Permutation polynomials

Gary L. Mullen; Qiang Wang

doi:10.1201/b15006

Permutation polynomials

Gary L. Mullen, Qiang Wang

Mathematics

Research output: Chapter in Book/Report/Conference proceeding › Chapter

17 Scopus citations

Abstract

For odd q, f(x) = x(q+1)/2 + ax is a PP of Fq if and only if a2- 1 is a nonzero square in Fq. Moreover, the polynomial f(x) + cx is a PP of Fq for (q - 3)/2 values of c ? Fq [1939]. The polynomial xr(f(xd))(q-1)/d is a PP of Fq if (r, q - 1) = 1, d q - 1, and f(xd) has no nonzero root in Fq.

Original language	English (US)
Title of host publication	Handbook of Finite Fields
Publisher	CRC Press
Pages	215-240
Number of pages	26
ISBN (Electronic)	9781439873823
ISBN (Print)	9781439873786
DOIs	https://doi.org/10.1201/b15006
State	Published - Jan 1 2013

All Science Journal Classification (ASJC) codes

General Computer Science
General Mathematics

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10.1201/b15006

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title = "Permutation polynomials",

abstract = "For odd q, f(x) = x(q+1)/2 + ax is a PP of Fq if and only if a2- 1 is a nonzero square in Fq. Moreover, the polynomial f(x) + cx is a PP of Fq for (q - 3)/2 values of c ? Fq [1939]. The polynomial xr(f(xd))(q-1)/d is a PP of Fq if (r, q - 1) = 1, d q - 1, and f(xd) has no nonzero root in Fq.",

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N2 - For odd q, f(x) = x(q+1)/2 + ax is a PP of Fq if and only if a2- 1 is a nonzero square in Fq. Moreover, the polynomial f(x) + cx is a PP of Fq for (q - 3)/2 values of c ? Fq [1939]. The polynomial xr(f(xd))(q-1)/d is a PP of Fq if (r, q - 1) = 1, d q - 1, and f(xd) has no nonzero root in Fq.

AB - For odd q, f(x) = x(q+1)/2 + ax is a PP of Fq if and only if a2- 1 is a nonzero square in Fq. Moreover, the polynomial f(x) + cx is a PP of Fq for (q - 3)/2 values of c ? Fq [1939]. The polynomial xr(f(xd))(q-1)/d is a PP of Fq if (r, q - 1) = 1, d q - 1, and f(xd) has no nonzero root in Fq.

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PB - CRC Press

ER -

Permutation polynomials

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