# A Note on the Primitive Roots and the Golomb Conjecture

Authors: Yiwei Hou; Hongyan Wang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we use the elementary methods and the estimates for character sums to prove the following conclusion. Letp be a prime large enough. Then, for any positive integer n with p1/2+ɛ≤n<p, there must exist two primitive roots α and β modulo p with 1<α,β≤n−1 such that the equation n=α+β holds, where 0<ɛ<1/2 is a fixed positive number. In other words, n can be expressed as the exact sum of two primitive roots modulo p.


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## Body

## 1. Introduction

Letp be a prime and Fq be a finite field of q (=ph,h≥1) elements with characteristic p. The Golomb conjecture (see [1]) can be summarized as follows: for any nonzero element c∈Fq, there exist two primitive elements α and β∈Fq such that the equation α+β=c holds.If we assume thatAp denotes the set of all primitive roots g modulo p with 1≤g≤p−1, then the Golomb conjecture in a reduced residue system modulo p can be described as that, for any integer 1≤n≤p−1, there exist two primitive roots α and β∈Ap such that the congruence α+β≡nmodp holds.This conjecture is not only basically solved but also carried on various generalizations. Interested readers can refer to the references [2–11]. For example, let p be an odd prime large enough. Then, for any integers a, b, and c with abc,p=1, there are at least two primitive roots α and βmodp such that the congruence aα+bβ≡cmodp holds (see Sun [2]).It is clear that if integer1<n<p and the primitive roots 1<α,β≤p−1 satisfy the congruence α+β≡nmodp, then α+β=n or α+β=n+p.A natural question is whether for a fixed1<n<p, there are two primitive roots 1<α,β≤n−1 of p such that(1)α+β=n?.Of course, for some positive integersn, equation (1) has no solutions. For example, n=1, n=2 and 3. So, we think that the problem in (1) is meaningful, and it is also closely related to the minimum primitive root modulo p.On the other hand, we also want to know how largen is (relative to p), so that equation (1) must have a solution.For the sake of convenience, for any odd primep and integer 1<n≤p−1, let Sn;p denote the number of all solutions of the equation α+β=n, where α and β are two primitive roots modulo p with 1<α,β≤n−1.In this paper, we shall use the elementary methods and the estimates for character sums to study the asymptotic properties ofSn;p and prove the following.Theorem 1.
Letp be an odd prime. Then, for any integer 1<n≤p−1, we have the asymptotic formula:(2)Sn;p=ϕ2p−1p2⋅n+Oϕ2p−1p3/2⋅4ωp−1⋅lnp,where, as usual, ϕk denotes the Euler function and ωk denotes the number of all distinct prime divisors of k.It is clear that, for any positive number0<ɛ<1/2, if prime p is large enough, then our theorem is nontrivial for all integers p1/2+ε≤n≤p−1. That is, the main term is much big than the error term in our theorem. So, from our theorem, we may immediately deduce the following:Corollary 1.
Letp be an odd prime large enough, 0<ɛ<1/2 be a fixed positive number. Then, for any positive integer n with p1/2+ε≤n≤p−1, there must exist two primitive roots α and β modulo p such that(3)α+β=n.Note: first, the conclusion in our theorem can also be generalized. That is, letp be an odd prime and k be a fixed positive integer. For any integers 1<n1<n2<n3<⋯<nk<p, if Sn1,n2,…,nk;p denotes the number of all solutions of the equations n1=α+β1, n2=α+β2,…,nk=α+βk, where α and all βi (i=1,2,…,k) are the primitive roots modulo p, then we have the following asymptotic formula:(4)Sn1,n2,…,nk;p=ϕk+1p−1pk+1⋅n1+Oϕk+1p−1pk+1/2⋅2k+1ωp−1⋅lnp.If0<ɛ<1/2 and p1/2+ɛ<n1<p, then this asymptotic formula is nontrivial.Second, the lower boundp1/2+ɛ of n in our corollary is very rough. How to improve the constant 1/2 is an interesting open problem.Conjecture 1.
Let0<δ<1 be a fixed positive number and p be a prime large enough. Then, for any positive integer pδ<n<p, there must exist two primitive roots α and β modulo p such that the equation α+β=n holds.


## 2. Several Lemmas

In order to complete the proof of the main result, we need several simple lemmas. For the sake of simplicity, we do not repeat some elementary number theory and analytic number theory results, which can be found in references [12–14]. First, we have the following.Lemma 1.
Letp be an odd prime. Then, for any integer a with a,p=1, we have the identity(5)ϕp−1p−1∑k|p−1μkϕk∑r=1k′er⋅indak=1,ifais a primitive rootmodp,0,ifais not a primitive rootmodp,where ey=e2πiy, ∑r=1k′ denotes the summation over all integers 1≤r≤k such that r is coprime to k, μn is the Möbius function, and inda denotes the index of a relative to some fixed primitive root gmodp.Proof.
See Proposition 2.2 in [13].Lemma 2.
Letp be an odd prime and χ1,…,χr be Dirichlet characters modulo p, at least one of which is nonprincipal character. Let fx be an integral coefficient polynomial of degree d. Then, for pairwise distinct integers a1,…,ar, we have the estimate(6)∑a=1p−1χ1a+a1χ2a+a2…χra+arefap≤r+d⋅p1/2.Proof.
In fact, this result is Lemma 17 in [15]. Some related works can also be found in [16–19].Lemma 3.
Letp be an odd prime. Then, for any integer 1<n<p and any two Dirichlet characters χ1 and χ2 (at least one of which is nonprincipal character) modulo p, we have the estimate(7)∑a=1n−1χ1aχ2n−a≪p1/2⋅lnp.Proof.
It is clear that, for any integerm, we have the trigonometric identity(8)∑r=0p−1emrp=p,ifp∣m,0,ifp∤m,and the estimate(9)∑b=1n−1e−rbp≪1sinπr/p≪pr.
From (8), (9), and Lemma 2, we have(10)∑a=1n−1χ1aχ2n−a=1p∑a=1p−1∑b=1n−1χ1aχ2n−a∑r=0p−1era−bp=1p∑r=1p−1∑a=1p−1χ1aχ2n−aerap∑b=1n−1e−rbp+n−1p∑a=1p−1χ1aχ2n−a≪pp⋅∑r=1p−1pr+n−1p⋅p≪p1/2⋅lnp.
This proves Lemma3.


## 3. Proof of the Theorem

Now, we shall complete the proof of our main result. For any integer1<n≤p−1, from the definition of Sn,p and Lemma 1, we have(11)Sn,p=ϕ2p−1p−12∑k|p−1μkϕk∑r=1k′∑h|p−1μhϕh∑s=1h′×∑a=1n−1er⋅indak⋅es⋅indn−ah.It is clear thatχt,ka=et⋅inda/k is a Dirichlet character modulo p. So, from the Polya and Vinogradov’s classical work (see [12]; Theorem 8.21 and Theorem 13.15), we have the estimate(12)∑a=1n−1χa≪p1/2lnp,where χ is any nonprincipal character modulo p.Now, from (11), (12), and Lemma 3, we have(13)Sn;p=ϕ2p−1p−12∑k|p−1μkϕk∑r=1k′∑h|p−1μhϕh∑s=1h′∑a=1n−1χr,kaχs,hn−a=ϕ2p−1p−12∑a=1n−11+ϕ2p−1p−12∑k|p−1k>1μkϕk∑r=1k′∑a=1n−1χr,ka+ϕ2p−1p−12∑k|p−1k>1μkϕk∑r=1k′∑h|p−1h>1μhϕh∑s=1h′∑a=1n−1χr,kaχs,hn−a+ϕ2p−1p−12∑h|p−1h>1μhϕh∑s=1h′∑a=1n−1χs,hn−a=ϕ2p−1p−12⋅n−1+Oϕ2p−1p3/2.∑k|p−1μk+Oϕ2p−1p3/2.∑k|p−1μk.∑h|p−1μh.lnp=ϕ2p−1p2⋅n+Oϕ2p−1p3/2⋅4ωp−1⋅lnp,where we have used the identity(14)∑k|mμk=∏pαm1+|μp|=2ωm,and ωm denotes the number of all distinct prime divisors of m.This completes the proof of our theorem.


## 4. Conclusion

The main result in this paper is a theorem, which is closely related to Golomb’s conjecture. It describes that when primep is large enough, for any integer p1/2+ɛ≤n≤p−1, there must exist two primitive roots α and β modulo p such that the equation n=α+β holds, where 0<ɛ<1/2 be a fixed positive number. At the same time, we also give a sharp asymptotic formula for the counting function of all such solutions α,β. In fact, our conclusion is much stronger than Golomb’s conjecture in the reduced residue system 1,2,3,…,p−1 modulo p. As a note of the corollary, we also proposed an open problem.



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