[Paper Review] On the Tu-Zeng Permutation Trinomial of Type $(1/4,3/4)$

Xiang‐dong Hou|arXiv (Cornell University)|Jun 17, 2019

Coding theory and cryptography10 references4 citations

TL;DR

This paper establishes the necessary and sufficient conditions for a specific class of trinomials over finite fields of even characteristic to be permutation polynomials. By analyzing a quartic equation derived from the functional equation of the trinomial and applying results on rational function composition and polynomial factorization, the authors prove that the Tu-Zeng trinomial of type $(1/4,3/4)$ is a permutation polynomial over $\bF_{q^2}$ if and only if $a = b$ and the cubic $X^3 + X + a^{-1}$ has no root in $\bF_q$.

ABSTRACT

Let $q$ be a power of $2$. Recently, Tu and Zeng considered trinomials of the form $f(X)=X+aX^{(1/4)q^2(q-1)}+bX^{(3/4)q^2(q-1)}$, where $a,b\in\Bbb F_{q^2}^*$. They proved that $f$ is a permutation polynomial of $\Bbb F_{q^2}$ if $b=a^{2-q}$ and $X^3+X+a^{-1-q}$ has no root in $\Bbb F_q$. In this paper, we show that the above sufficient condition is also necessary.

Motivation & Objective

• To close a gap in the classification of permutation trinomials with Niho exponents by resolving the case $(r,s_1,s_2) = (1,1/4,3/4)$ in characteristic 2.
• To determine the exact conditions under which the trinomial $f(X) = X^4(1 + aX^{q-1} + bX^{3(q-1)})$ permutes $\bF_{q^2}$, where $a \in \bF_q^*$ and $b \in \bF_{q^2}^*$.
• To prove that the sufficient condition previously given by Tu and Zeng—namely, $b = a^{2-q}$ and $X^3 + X + a^{-1-q}$ having no root in $\bF_q$—is also necessary.
• To establish that $b$ must lie in $\bF_q^*$ when $f$ is a permutation polynomial, which is a key technical step in the proof.
• To complete the characterization by showing that under $b \in \bbF_q^*$, the polynomial $f$ is a PP if and only if $a = b$ and $X^3 + X + a^{-1}$ is irreducible over $\bF_q$.

Proposed method

• Transform the permutation condition into a functional equation: $f(x) = y$ has a unique solution $x \in \bbF_q$ for each $y \in \bbF_q$, leading to a quartic equation in $X$ over $\bbF_q(Y)$.
• Apply Leonard and Williams’ theorem on quartic factorization over finite fields of even characteristic, linking the uniqueness of roots to the irreducibility of an associated cubic.
• Use Hou and Iezzi’s theorem on composition of rational functions to deduce that a certain rational function must factor as a composition, enabling coefficient-wise comparison.
• Derive a system of polynomial equations in $a$, $b$, and auxiliary variables by equating coefficients in the compositional factorization.
• Perform symbolic computation to analyze the system, filtering solutions under constraints such as $a \in \bbF_q^*$ and $b \in \bbF_{q^2}^*$, and eliminate extraneous solutions via contradiction.
• Use trace identities and field-theoretic properties (e.g., $\operatorname{Tr}_{q/2}(k) = 1$) to construct auxiliary elements and test consistency of candidate solutions.

Experimental results

Research questions

• RQ1 Is the sufficient condition for the Tu-Zeng trinomial of type $(1/4,3/4)$ to be a permutation polynomial over $\bF_{q^2}$ also necessary?
• RQ2 Under what conditions on $a$ and $b$ does the trinomial $f(X) = X^4(1 + aX^{q-1} + bX^{3(q-1)})$ permute $\bF_{q^2}$?
• RQ3 Must the coefficient $b$ lie in the subfield $\bF_q^*$ if $f$ is a permutation polynomial over $\bF_{q^2}$?
• RQ4 What is the precise relationship between the irreducibility of $X^3 + X + a^{-1}$ over $\bF_q$ and the permutation property of $f$?
• RQ5 Can the functional equation derived from $f(x) = y$ be used to derive a compositional factorization of a rational function, and how does this constrain the parameters $a$ and $b$?

Key findings

• The sufficient condition for $f$ to be a permutation polynomial—namely, $b = a^{2-q}$ and $X^3 + X + a^{-1-q}$ having no root in $\bF_q$—is also necessary.
• The coefficient $b$ must lie in $\bF_q^*$ if $f$ is a permutation polynomial over $\bF_{q^2}$, which is a nontrivial constraint derived through symbolic analysis of the functional equation.
• When $b \in \bbF_q^*$, the trinomial $f$ is a permutation polynomial if and only if $a = b$ and the cubic $X^3 + X + a^{-1}$ is irreducible over $\bF_q$.
• The proof relies on a compositional factorization of a rational function in $\bF_q(Y)$, which is derived from the uniqueness of solutions to $f(x) = y$ for each $y \in \bbF_q$.
• The analysis of the resulting polynomial system, particularly the elimination of solutions with $1 + a^2 + a^3 = 0$, leads to a contradiction, confirming that only the stated parameter values are valid.
• The final characterization is equivalent to the condition that $X^3 + X + 1/a$ has no root in $\bF_q$, which is equivalent to the irreducibility of the cubic over $\bF_q$.