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

About 7 years old · traced to 2019

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.

References

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.

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A 2026 paper proves one coefficient case, while an unverified posted argument claims a complete proof in every characteristic and coefficient case.

The conjecture asserts that no polynomial of the form $x^{q+1}+bx^q+cx+d$ permutes $\mathbb{F}_{q^2}$ for odd $q$ . Sharma and Basnet formulate the unresolved case as Conjecture $3.10$ .

Known results

Sharma and Basnet (2026) prove nonexistence when $b^q=c$ , for both even and odd $q$ .
For odd $q$ with $b^q\ne c$ , their SageMath computations find no examples but do not prove nonexistence.

Posted attempt

A posted argument claims a complete proof for every prime power $q$ and all $b,c,d\in\mathbb{F}_{q^2}$ : it selects a nonzero $v$ with $bv^q+cv\in\mathbb{F}_q$ , then uses surjectivity of a trace expression to produce distinct $x$ and $x+v$ with equal images. The argument has not been independently verified.

Current status (as of August 2026): The case $b^q=c$ is proved impossible, the case $b^q\ne c$ remains formally open, and a complete posted proof claim is unverified.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

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 $b^q\ne c$ .