Nonexistence conjecture for permutation polynomials of the form x^{q+1}+bx^q+cx+d

From papers

Let $q$ be an odd prime power and let $b,c,d\in\mathbb{F}_{q^2}$ . Define

f(x)=x^{q+1}+bx^q+cx+d.

Nonexistence conjecture. The polynomial $f(x)$ is not a permutation polynomial over $\mathbb{F}_{q^2}$ .

Computational evidence suggests this for odd $q$ , including the unresolved case $b^q\neq c$ ; the preceding results rule out the case $b^q=c$ . A proof or theoretical explanation for the general claim remains open.

Progress summary

Open

A 2026 paper settles one coefficient case and reports computer evidence for the rest, but the general conjecture remains unproved.

The conjecture asserts that no polynomial of the stated form permutes the quadratic extension field when $q$ is odd. No proposer or earlier date is identified in the retrieved source.

Known results

When $b^q=c$ , the polynomial is never a permutation polynomial over $\mathbb{F}_{q^2}$ .
For $b^q\ne c$ , computations found no permutation examples, but this does not constitute a proof.

2026 computational report

Bidushi Sharma and Dhiren Kumar Basnet formulate the general nonexistence conjecture for odd $q$ and report SageMath evidence supporting it. They explicitly state that the case $b^q\ne c$ remains unproved; no counterexample, proof, or verification was found in the retrieved sources.

Current status (as of August 2026): The case $b^q=c$ is settled negatively, while the conjecture for $b^q\ne c$ remains open with only computational evidence.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Bidushi Sharma and Dhiren Kumar Basnet, “A Weil Sum Approach to Permutation Polynomials over Quadratic Extensions of Finite Fields”, arXiv:2606.14529 (2026).

Additional references

5 papers in this index state this conjecture (2019–2026). The statement above is taken from the most recent of them; the others are arXiv:2604.25017, arXiv:2508.16043, arXiv:2006.02998, arXiv:1910.11989.

1 week ago

The conjecture holds for every prime power $q$ , not only odd prime powers.

Let

K=\mathbb F_q,\qquad E=\mathbb F_{q^2},

and choose arbitrary $b,c,d\in E$ . Write

N(x)=x^{q+1},\qquad L(x)=bx^q+cx,\qquad f(x)=N(x)+L(x)+d.

The norm $N:E\to K$ is $K$ -valued, and $L:E\to E$ is $K$ -linear.

Let $\pi:E\to E/K$ be the quotient map of $K$ -vector spaces. Because $E$ has dimension two and $E/K$ has dimension one, the map

\pi\circ L:E\longrightarrow E/K

has a nonzero kernel vector $v$ . Thus

v\ne0,\qquad L(v)\in K.

For every $x\in E$ ,

\begin{aligned} f(x+v)-f(x) &=x^qv+xv^q+v^{q+1}+L(v)\\ &=\operatorname{Tr}_{E/K}(x^qv)+N(v)+L(v). \end{aligned}

The finite-field extension $E/K$ is separable, so its trace map is a nonzero, hence surjective, $K$ -linear map $E\to K$ . Since $v\ne0$ , the map $x\mapsto x^qv$ is a $K$ -linear bijection. Consequently

x\longmapsto\operatorname{Tr}_{E/K}(x^qv)

is surjective onto $K$ .

Both $N(v)$ and $L(v)$ lie in $K$ , so choose $x\in E$ satisfying

\operatorname{Tr}_{E/K}(x^qv)=-N(v)-L(v).

Then

f(x+v)=f(x),\qquad x+v\ne x.

Therefore $f$ is never injective and hence is never a permutation polynomial of $\mathbb F_{q^2}$ . The argument covers all coefficients and all characteristics, including the previously unresolved case $q$ odd and $b^q\ne c$ .

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Shivam Patel · 1 day ago