4 Proof of the main theorem

A Carlitz type result for linearized polynomials

Bence Csajb´ok, Giuseppe Marino, Olga Polverino^∗

Abstract

For an arbitrary q-polynomial f over Fqⁿ we study the problem of finding those q-polynomials g over Fqⁿ for which the image sets of f(x)/x and g(x)/x coincide. For n ≤ 5 we provide sufficient and necessary conditions and then apply our result to study maximum scattered linear sets of PG(1, q⁵).

1 Introduction

LetFqⁿ denote the finite field of qⁿ elements where q =p^h for some prime p. For n >1 and s|n the trace and norm over Fq^s of elements ofFqⁿ are defined as Tr_qⁿ_/q^s(x) =x+x^qs+. . .+x^qn−s and N_qⁿ_/q^s(x) =x^{1+qs+...+qn−s}, respectively. Whens= 1 then we will simply write Tr(x) and N(x). Every function f: Fqⁿ → Fqⁿ can be given uniquely as a polynomial with coeffi- cients in Fqⁿ and of degree at most qⁿ−1. The function f is Fq-linear if and only if it is represented by a q-polynomial, that is,

f(x) =

n−1

X

i=0

aix^qi (1)

with coefficients in Fqⁿ. Such polynomials are also called linearized. If f is given as in (1), then its adjoint (w.r.t. the symmetric non-degenerate bilinear form defined by hx, yi= Tr(xy)) is

fˆ(x) :=

n−1

X

i=0

a^q_iⁿ⁻ⁱx^qn−i,

∗The research was supported by the Italian National Group for Algebraic and Geomet- ric Structures and their Applications (GNSAGA - INdAM). The first author is supported by the J´anos Bolyai Research Scholarship of the Hungarian Academy of Sciences. The first author acknowledges the support of OTKA Grant No. K 124950.

i.e. Tr(xf(y)) = Tr(yf(x)) for anyˆ x, y∈Fqⁿ.

The aim of this paper is to study what can be said about twoq-polynomials f and g overFqⁿ if they satisfy

Im

f(x) x

=Im g(x)

x

, (2)

where by Im(f(x)/x) we mean the image of the rational function f(x)/x, i.e. {f(x)/x:x∈F^∗qⁿ}.

For a given q-polynomialf, the equality (2) clearly holds with g(x) = f(λx)/λ for each λ∈F^∗_qⁿ. It is less obvious that (2) holds also for g(x) = f(λx)/λ, see [3, Lemma 2.6] and the first part of [8, Section 3].ˆ

When one of the functions in (2) is a monomial then the answer to the question posed above follows from McConnel’s generalization [24, Theorem 1] of a result due to Carlitz [7] (see also Bruen and Levinger [6]).

Theorem 1.1. [24, Theorem 1] Let p denote a prime,q =p^h, and 1< da divisor ofq−1. Also, let F:Fq →Fq be a function such thatF(0) = 0 and F(1) = 1. Then

(F(x)−F(y))

q−1

d = (x−y)

q−1 d

for allx, y∈Fq if and only ifF(x) =x^pj for some0≤j < hand d|p^j−1.

Indeed, when the function F of Theorem 1.1 is Fq-linear, we easily get the following corollary (see Section 2 for the proof, or [16, Corollary 1.4] for the case whenq is an odd prime).

Corollary 1.2. Let g(x) and f(x) = αx^qk, q = p^h, be q-polynomials over Fqⁿ satisfying Condition (2). Denotegcd(k, n) by t. Theng(x) =βx^qs with gcd(s, n) =tfor some β with N_qⁿ_/q^t(α) = N_qⁿ_/q^t(β).

Another case for which we know a complete answer to our problem is when f(x) = Tr(x).

Theorem 1.3 ([8, Theorem 3.7]). Let f(x) = Tr(x) and let g(x) be a q- polynomial overFqⁿ such that

Im(f(x)/x) =Im(g(x)/x).

Theng(x) = Tr(λx)/λ for some λ∈F^∗qⁿ.

Note that in Theorem 1.3 we have ˆf(x) = f(x) and the only solutions for g are g(x) = f(λx)/λ, while in Corollary 1.2 we have (up to scalars) ϕ(n) different solutions forg, whereϕ is the Euler’s totient function.

The problem posed in (2) is also related to the study of the directions determined by an additive function. Indeed, whenf is additive, then

Im(f(x)/x) =

f(x)−f(y)

x−y :x6=y, x, y ∈Fqⁿ

,

is the set of directions determined by the graph of f, i.e. by the point set G_f := {(x, f(x)) : x ∈ Fqⁿ} ⊂ AG(2, qⁿ). Hence, in this setting, the problem posed in (2) corresponds to finding the Fq-linear functions whose graph determines the same set of directions. The size of Im(f(x)/x) (for anyf, not necessarily additive) was studied extensively. Whenf isFq-linear the following result holds.

Result 1.4 ([1, 2]). Let f be aq-polynomial over Fqⁿ, with maximum field of linearity Fq. Then

qⁿ⁻¹+ 1≤ |Im(f(x)/x)| ≤ qⁿ−1 q−1 .

The classical examples which show the sharpness of these bounds are the monomial functions x^qs, with gcd(s, n) = 1, and the Tr(x) function.

However, these bounds are also achieved by other polynomials which are not ”equivalent” to these examples (see Section 2 for more details).

Two Fq-linear polynomialsf(x) and h(x) ofFqⁿ[x] areequivalent if the two graphsG_f andG_hare equivalent under the action of the group ΓL(2, qⁿ), i.e. if there exists an element ϕ ∈ ΓL(2, qⁿ) such that G_f^ϕ = G_h. In such a case, we say that f and h are equivalent (via ϕ) and we write h = fϕ. It is easy to see that in this way we defined an equivalence relation on the set of q-polynomials over Fqⁿ. If f and g are two q-polynomials such that Im(f(x)/x) = Im(g(x)/x), then Im(f_ϕ(x)/x) = Im(g_ϕ(x)/x) for any admissible ϕ∈ ΓL(2, qⁿ) (see Proposition (2.6)). This means that the problem posed in (2) can be investigated up to equivalence.

For n ≤4, the only solutions for g in Problem (2) are the trivial ones, i.e. eitherg(x) =f(λx)/xorg(x) = ˆf(λx)/x(cf. Theorem 2.8).

For the casen= 5, in Section 4, we prove the following main result.

Theorem 1.5. Let f(x) andg(x) be twoq-polynomials overFq⁵, with max- imum field of linearity Fq, such that Im(f(x)/x) = Im(g(x)/x). Then

either there exists ϕ∈ ΓL(2, q⁵) such that fϕ(x) =αx^qi and gϕ(x) = βx^qj with N(α) = N(β) for some i, j ∈ {1,2,3,4}, or there exists λ ∈ F^∗_q5 such thatg(x) =f(λx)/λor g(x) = ˆf(λx)/λ.

Finally, the relation between Im(f(x)/x) and the linear sets of rank n of the projective line PG(1, qⁿ) will be pointed out in Section 5. As an application of Theorem 1.5 we get a criterium of PΓL(2, q⁵)-equivalence for linear sets in PG(1, q⁵) and this allows us to prove that the family of (maximum scattered) linear sets of ranknof size (qⁿ−1)/(q−1) in PG(1, qⁿ) found by Sheekey in [27] contains members which are not-equivalent to the previously known linear sets of this size.

2 Background and preliminary results

Let us start this section by the following immediate corollary of Result 1.4.

Proposition 2.1. If Im(f(x)/x) = Im(g(x)/x) for two q-polynomials f andg over Fqⁿ, then their maximum fields of linearity coincide.

Proof. Let Fq^m and Fq^k be the maximum fields of linearity of f and g, respectively. Suppose to the contrary m < k. Then|Im(g(x)/x)| ≤(qⁿ− 1)/(q^k−1) < q^n−k+1+ 1 ≤q^n−m+ 1≤ |Im(f(x)/x)|, a contradiction by Corollary 1.2.

Now we are able to prove Corollary 1.2.

Proof. The maximum field of linearity of f(x) is Fq^t, thus, by Proposition 2.1,g(x) has to be aq^t-polynomial as well. Then fort >1 the result follows from the t = 1 case (after substituting q for q^t and n/t for n) and hence we can assume that f(x) and g(x) are strictly Fq-linear. By (2), we note that g(1) = αz^q₀^k−1, for some z₀ ∈ F^∗qⁿ. Let F(x) := g(x)/g(1), then F is a q-polynomial over Fqⁿ, withF(0) = 0 and F(1) = 1. Also, from (2), for each x∈F^∗qⁿ there existsz∈F^∗qⁿ such that

F(x)

x =

z z0

q^k−1

.

This means that for each x ∈ F^∗qⁿ we get N_F(x)

x

= 1. By Theorem 1.1 (applied to the q-polynomial F with d= q−1 |qⁿ−1 and using the fact that F is additive) it follows that F(x) =x^pj for some 0≤ j < nh. Then

Proposition 2.1 yields p^j =q^s with gcd(s, n) = 1. We get the first part of the statement by puttingβ=g(1). Then from the assumption (2) it is easy to deduce N(α) = N(β).

We will use the following definition.

Definition 2.2. Let f and g be two equivalent q-polynomials over Fqⁿ via the element ϕ∈ΓL(2, qⁿ) represented by the invertible matrix

a b c d

and with companion automorphismσ of Fqⁿ. Then

x g(x)

:x∈Fqⁿ

=

a b c d

x^σ f(x)^σ

:x∈Fqⁿ

. (3)

Let

K_f^ϕ(x) =ax^σ+bf(x)^σ and

H_f^ϕ(x) =cx^σ+df(x)^σ.

Proposition 2.3. Let f and g be q-polynomials over Fqⁿ such that g=fϕ

for someϕ∈ΓL(2, qⁿ). Then K_f^ϕ is invertible and g(x) =H_f^ϕ((K_f^ϕ)⁻¹(x)).

Proof. It easily follows from (3).

From (3) it is also clear that Im

f_ϕ(x) x

=

c+dz^σ

a+bz^σ:z∈Im

f(x) x

(4) and hence

|Im(f_ϕ(x)/x)|=|Im(f(x)/x)|. (5) From Equations (5) and Result 1.4 the next result easily follows.

Proposition 2.4. If two q-polynomials over Fqⁿ are equivalent, then their maximum fields of linearity coincide.

Note that|Im(g(x)/x)|=|Im(f(x)/x)|does not imply the equivalence of f and g. In fact, in the last section we will list the known examples of q-polynomials f which are not equivalent to monomials but the size of Im(f(x)/x) is maximal. To find such functions was also proposed in [16]

and, as it was observed by Sheekey, they determine certain MRD-codes [27].

Let us give the following definition.

Definition 2.5. An elementϕ∈ΓL(2, qⁿ)represented by the invertible ma- trix

a b c d

and with companion automorphismσ ofFqⁿ is said to beadmis- sible w.r.t. a given q-polynomial f over Fqⁿ if either b= 0 or −(a/b)^σ−1 ∈/ Im(f(x)/x).

The following results will be useful later in the paper.

Proposition 2.6. If Im(f(x)/x) = Im(g(x)/x) for some q-polynomials over Fqⁿ, thenIm(f_ϕ(x)/x) =Im(g_ϕ(x)/x) holds for each admissible ϕ∈ ΓL(2, qⁿ).

Proof. From Im(f(x)/x) = Im(g(x)/x) it follows that any ϕ ∈ ΓL(2, qⁿ) admissible w.r.t. f is admissible w.r.t. g as well. Hence K_f^ϕ and K_g^ϕ are both invertible and we may constructfϕ and gϕ as indicated in Proposition 2.3. The statement now follows from Equation (4).

Proposition 2.7. Let f and g be q-polynomials over Fqⁿ and take some ϕ∈ΓL(2, qⁿ) with companion automorphism σ. Then gϕ(x) =fϕ(λ^σx)/λ^σ for some λ∈F^∗qⁿ if and only if g(x) =f(λx)/λ.

Proof. First we prove the ”if” part. Since g(x) = f(λx)/λ = (ω_1/λ◦f ◦ ω_λ)(x), where ω_α denotes the scalar map x ∈ Fqⁿ 7→ αx ∈ Fqⁿ, direct computations show thatH_g^ϕ =ω_1/λσ ◦H_f^ϕ◦ω_λ and K_g^ϕ =ω_1/λσ ◦K_f^ϕ◦ω_λ. Thengϕ =ω_1/λ^σ ◦fϕ◦ωλ^σ and the first part of the statement follows. The

”only if” part follows from the ”if” part applied tog_ϕ(x) =f_ϕ(λ^σx)/λ^σ and ϕ⁻¹; and from (f_ϕ)_ϕ⁻¹ =f and (g_ϕ)_ϕ⁻¹ =g.

Next we summarize what is known about Problem (2) forn≤4.

Theorem 2.8. SupposeIm(f(x)/x) =Im(g(x)/x)for someq-polynomials over Fqⁿ, n ≤ 4, with maximum field of linearity Fq. Then there exist ϕ∈GL(2, qⁿ) and λ∈F^∗qⁿ such that the following holds.

• If n= 2 then fϕ(x) =x^q and g(x) =f(λx)/λ.

• If n= 3 then either

f_ϕ(x) = Tr(x) and g(x) =f(λx)/λ or

fϕ(x) =x^q andg(x) =f(λx)/λ or g(x) = ˆf(λx)/λ.

• If n= 4 then g(x) =f(λx)/λor g(x) = ˆf(λx)/λ.

Proof. In then= 2 casef(x) =ax+bx^q,b6= 0. Letϕbe represented by the matrix

1 0

−a/b 1/b

. Then ϕ∈ GL(2, q²) mapsf(x) to x^q. Then Propo- sition 2.6 and Corollary 1.2 give gϕ(x) = fϕ(µx)/µ and hence Proposition 2.7 gives g(x) =f(λx)/λfor some λ∈Fqⁿ. If n= 3 then according to [20, Theorem 5] and [8, Theorem 1.3] there existsϕ∈GL(2, q³) such that either fϕ(x) = Tr(x) orfϕ(x) = x^q. In the former case Proposition 2.6 and The- orem 1.3 giveg_ϕ(x) =f_ϕ(µx)/µand the assertion follows from Proposition 2.7. In the latter case Proposition 2.6 and Corollary 1.2 giveg_ϕ(x) =αx^qi where i ∈ {1,2} and N(α) = 1. If i = 1, then gϕ(x) = fϕ(µx)/µ where µ^q−1=α and the assertion follows from Proposition 2.7. Let nowi= 2 and denote by

A B

C D

the matrix ofϕ⁻¹. Also, let ∆ denote the determinant of this matrix and recall that fϕ(x) = x^q, with ϕ ∈ GL(2, q³). Then by Proposition 2.3

K_f^ϕ−1

ϕ (x) =Ax+Bx^q is invertible and its inverse is the map

ψ(x) := A^q+q2x−A^q2Bx^q+B^1+qx^q2 N(A) + N(B) . Also, by Proposition 2.3 we have

(f_ϕ)_ϕ−1(x) =Cψ(x) +Dψ(x)^q, which gives f(x) = (fϕ)_ϕ−1(x).

Using similar arguments, since N(α) = 1, direct computations show g(x) = (g_ϕ)_ϕ⁻¹(x) = (A^q+q2C+B^q+q2D)x−B^q2∆α^q2+1x^q+A^q∆αx^q2

N(A) + N(B) ,

and henceg(x) = ˆf(λx)/λfor each λ∈F^∗_q³ withλ^q−1= ∆^1−q/α^q. The casen= 4 is [8, Proposition 4.2].

Remark 2.9. Theorem 2.8 yields that there is a unique equivalence class of q-polynomials, with maximum field of linearityFq, whenn= 2. For n= 3 there are two non-equivalent classes and they correspond to the classical examples: Tr(x) and x^q. Whereas, for n = 4, from [8, Sec. 5.3] and [4, Table p. 54], there exist at least eight non-equivalent classes. The possible sizes for the sets of directions determined by these strictlyFq-linear functions areq³+ 1,q³+q²−q+ 1,q³+q²+ 1 andq³+q²+q+ 1 and each of them

is determined by at least two non-equivalent q-polynomials. Also, by [13, Theorem 3.4], iff is a q-polynomial over Fq⁴ for which the set of directions is of maximum size thenf is equivalent either tox^qor toδx^q+x^q3, for some δ∈F^∗_q4 withN(δ)6= 1 (see [22]).

3 Preliminary results about Tr(x) and the mono- mial q -polynomials over F

Letq be a power of a prime p. We will need the following results.

Proposition 3.1. Letf(x) =P4

i=0a_ix^qi andg(x) = Tr(x)beq-polynomials over Fq⁵. Then there is an elementϕ∈ΓL(2, q⁵) such thatIm(fϕ(x)/x) = Im(g(x)/x)if and only ifa1a2a3a46= 0,(a1/a2)^q=a2/a3,(a2/a3)^q =a3/a4

andN(a₁) = N(a₂).

Proof. Letϕ∈ΓL(2, q⁵) such thatIm(f_ϕ(x)/x) =Im(g(x)/x). By Propo- sition 2.4, the maximum field of linearity of f is Fq and by Theorem 1.3 there exists λ∈F^∗_q5 such that fϕ(x) = Tr(λx)/λ. This is equivalent to the existence ofa, b, c, d,ad−bc6= 0 and σ:x7→x^ph such that

y Tr(y)

:y∈Fq⁵

=

a b c d

x^σ f(x)^σ

:x∈Fq⁵

. Thencx^σ+df(x)^σ ∈Fq for each x∈Fq⁵. Letz=x^σ. Then

cz+d

4

X

i=0

a^σ_iz^qi =c^qz^q+d^q

4

X

i=0

a^σq_i z^qi+1,

for each z. As polynomials of z the left and right-hand sides of the above equation coincide moduloz^q5 −z and hence comparing coefficients yield

c+da^σ₀ =d^qa^σq₄ , da^σ₁ =c^q+d^qa^σq₀ ,

da^σ_k+1=d^qa^σq_k ,

for k = 1,2,3. If d = 0, then c = 0, a contradiction. Since d6= 0, if one of a1, a2, a3, a4 is zero, then all of them are zero and hence f is Fq⁵-linear.

This is not the case, so we havea1a2a3a4 6= 0. Then the last three equations yield

a₁ a₂

q

= a₂ a₃,

a₂ a3

q

= a₃ a4

,

and by taking the norm of both sides inda^σ₂ =d^qa^σq₁ we get N(a₁) = N(a₂).

Now assume that the conditions of the assertion hold. It follows that a3 = a^q+1₂ /a^q₁ and a4 = a^q+1₃ /a^q₂ = a^q₂^2+q+1/a^q₁^2+q. Let αi = ai/a1 for i= 0,1,2,3,4. Thenα₁ = 1, N(α₂) = 1,α₃ =α^q+1₂ and α₄ =α^1+q+q₂ ². We haveα₂ =λ^q−1 for someλ∈F^∗_q5. If

a b c d

=

1 0 1−λ^1−q4a0/a1 λ^1−q4/a1

,

then

a b c d

x f(x)

=

x

x+λ^1−q4x^q+λ^q−q4x^q2 +λ^q2−q4x^q3+λ^q3−q4x^q4

=

x Tr(xλ^q4)/λ^q4

, i.e. f_ϕ(x) = Tr(λ^q4x)/λ^q4, whereϕ is defined by the matrix

a b c d

. Proposition 3.2. Let f(x) = P4

i=0a_ix^qi, with a₁a₂a₃a₄ 6= 0. Then there is an elementϕ∈ΓL(2, q⁵) such thatIm(fϕ(x)/x) =Im(x^q/x) if and only if one of the following holds:

1. (a1/a2)^q=a2/a3, (a2/a3)^q=a3/a4 and N(a1)6= N(a₂), or 2. (a₄/a₁)^q2 =a₁/a₃,(a₁/a₂)^q2 =a₃/a₄ and N(a₁)6= N(a₃).

In both cases, if the condition on the norms does not hold, thenIm(f_ϕ(x)/x) = Im(Tr(x)/x).

Proof. We first note that the monomialsx^qi andx^q5−i are equivalent via the mapψ:=

0 1 1 0

. Hence, by Corollary 1.2, the statement holds if and only if there exista, b, c, d,ad−bc6= 0, σ:x7→x^ph and i∈ {1,2}such that

y y^qi

:y∈Fq⁵

=

a b c d

x^σ f(x)^σ

:x∈Fq⁵

. (6)

If Condition 1 holds then let αj =aj/a1 for j = 0,1,2,3,4. So α1 = 1, N(α₂)6= 1,α₃=α^q+1₂ ,α₄=α^1+q+q₂ ² and it turns out that

1 α^q₂⁴ α^1+q+q₂ ^2+q3 1

!

1 0

−α₀ 1/a₁ x f(x)

=

= 1 α^q₂⁴ α^1+q+q₂ ^2+q3 1

!

x

x^q+α2x^q2 +α3x^q3 +α4x^q4

. Hence (6) is satisfied withi= 1,σ:x7→x and

a b c d

= 1 α^q₂⁴

α^1+q+q₂ ^2+q3 1

!

1 0

−α₀ 1/a1

.

If Condition (2) holds then letα_j =a_j/a₃ forj= 0,1,2,3,4. Soα₃ = 1, N(α₁) 6= 1, α₂ = α₁^1+q+q3, α₄ = α^1+q₁ ³ and (6) is satisfied with i = 2, σ:x7→x and

a b c d

= α^1+q+q₁ ^3+q4 1 1 α^q₁²

!

1 0

−α₀ 1/a₃

. Suppose now that (6) holds and put z=x^σ. Then

(za+b

4

X

j=0

a^σ_jz^qj)^qi =cz+d

4

X

j=0

a^σ_jz^qj

for eachz∈Fq⁵ and hence, as polynomials inz, the left-hand side and right- hand side of the above equation coincide modulo z^q5 −z. The coefficients ofz,z^qi and z^qk with i∈ {1,2} and k∈ {1,2,3,4} \ {i}give

b^qia^σq_−iⁱ =c+da^σ₀, a^qi+b^qia^σq₀ ⁱ =da^σ_i,

b^qia^σq_k−iⁱ =da^σ_k,

respectively, where the indices are considered modulo 5. Note that db6= 0 since otherwise also a = c = 0 and hence ad−bc = 0. With {r, s, t} = {1,2,3,4} \ {i}, the last three equations yield:

ar−i

as−i

qⁱ

= a_r as

, as−i

at−i

qⁱ

= a_s a_t.

First assume i= 1. Then we have a1

a₂ q

= a2

a₃ and a2

a₃ q

= a3

a₄.

If N(a1) = N(a2), from Proposition 3.1 and Equation (5) it follows that

|Im(x^q/x)| = |Im(Tr(x)/x)|. Since |Im(x^q/x)| = (qⁿ−1)/(q −1) and

|Im(Tr(x)/x)|=qⁿ⁻¹+ 1, we get a contradiction.

Now assume i= 2. Then we have (a4/a1)^q2 =a1/a3 and a₁

a2

q²

= a₃ a4

. (7)

Multiplying these two equations yieldsa^q₄²⁺¹=a1a^q₂² and hence

a2 =a^1+q+q₁ ³/a^q₃^3+q. (8) By (7) this implies

a4 =a^q₁³⁺¹/a^q₃³. (9) If N(a1) = N(a3), then also N(a1) = N(a2) = N(a3) = N(a4). We show that in this caseIm(f_ϕ(x)/x) =Im(Tr(x)/x), so we must have N(a₁)6= N(a₃).

According to Proposition 3.1 it is enough to show (a1/a2)^q = a2/a3 and (a2/a3)^q=a3/a4. By (7) we have (a1/a2)^q= (a3/a4)^q4, which equalsa2/a3

if and only if (a₂/a₃)^q = a₃/a₄, i.e. a^1+q₃ = a₄a^q₂. Taking into account (8) and (9), this equality follows from N(a1) = N(a3).

4 Proof of the main theorem

In this section we prove Theorem 1.5. In order to do this, we use the following two results and the technique developed in [8].

Lemma 4.1 ([8, Lemma 3.4]). Let f and g be two linearized polynomials overFqⁿ. IfIm(f(x)/x) =Im(g(x)/x), then for each positive integerdthe following holds

X

x∈F^∗_qn

f(x) x

d

= X

x∈F^∗_qn

g(x) x

d

.

Lemma 4.2(see for example [8, Lemma 3.5]). For any prime powerq and integer dwe have P

x∈F^∗qx^d=−1 if q−1|dand P

x∈F^∗qx^d= 0 otherwise.

Proposition 4.3. Let f(x) = P4

i=0aix^qi and g(x) = P4

i=0bix^qi be two q-polynomials over Fq⁵ such that Im(f(x)/x) = Im(g(x)/x). Then the following relations hold between the coefficients of f and g:

a₀ =b₀, (10)

a1a^q₄ =b1b^q₄, (11) a₂a^q₃² =b₂b^q₃², (12) a^q+1₁ a^q₃² +a2a^q+q₄ ² =b^q+1₁ b^q₃²+b2b^q+q₄ ², (13) a1a^q+q₂ ³ +a^1+q₃ ³a^q₄=b1b^q+q₂ ³+b^1+q₃ ³b^q₄, (14) a^1+q+q₁ ²a^q₂³+a^1+q₂ a^q₃^2+q3 +a^q₁a^1+q₃ ^2+q3 +a^q₁²a₂a^q₃³a^q₄+a^1+q+q₂ ³a^q₄²+ (15)

a^q₁a^q₂³a3a^q₄² +a1a^q₂a^q₃²a^q₄³ +a^1+q₁ ²a^q+q₄ ³ +a3a^q+q₄ ^2+q3 = b^1+q+q₁ ²b^q₂³+b^1+q₂ b^q₃^2+q3 +b^q₁b^1+q₃ ^2+q3+b^q₁²b₂b^q₃³b^q₄+b^1+q+q₂ ³b^q₄²+

b^q₁b^q₂³b₃b^q₄² +b₁b^q₂b^q₃²b^q₄³ +b^1+q₁ ²b^q+q₄ ³+b₃b^q+q₄ ^2+q3,

N(a1) + N(a2) + N(a3) + N(a4) + Tr(a^q₁a^q₂^2+q3+q4a3+a^q+q₁ ³a^q₂⁴a^1+q₃ ²+ (16) a^q+q₁ ²a^q₂^3+q4a₄+a^q+q₁ ^2+q4a^q₃³a₄+a^q₂a^q₃^2+q3+q4a₄+a^q₁²a^q₃^3+q4a^1+q₄ +

a^q+q₂ ³a^q₃⁴a^1+q₄ ² +a^q₁²a^q₂⁴a^1+q+q₄ ³) =

N(b1) + N(b2) + N(b3) + N(b4) + Tr(b^q₁b^q₂^2+q3+q4b3+b^q+q₁ ³b^q₂⁴b^1+q₃ ²+ b^q+q₁ ²b^q₂^3+q4b₄+b^q+q₁ ^2+q4b^q₃³b₄+b^q₂b^q₃^2+q3+q4b₄+b^q₁²b^q₃^3+q4b^1+q₄ +

b^q+q₂ ³b^q₃⁴b^1+q₄ ² +b^q₁²b^q₂⁴b^1+q+q₄ ³).

Proof. Equations (10)–(14) follow from [8, Lemma 3.6]. To prove (15) we will use Lemma 4.1 withd=q³+q²+q+ 1. This gives us

X

1≤i,j,m,n≤4

a_ia^q_ja^q_m²a^q_n³ X

x∈F^∗_q5

x^{qi−1+qj+1−q+qm+2−q2+qn+3−q3} =

X

1≤i,j,m≤4

b_ib^q_jb^q_m²b^q_n³ X

x∈F^∗_q5

x^{qi−1+qj+1−q+qm+2−q2+qn+3−q3}.

By Lemma 4.2 we have P

x∈F^∗_q5x^{qi−1+qj+1−q+qm+2−q2+qn+3−q3} =−1 if and only if

qⁱ+q^j+1+q^m+2+qⁿ⁺³ ≡1 +q+q²+q³ (mod q⁵−1), (17)

and zero otherwise. Suppose that the former case holds. The right-hand side of (17) is smaller than the left-hand side, thus

qⁱ+q^j+1+q^m+2+qⁿ⁺³ = 1 +q+q²+q³+k(q⁵−1),

for some positive integer k. We have qⁱ+q^j+1+q^m+2+qⁿ⁺³ ≤q⁴+q⁵+ q⁶+q⁷ <1 +q+q²+q³+ (q²+q+ 2)(q⁵−1) and hencek≤q²+q+ 1. If i= 1, thenq² |1−kand hencek= 1,j=m= 1 and n= 2, or k=q²+ 1, n= 4 and either j = 2 and m= 3, or j = 4 and m = 1. If i >1, then q² dividesq+ 1−kand hencek=q+ 1, ork=q²+q+ 1. In the former case i= j = n= 2 and m = 4, or i= j = 2 and n= m = 3, or i= 3, j = 1, m= 4 and n= 2, or i= 3, j = 1 and m =n= 3, orm = 1, i= 2, j = 4 and n= 3. In the latter casei= 3 and n=m=j= 4. Then (15) follows.

To prove (16) we follow the previous approach withd=q⁴+q³+q²+q+1.

We obtain

Xa_ia^q_ja^q_m²a^q_n³a^q_r⁴ =X

b_ib^q_jb^q_m²b^q_n³b^q_r⁴,

where the summation is on the quintuples (i, j, m, n, r) with elements taken from{1,2,3,4} such thatL_i,j,m,n,r := (qⁱ−1) + (q^j+1−q) + (q^m+2−q²) + (qⁿ⁺³−q³) + (q^r+4−q⁴) is divisible byq⁵−1. Then

Li,j,m,n,r ≡Ki,j⁰,m⁰,n⁰,r⁰ (mod q⁵−1), where

K_i,j⁰_,m⁰_,n⁰_,r⁰ = (qⁱ−1) + (q^j0−q) + (q^m0 −q²) + (qⁿ⁰ −q³) + (q^r0−q⁴), such thatj⁰≡j+ 1, m⁰ ≡m+ 2, n⁰ ≡n+ 3, r⁰ ≡r+ 4 (mod 5) with

j⁰∈ {0,2,3,4}, m⁰∈ {0,1,3,4}, n⁰ ∈ {0,1,2,4}, r⁰ ∈ {0,1,2,3}.

(18) For q = 2 and q = 3 we can determine by computer those quintuples (i, j⁰, m⁰, n⁰, r⁰) for which K_i,j⁰_,m⁰_,n⁰_,r⁰ is divisible by q⁵−1 and hence (16) follows. So we may assume q >3. Then

3−q²−q³−q⁴ = (q−1) + (1−q) + (1−q²) + (1−q³) + (1−q⁴)≤ K_i,j⁰_,m⁰_,n⁰_,r⁰ ≤

(q⁴−1) + (q⁴−q) + (q⁴−q²) + (q⁴−q³) + (q³−q⁴) = 3q⁴−1−q−q², and hence L_i,j,m,n,r is divisible by q⁵−1 if and only if K_i,j⁰_,m⁰_,n⁰_,r⁰ = 0. It follows that

qⁱ+q^j0 +q^m0 +qⁿ⁰+q^r0 = 1 +q+q²+q³+q⁴. (19)

For h ∈ {0,1,2,3,4} let ch denote the number of elements in the multiset {i, j⁰, m⁰, n⁰, r⁰}which equalsh. SoP4

h=0c_hq^h= 1 +q+q²+q³+q⁴ for some 0≤c_h ≤5 with P4

h=0c_h = 5. We cannot havec₀ = 5 sinceq >1. Ifc_i = 5 for some 1 ≤ i ≤ 4 then the left hand side of (19) is not congruent to 1 moduloq, a contradiction. It follows thatc_h≤4. Thus forq >3 (19) holds if and only ifc_h = 1 forh= 0,1,2,3,4 and we have to find those quintuples (i, j⁰, m⁰, n⁰, r⁰) for which i∈ {1,2,3,4},{i, j⁰, m⁰, n⁰, r⁰}={0,1,2,3,4} and (18) are satisfied. This can be done by computer and the 44 solutions yield (16).

Proof of Theorem 1.5

Since f has maximum field of linearity Fq, we cannot havea₁ =a₂ =a₃ = a₄ = 0. If three of {a₁, a₂, a₃, a₄} are zeros, then f(x) = a₀x+a_ix^qi, for somei∈ {1,2,3,4}. Hence withϕrepresented by

1 0

−a₀/ai 1/ai

we have f_ϕ(x) = x^qi. Then Proposition 2.6 and Corollary 1.2 give g_ϕ(x) = βx^qj where N(β) = 1 and j ∈ {1,2,3,4}. Now, we distinguish three main cases according to the number of zeros among {a₁, a2, a3, a4}.

Two zeros among {a1, a2, a3, a4}

Applying Proposition 4.3 we obtaina0 =b0. The two non-zero coefficients can be chosen in six different ways, however the casesa1a26= 0 anda1a36= 0 correspond to a₃a₄ 6= 0 and a₂a₄ 6= 0, respectively, since Im(f(x)/x) = Im( ˆf(x)/x). Thus, after possibly interchangingf with ˆf, we may consider only four cases.

First let f(x) =a0x+a1x^q+a4x^q4,a1a4 6= 0 .

Applying Proposition 4.3 we obtain 0 =b₂b^q₃². Sinceb₁b₄ 6= 0, from (13) we getb₂ =b₃ = 0 and hence (16) gives

N(a₁) + N(a₄) = N(b₁) + N(b₄).

Also, from (11) we have N(a₁) N(a₄) = N(b₁) N(b₄). It follows that either N(a₁) = N(b₁) and N(a₄) = N(b₄), or N(a₁) = N(b₄) and N(a₄) = N(b₁).

In the first case b1 = a1λ^q−1 for some λ∈ F^∗_q5 and by (11) we get g(x) = f(λx)/λ. In the latter case b1 = a^q₄λ^q−1 for some λ∈ F^∗_q5 and by (11) we getg(x) = ˆf(λx)/λ.

Now consider f(x) =a₁x^q+a₃x^q3,a₁a₃ 6= 0 .

Applying Proposition 4.3 and arguing as above we have either b₂ =b₄ = 0 orb1=b3= 0. In the first case from (15) we obtain

a^q₁a^1+q₃ ^2+q3 =b^q₁b^1+q₃ ^2+q3

and together with (13) this yields N(a₁) = N(b₁) and N(a₃) = N(b₃). In this caseg(x) =f(λx)/λ for someλ∈F^∗_q5. If b1=b3 = 0, then in ˆg(x) the coefficients ofx^q2 andx^q4 are zeros thus applying the result obtained in the former case we get λˆg(x) =f(λx) and hence after substituting y=λxand taking the adjoints of both sides we obtaing(y) = ˆf(µy)/µ, where µ=λ⁻¹. The cases f(x) =a1x^q+a2x^q2 and f(x) =a2x^q2 +a3x^q3 can be han- dled in a similar way, applying Equations (11)–(16) of Proposition 4.3.

One zero among {a₁, a₂, a₃, a₄}

SinceIm(f(x)/x) =Im( ˆf(x)/x), we may assume a₁ = 0 or a₂= 0.

First supposea₁= 0. Then by (11) eitherb₁ = 0 orb₄ = 0. In the former case putting together Equations (12), (13), (14) we get N(ai) = N(bi) for i ∈ {2,3,4} and hence there exists λ∈ F^∗_q5 such that g(x) = f(λx)/λ. If a₁ = b₄ = 0, then in ˆg(x) the coefficient of x^q is zero thus applying the previous result we getg(x) = ˆf(µx)/µ, whereµ=λ⁻¹.

Now suppose a2 = 0. Then by (12) either b2 = 0 or b3 = 0. Using the same approach but applying (11), (13) and (14) we obtain g(x) =f(λx)/λ org(x) = ˆf(λx)/λ.

Case a₁a₂a₃a₄ 6= 0

We will apply (10)-(15) of Proposition 4.3. Note that Equations (11) and (12) yielda₁a₂a₃a₄ 6= 0⇔b₁b₂b₃b₄ 6= 0. Multiplying (13) bya₂ and apply- ing (12) yield

a²₂a^q+q₄ ²−a2(b^q+1₁ b^q₃² +b2b^q+q₄ ²) +a^q+1₁ b^q₃²b2 = 0.

Taking (11) into account, this is equivalent to

(a₂a^q+q₄ ²−b^q+1₁ b^q₃²)(a₂a^q+q₄ ² −b₂b^q+q₄ ²) = 0.

Multiplying (14) bya1 and applying (11) yield

a²₁a^q+q₂ ³−a1(b1b^q+q₂ ³+b^1+q₃ ³b^q₄) +a^1+q₃ ³b^q₄b1= 0.

Taking (12) into account, this is equivalent to

(a1a^q+q₂ ³−b1b^q+q₂ ³)(a1a^q+q₂ ³ −b^1+q₃ ³b^q₄) = 0.

We distinguish four cases:

Case 1. a₂a^q+q₄ ² =b^q+1₁ b^q₃² anda₁a^q+q₂ ³ =b₁b^q+q₂ ³, Case 2. a₂a^q+q₄ ² =b^q+1₁ b^q₃² anda₁a^q+q₂ ³ =b^1+q₃ ³b^q₄, Case 3. a2a^q+q₄ ² =b2b^q+q₄ ² anda1a^q+q₂ ³ =b1b^q+q₂ ³, Case 4. a2a^q+q₄ ² =b2b^q+q₄ ² anda1a^q+q₂ ³ =b^1+q₃ ³b^q₄.

We show that these four cases produce the relations:

N b₁

a4

= a₁a^q+q₂ ³ a^q₄a^q₃³⁺¹

= b₁b^q+q₂ ³ b^q₄b^q₃³⁺¹

, (20)

N b₁

a₄

= 1, (21)

N b₁

a₁

= 1, (22)

N b₁

a1

= a^q₃³⁺¹a^q₄ a1a^q+q₂ ³

= b₁b^q+q₂ ³ b^q₃³⁺¹b^q₄

, (23)

respectively.

To see (20) observe that from a2a^q+q₄ ² =b^q+1₁ b^q₃² and (11) we get N

b₁ a₄

= b^q+1₁ a^q+q₄ ²

!q²+1

b^q₁⁴

a₄ = a^q₂²⁺¹ b^q₃^2+q4

!a^q₁⁴

b₄ = a₁a^q+q₂ ³ b^q₄b^q₃³⁺¹

, (24) where the last equation follows from N(b₁/a₄)^q = N(b₁/a₄). Hence by a1a^q+q₂ ³ =b1b^q+q₂ ³ and (14) we get (20).

Equation (21) immediately follows from (24) taking a₁a^q+q₂ ³ = b^1+q₃ ³b^q₄ into account.