Let q=p^m and Q=p^e. For integers n and ell≥ 1, let g_n,q^(ell) be the multivariate polynomial obtained from g_n,q by composing with sigma_ell(tt X_1,…,tt X_k)=tt X_1^ell+…+tt X_k^ell. A polynomial is a permutation polynomial, or PP, over a finite field if it induces a bijection on that field. Permutation criterion. The polynomial g_n,q^(ell)∈F_Q[tt X_1,…,tt X_k] is a PP over F_Q if and only if gcd(ell,Q-1)=1 and g_n,q(tt X) is a PP over F_Q. This gives a conjectural characterization of when the multivariate polynomial has permutation behaviour over a finite field of even characteristic; the surrounding results establish several cases where the polynomial cannot be a PP, but the full equivalence remains open.

The permutation criterion for multivariate polynomial $g_{n,q}^{(\ell)}$ over finite fields

Let $q=p^m$ and $Q=p^e$ . For integers $n$ and $\ell\geq 1$ , let $g_{n,q}^{(\ell)}$ be the multivariate polynomial obtained from $g_{n,q}$ by composing with $\sigma_\ell({\tt X}_1,\dots,{\tt X}_k)={\tt X}_1^\ell+\cdots+{\tt X}_k^\ell$ . A polynomial is a permutation polynomial, or PP, over a finite field if it induces a bijection on that field. Permutation criterion. The polynomial

g_{n,q}^{(\ell)}\in\mathbb{F}_Q[{\tt X}_1,\dots,{\tt X}_k]

is a PP over $\mathbb{F}_Q$ if and only if

\gcd(\ell,Q-1)=1

and $g_{n,q}({\tt X})$ is a PP over $\mathbb{F}_Q$ . This gives a conjectural characterization of when the multivariate polynomial has permutation behaviour over a finite field of even characteristic; the surrounding results establish several cases where the polynomial cannot be a PP, but the full equivalence remains open.

References

Primary source

Neranga Fernando and Bhitali Kousik, “A further study of polynomial g_n,q over finite fields”, arXiv:2606.01037 (2026).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A 2026 paper leaves the criterion open, while an unverified reader-written argument claims a complete proof.

Fernando and Kousik formulate the criterion in their 2026 study of $g_{n,q}$ over finite fields. It asserts that the multivariate polynomial is a PP exactly when $g_{n,q}$ is a PP and $\gcd(\ell,Q-1)=1$ .

Known results

If $\gcd(\ell,Q-1)=1$ , Theorem 5.9 proves equivalence of the multivariate and univariate PP properties.
If $\gcd(\ell,Q-1)\ne1$ , Theorem 5.11 rules out local permutation behaviour.
For two variables, Proposition 5.18 proves the power-sum condition is necessary and sufficient.
Theorem 5.21 excludes further cases with $\gcd(\ell,Q-1)>1$ ; Remark 5.22 handles non-PP univariate $g_{n,q}$ .

Posted attempt

A reader-written Fourier-convolution argument claims a complete proof for every outer polynomial and every finite field, including the missing implication for $\sigma_\ell$ . The attempt has not been independently verified.

Current status (as of August 2026): the sufficient direction and several special cases are proved, but a complete proof exists only as an unverified posted attempt, so the criterion is not resolved.

Sources

ProofThis solution needs a summarySee full solutionHide full solution

The following stronger statement holds over every finite field and for every outer polynomial.

Theorem. Let $F=\mathbb F_Q$ , let $k,\ell\geq1$ , and let $f\in F[X]$ . Put

\Sigma_\ell(x_1,\ldots,x_k)=x_1^\ell+\cdots+x_k^\ell.

Then $f\circ\Sigma_\ell$ is a permutation polynomial in $k$ variables, meaning that each element of $F$ has exactly $Q^{k-1}$ preimages, if and only if $f$ permutes $F$ and

\gcd(\ell,Q-1)=1.

Proof. Suppose $f\circ\Sigma_\ell$ is balanced. Every $a\in F$ therefore occurs as a value of $f\circ\Sigma_\ell$ , and hence as a value of $f$ . Thus $f:F\to F$ is surjective and therefore bijective. Applying $f^{-1}$ to the outputs shows that $\Sigma_\ell$ itself has exactly $Q^{k-1}$ preimages above every $t\in F$ .

Define

\mu(t)=\#\{x\in F:x^\ell=t\}.

On the additive group of $F$ , the number of solutions of

x_1^\ell+\cdots+x_k^\ell=t

is the $k$ -fold convolution $\mu^{*k}(t)$ . Therefore

\mu^{*k}(t)=Q^{k-1}\qquad(t\in F).

For every nontrivial complex additive character $\chi$ of $F$ , Fourier transformation gives

\left(\sum_{t\in F}\mu(t)\chi(t)\right)^k = \sum_{t\in F}\mu^{*k}(t)\chi(t) = Q^{k-1}\sum_{t\in F}\chi(t) =0.

Hence

\sum_{t\in F}\mu(t)\chi(t)=0

for every nontrivial additive character. The trivial Fourier coefficient equals

\sum_{t\in F}\mu(t)=Q.

Fourier inversion now gives $\mu(t)=1$ for every $t\in F$ . Thus $x\mapsto x^\ell$ permutes $F$ . Because $F^\times$ is cyclic of order $Q-1$ , this is equivalent to $\gcd(\ell,Q-1)=1$ .

Conversely, assume $\gcd(\ell,Q-1)=1$ . Then $x\mapsto x^\ell$ permutes $F$ . Given $t\in F$ and any $x_1,\ldots,x_{k-1}\in F$ , there is exactly one $x_k\in F$ with

x_1^\ell+\cdots+x_k^\ell=t.

Thus $\Sigma_\ell$ is balanced, and composing with a permutation $f$ preserves balancedness.

Taking $f=g_{n,q}$ proves the conjecture and the missing implication identified in Remark 5.24.