DeepAI

# Solving X^q+1+X+a=0 over Finite Fields

Solving the equation P_a(X):=X^q+1+X+a=0 over finite field Q, where Q=p^n, q=p^k and p is a prime, arises in many different contexts including finite geometry, the inverse Galois problem <cit.>, the construction of difference sets with Singer parameters <cit.>, determining cross-correlation between m-sequences <cit.> and to construct error-correcting codes <cit.>, as well as to speed up the index calculus method for computing discrete logarithms on finite fields <cit.> and on algebraic curves <cit.>. Subsequently, in <cit.>, the Q-zeros of P_a(X) have been studied: in <cit.> it was shown that the possible values of the number of the zeros that P_a(X) has in Q is 0, 1, 2 or p^(n, k)+1. Some criteria for the number of the Q-zeros of P_a(x) were found in <cit.>. However, while the ultimate goal is to identify all the Q-zeros, even in the case p=2, it was solved only under the condition (n, k)=1<cit.>. We discuss this equation without any restriction on p and (n,k). New criteria for the number of the Q-zeros of P_a(x) are proved. For the cases of one or two Q-zeros, we provide explicit expressions for these rational zeros in terms of a. For the case of p^(n, k)+1 rational zeros, we provide a parametrization of such a's and express the p^(n, k)+1 rational zeros by using that parametrization.

• 11 publications
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01/04/2021

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## 1 Introduction

Let and be any positive integers with . Let and where is a prime. We consider the polynomial

 Pa(X):=Xq+1+X+a,a∈F∗Q.

Notice the more general polynomial forms with and can be transformed into this form by the substitution . It is clear that have no multiple roots.

These polynomials have arisen in several different contexts including finite geometry, the inverse Galois problem [1], the construction of difference sets with Singer parameters [8], determining cross-correlation between -sequences [11, 14] and to construct error-correcting codes [4]. These polynomials are also exploited to speed up (the relation generation phase in) the index calculus method for computation of discrete logarithms on finite fields [12, 13] and on algebraic curves [21].

Let denote the number of zeros in of polynomial and denote the number of such that has exactly zeros in . In 2004, Bluher [2] proved that takes either of 0, 1, 2 and where and computed for every . She also stated some criteria for the number of the -zeros of .

The ultimate goal in this direction of research is to identify all the -zeros of . Subsequently, there were much efforts for this goal, specifically for a particular instance of the problem over binary fields i.e. . In 2008 and 2010, Helleseth and Kholosha [16, 17] found new criteria for the number of -zeros of . In the cases when there is a unique zero or exactly two zeros and

is odd, they provided explicit expressions of these zeros as polynomials of

[17]. In 2014, Bracken, Tan and Tan [5] presented a criterion for in when and is even. Very recently, Kim and Mesnager [15] completely solved this equation over when . They showed that the problem of finding zeros in of in fact can be divided into two problems with odd : to find the unique preimage of an element in under a MCM polynomial and to find preimages of an element in under a Dickson polynomial. By completely solving these two independent problems, they explicitly calculated all possible zeros in of , with new criteria for which is equal to , or as a by-product.

Very recently, new criteria for which has , , or roots were stated by [22] for any characteristic.

We discuss the equation , without any restriction on and . After defining a sequence of polynomials and considering its properties in Section 2, it is shown in Section 3 that if then there exists a quadratic equation that the rational zeros must satisfy. In Section 4, we state some useful properties of the polynomials which appear as the coefficients of that quadratic equation. In Section 5, new criteria for the number of the -zeros of are proved. For the cases of one or two -zeros, we provide explicit expressions for these rational zeros in terms of . We also provide a parametrization of the ’s for which has rational zeros. Based that parametrization, all the rational zeros are also expressed. For the case of rational zeros, some results to explicitly express these rational zeros in terms of are further presented in Section 6. Finally, we conclude in Section 7.

## 2 Preliminaries

Given positive integers and , define a polynomial

 Tklk(X):=X+Xpk+⋯+Xpk(l−2)+Xpk(l−1).

Usually we will abbreviate as . For , is the absolute trace of . The zeros of this polynomial are studied in [KCLGM2019]. In particular, we need the following.

###### Proposition 1

For any positive integers and ,

 {x∈¯¯¯¯¯¯Fp∣Tkrk(x)=0}={u−upk∣u∈Fpkr}.
###### Proof

Evidently, . The linear mapping has the kernel and so . On the other hand, can not have a kernel of greater cardinality than its degree .∎

Define the sequence of polynomials as follows:

 A1(X)=1,A2(X)=−1, (1) Ar+2(X)=−Ar+1(X)q−XqAr(X)q2 for r≥1.

The following lemma gives another identity which can be used as an alternative definition of and an interesting property of this polynomial sequence which will be importantly applied afterwards.

###### Lemma 1

For any , the following are true.

1.  Ar+2(X)=−Ar+1(X)−XqrAr(X). (2)
2.  Ar+1(X)q+1−Ar(X)qAr+2(X)=Xq(qr−1)q−1. (3)
###### Proof

We will prove these identities by induction on . It is easy to check that they hold for . Suppose that they hold for all indices less than . Then, we have

 Ar+3(X) =−Ar+2(X)q−XqAr+1(X)q2 =(Ar+1(X)+XqrAr(X))q+Xq(Ar(X)+Xqr−1Ar−1(X))q2 =(Aqr+1(X)+XqAq2r(X))+Xqr+1(Aqr(X)+XqAq2r−1(X)) =−Ar+2(X)−Xqr+1Ar+1(X),

which proves (2) for all . Also, using the proved equality (2), we have

 Ar+2(X)q+1−Ar+1(X)qAr+3(X) =Ar+2(X)q+1+Ar+1(X)q(Ar+2(X)+Xqr+1Ar+1(X)) =Xqr+1(Ar+1(X)q+1−Ar(X)qAr+2(X))+Ar+2(X)(Ar+2(X)q+Ar+1(X)q+Xqr+1Ar(X)q) Xqr+1(Ar+1(X)q+1−Ar(X)qAr+2(X)) =Xqr+1Xq(qr−1)q−1=Xq(qr+1−1)q−1,

which proves (3) for all .∎

The zero set of can be completely determined for all :

###### Proposition 2

For any ,

 {x∈¯¯¯¯¯¯Fp∣Ar(x)=0}={(u−uq)q2+1(u−uq2)q+1,  u∈Fqr∖Fq2}.

Proof. Given any , there exists at least one element such that and . Then, for any , we have

 Ar(x)=(−1)r+1∑rj=1vqjvq+vq2r−1∏j=2(vv+vq)qj,

where for it is assumed that the product over the empty set is equal to 1. Indeed, this can be proved by induction on as follows. For and , we have

 A2(x)=−1=(−1)3∑2j=1vqjvq+vq2

and

 A3(x)=1−xq=1−vq+q3(v+vq)q+q2=(−1)4∑3j=1vqjvq+vq2(vv+vq)q2.

Assuming this identity holds for all indices less than , we have

 Ar(x) −Ar−1(x)−xqr−2Ar−2(x) =(−1)r+1∑r−1j=1vqjvq+vq2r−2∏j=2(vv+vq)qj−(−1)r+1vqr∑r−2j=1vqj(v+vq)qr−1+qr−2∏j=2(vv+vq)qj =(−1)r+1(v+vq)qr−1∑r−1j=1vqj−vqr∑r−2j=1vqjvqr−1(v+vq)qr−1∏j=2(vv+vq)qj =(−1)r+1∑rj=1vqjvq+vq2r−1∏j=2(vv+vq)qj.

Thus if and only if and , which by Proposition 1 is equivalent to for some .

Therefore, if and only if for some . ∎

## 3 Quadratic equation satisfied by rational zeros of Pa(X)

Letting , define polynomials

 F(X):=Am(X), G(X):=−Am+1(X)−XAqm−1(X).

We will show that if then the -zeros of satisfy a quadratic equation and therefore necessarily .

###### Lemma 2

Let . If for then

 F(a)x2+G(a)x+aFq(a)=0. (4)
###### Proof

If for , then and thus we get

 xq=−x−ax. (5)

Now, we prove that for any

 xqr(Ar(a)x−aAr−1(a)q)−Ar+1(a)x+aAr(a)q=0 (6)

with the assumption . In fact, if then the left side of (6) is and so it holds for . Suppose that it holds for . Raising th power to both sides of (6) and substituting (5), we have

This shows that (6) holds for and so for all .

Taking in (6) and using the fact that when , we obtain the result of the lemma.∎

## 4 Some equalities involving F and G

To determine the rational zeros of when , we will need the following properties of the polynomials and which appear as coefficients of the quadratic equation (4).

###### Proposition 3

For any , the following are true.

1.  (G(x)−2F(x))q=−G(x). (7)
2.  G(x)2−4xF(x)q+1∈Fq. (8)
3.  G(x)=−xqFq2(x)+Fq(x)+xF(x). (9)
###### Proof

The first item follows from

 (G(x)−2F(x))q =G(x)q−2F(x)q=−Am+1(x)q−xqAm−1(x)q2−2Am(x)q (Am(x)+xqm−1Am−1(x))q−xqAm−1(x)q2−2Am(x)q =xAm−1(x)q−xqAm−1(x)q2−Am(x)q (% since xqm=x) xAm−1(x)q+Am+1(x)=−G(x).

The second item is proved as follows. Let . Then

 Eq−E=(Am+1(x)q+xqAm−1(x)q2)2 −4xqAm(x)q(q+1) −(Am+1(x)+xAm−1(x)q)2+4xAm(x)q+1.

Consider . By substituting this and using (1), we have

 Eq−E =(−Am(x)q−xAm−1(x)q+xqAm−1(x)q2)2−4xqAm(x)q(q+1) −(−Am(x)q−xqAm−1(x)q2+xAm−1(x)q)2+4xAm(x)q+1 =4Am(x)q(xAm−1(x)q−xqAm−1(x)q2)−4xqAm(x)q(q+1)+4xAm(x)q+1 =4Am(x)q(xAm−1(x)q−xqAm−1(x)q2−xqAm(x)q2+xAm(x)).

By the way, since

 xqAm−1(x)q2+xqAm(x)q2 =xq(Am−1(x)+Am(x))q2 −xq(xqm−2Am−2(x))q2=−xq+1Am−2(x)q2,

we get that is,

Finally, the third item is verified as follows:

 G(x) =−Am+1(x)−xAm−1(x)qAm(x)q+xqAm−1(x)q2−xAm−1(x)q Am(x)q+xqAm−1(x)q2+x(xqAm−2(x)q2+Am(x)) =xq(Am−1(x)+xqm−2Am−2(x))q2+Am(x)q+xAm(x) =−xqAm(x)q2+Am(x)q+xAm(x).

For even, we will further need the following proposition.

###### Proposition 4

Let . Let with . Let and . The followings hold.

1.  Trn1(E)=mH. (10)
2.  Tk(E)=G(a)+F(a)qG(a)+kdH. (11)
###### Proof

Regarding the fact that as , we have

 E =aF(a)q+1G(a)2aAm−1(a)qAm+1+Nrnd(a)G(a)2=(Am+1(a)+G(a))Am+1+Nrnd(a)G(a)2 =Am+1(a)G(a)+(Am+1(a)G(a))2+Nrnd(a)G(a)2.

Hence, immediately (10) follows from the facts and (which follows from (13) as ). And also

 Tk(E) =Am+1(a)G(a)+(Am+1(a)G(a))q+kdHAm+1(a)+Am+1(a)qG(a)+kdH =G(a)+aAm−1(a)q+Am+1(a)qG(a)+kdH G(a)+aAm−1(a)q+(Am(a)+aqm−1Am−1(a))qG(a)+kdH =G(a)+F(a)qG(a)+kdH.

## 5 Rational zeros of Pa(X)

By exploiting the results of previous sections, now we can completely solve the equation in arbitrary finite fields.

### 5.1 Na=pd+1

###### Lemma 3

Let . The following are equivalent.

1. i.e. has exactly zeros in .

2. , or equivalently by Proposition 2, there exists such that .

3. There exists such that . Then the zeros in of are and for .

###### Proof

(Item 1 Item 2)

We already showed that if , then i.e. .

If i.e. there exists such that , then the set given by

 ⋃α∈Fq{−(u+α)q2−q1+(u−uq)q−1}⋃{−11+(u−uq)q−1}

is the set of all zeros of . In fact, the cardinality of this set is exactly as is not in . Also we have

 Pa(−11+(u−uq)q−1) =−11+(u−uq)q−1(1−11+(u−uq)q−1)q+(u−uq)q2+1(u−uq2)q+1 =−(u−uq)u−uq2((u−uq)qu−uq2)q+(u−uq)q2+1(u−uq2)q+1=0

and

 =−(u−uq)(u−uq2)q+1(u+α)q((u−uq2)(u+α)q−(u−uq)(u+α)q2)q+(u−uq)q2+1(u−uq2)q+1 =−(u−uq)(u−uq2)q+1(u+α)q((u−uq2)(uq+α)−(u−uq)(uq2+α))q+(u−uq)q2+1(u−uq2)q+1 =−(u−uq)(u−uq2)q+1(u+α)q((u−uq)q(u+α))q+(u−uq)q2+1(u−uq2)q+1=0.

Thus splits in . Corollary 7.2 of [2] states that splits in if and only if has exactly zeros in .

(Item 1 Item 3)

To begin with, define , , and .

Now, we will show that the mapping

 Ψ:u∈S0⟼(u−uq)q2+1(u−uq2)q+1∈S,

which is well-defined by Proposition 2 and the equivalence between Item 1 and Item 2, is surjective.

Regarding , we can write where , ,