if and only if

$$ \begin{align*} \gcd\, (3^m+1,3^s+2(-1)^s)=1. \end{align*} $$ gcd left parenthesis 3 Superscript m Baseline plus 1 comma 3 Superscript s Baseline plus 2 left parenthesis negative 1 right parenthesis Superscript s Baseline right parenthesis equals 1 period

If this greatest common divisor is greater than $1$ 1 , then (1.1) has exactly

$$ \begin{align*} \gcd\, (3^m+1,3^s+2(-1)^s)-1 \end{align*} $$ gcd left parenthesis 3 Superscript m Baseline plus 1 comma 3 Superscript s Baseline plus 2 left parenthesis negative 1 right parenthesis Superscript s Baseline right parenthesis minus 1

roots in $\mathbb {F}_{3^m}$ double struck upper F Subscript 3 Sub Superscript m .

Proof. This follows at once from Theorem 1.1 with $q=3^m$ q equals 3 Superscript m .

Remark 4.2. The precise cyclic code depends on the exponent family under consideration. However, whenever the analysis reduces to the no-root condition for (1.1), Corollary 4.1 gives the exact answer. The Cayley transform therefore changes the finite-field solvability question appearing in the coding-theoretic argument into an elementary calculation in a cyclic group.

The following consequence strengthens the single counterexample obtained by taking $(m,s)=(83,11)$ left parenthesis m comma s right parenthesis equals left parenthesis 83 comma 11 right parenthesis .

Corollary 4.3. Let $s=11$ s equals 11 and put $ m=83+166t$ m equals 83 plus 166 t with $t\ge 0. $ t greater than or equals 0 period If

$$ \begin{align*} t\not\equiv 1 \pmod 3 \quad\text{and}\quad t\not\equiv 5 \pmod {11}, \end{align*} $$ t not identical to 1 left parenthesis mod 3 right parenthesis and t not identical to 5 left parenthesis mod 11 right parenthesis comma

then $m$ m is odd and

$$ \begin{align*} \gcd\, (m,3)=\gcd\, (m,11)=1. \end{align*} $$ gcd left parenthesis m comma 3 right parenthesis equals gcd left parenthesis m comma 11 right parenthesis equals 1 period

Equation (1.1) has at least $498$ 498 roots in $\mathbb {F}_{3^m}$ double struck upper F Subscript 3 Sub Superscript m . Consequently, the conjectural no-root statement from the conclusion of [Reference Zhao, Luo and Sun7] fails for infinitely many pairs $(m,s)$ left parenthesis m comma s right parenthesis satisfying its hypotheses.

Proof. A direct computation gives

$$ \begin{align*} 3^{11}\equiv 2 \pmod {499} \quad\text{and}\quad 3^{83}\equiv -1 \pmod {499}. \end{align*} $$ 3 Superscript 11 Baseline identical to 2 left parenthesis mod 499 right parenthesis and 3 Superscript 83 Baseline identical to negative 1 left parenthesis mod 499 right parenthesis period

Squaring the second congruence gives

$$ \begin{align*} 3^{166}\equiv 1 \pmod {499}. \end{align*} $$ 3 Superscript 166 Baseline identical to 1 left parenthesis mod 499 right parenthesis period

Hence, for every $m$ m of the form $m=83+166t$ m equals 83 plus 166 t ,

$$ \begin{align*} 3^m = 3^{83}(3^{166})^t \equiv -1 \pmod {499}. \end{align*} $$ 3 Superscript m Baseline equals 3 Superscript 83 Baseline left parenthesis 3 Superscript 166 Baseline right parenthesis Superscript t Baseline identical to negative 1 left parenthesis mod 499 right parenthesis period

Therefore,

$$ \begin{align*} 499\mid 3^{11}-2 \quad\text{and}\quad 499\mid 3^m+1. \end{align*} $$ 499 StartAbsoluteValue 3 Superscript 11 Baseline minus 2 and 499 EndAbsoluteValue 3 Superscript m plus 1 period

Since $s=11$ s equals 11 is odd, Theorem 1.1 gives

$$ \begin{align*} N_{m,11} = \gcd\, (3^m+1,3^{11}-2)-1 \ge 499-1 = 498. \end{align*} $$ upper N Subscript m comma 11 Baseline equals gcd left parenthesis 3 Superscript m Baseline plus 1 comma 3 Superscript 11 Baseline minus 2 right parenthesis minus 1 greater than or equals 499 minus 1 equals 498 period

It remains to check the hypotheses in the conjectural statement. Since $83$ 83 is odd and $166$ 166 is even, every integer $m=83+166t$ m equals 83 plus 166 t is odd. In addition,

$$ \begin{align*} m=83+166t\equiv 2+t \pmod 3, \end{align*} $$ m equals 83 plus 166 t identical to 2 plus t left parenthesis mod 3 right parenthesis comma

so $\gcd \, (m,3)=1$ gcd left parenthesis m comma 3 right parenthesis equals 1 precisely when $t\not \equiv 1\pmod 3$ t not identical to 1 left parenthesis mod 3 right parenthesis . Similarly,

$$ \begin{align*} m=83+166t\equiv 6+t \pmod {11}, \end{align*} $$ m equals 83 plus 166 t identical to 6 plus t left parenthesis mod 11 right parenthesis comma

so $\gcd \, (m,11)=1$ gcd left parenthesis m comma 11 right parenthesis equals 1 precisely when $t\not \equiv 5\pmod {11}$ t not identical to 5 left parenthesis mod 11 right parenthesis . For example, all integers $t=33r$ t equals 33 r with $r\ge 0$ r greater than or equals 0 satisfy both restrictions. This gives infinitely many admissible pairs and the assertion follows.

Corollary 4.4. Let $m\ge 1$ m greater than or equals 1 be odd. Then, the equation

$$ \begin{align*} x^4-x^2+1=0 \end{align*} $$ x Superscript 4 Baseline minus x squared plus 1 equals 0

has no root in $\mathbb {F}_{3^m}$ double struck upper F Subscript 3 Sub Superscript m .

Proof. Take $s=1$ s equals 1 in Theorem 1.1. Then,

$$ \begin{align*} N_{m,1} = \gcd\, (3^m+1,3-2)-1 = \gcd\, (3^m+1,1)-1 = 0. \end{align*} $$ upper N Subscript m comma 1 Baseline equals gcd left parenthesis 3 Superscript m Baseline plus 1 comma 3 minus 2 right parenthesis minus 1 equals gcd left parenthesis 3 Superscript m Baseline plus 1 comma 1 right parenthesis minus 1 equals 0 period

This gives the claim.

Remark 4.5. Theorem 1.1 shows that the root count depends only on

$$ \begin{align*} \gcd\, (3^m+1,3^s+2(-1)^s). \end{align*} $$ gcd left parenthesis 3 Superscript m Baseline plus 1 comma 3 Superscript s Baseline plus 2 left parenthesis negative 1 right parenthesis Superscript s Baseline right parenthesis period

Thus, the condition $\gcd \, (m,s)=1$ gcd left parenthesis m comma s right parenthesis equals 1 alone does not control the root count. Together with Corollary 4.3, this gives a negative answer to the conjectural no-root statement in [Reference Zhao, Luo and Sun7].

Footnotes

The author acknowledges support from the ‘PREDOCT2022-006’ program at Universidad Rey Juan Carlos. In addition, he was partially supported by the grant PID2023-151238OA-I00, financed by MICIU/AEI/10.13039/501100011033 and by EU, ERDF.

References

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