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

Abstract

Let

with

m $m$ m

odd and let

s ≥ 1 $s\geq 1$ s greater than or equals 1

. We determine exactly the number of roots in

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

of

x 3 s + 1 − x 2 + 1 = 0. $ x^{3^s+1}-x^2+1=0. $ x Superscript 3 Super Superscript s Superscript plus 1 Baseline minus x squared plus 1 equals 0 period

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

F q 2 × $\mathbb {F}_{q^2}^{\times }$ double struck upper F Subscript q squared Superscript times

, where the question becomes a calculation in a cyclic group. We prove that

# { x ∈ F q : x 3 s + 1 − x 2 + 1 = 0 } = gcd ( 3 m + 1 , 3 s + 2 ( − 1 ) s ) − 1. $ \#\{x\in \mathbb {F}_q:x^{3^s+1}-x^2+1=0\} = \gcd \, (3^m+1,\,3^s+2(-1)^s)-1. $ number sign StartSet x element of double struck upper F Subscript q Baseline colon x Superscript 3 Super Superscript s Superscript plus 1 Baseline minus x squared plus 1 equals 0 EndSet equals 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

Consequently, the corresponding no-root condition is equivalent to

gcd ( 3 m + 1 , 3 s + 2 ( − 1 ) s ) = 1. $ \gcd \, (3^m+1,\,3^s+2(-1)^s)=1. $ 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

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 ) $(m,s)$ left parenthesis m comma s right parenthesis

satisfying the proposed hypotheses for which the equation has roots.