Simultaneous Elements of Prescribed Multiplicative Orders 2

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results for admissible rationals k-tuples u1, u2, . . . , uk ∈ Q of simultaneous elements of independent (or pseudo independent) multiplicative orders modulo a prime p ≥ 2 are considered in [26], and [14]. An important part of this problem is the characterization of admissible rationals k-tuples. There are various partial characterizations of admissible rationals k-tuples. For example, an important criterion states that a rationals k-tuple is admissible if and only if e1 e2 ek u1 u2 · · · uk 6= −1, (1) where e1, e2,...,ek ∈ Z are integers. A more general characterization and the proof appears in [26, Proposition 14], [14, Lemma 5.1]. Other ad hoc techniques are explained in [15]. The characterization of admissible triple of rational numbers a, b, c ∈ Q−{−1, 0, 1} with simultaneous equal multiplicative orders

ordp a = ordp b = ordp c, (2) known as the Schinzel-Wojcik problem, is an open problem, see [26, p. 2], and [8]. The speciﬁc relationship between the orders ordp u | d and ordp v | d of a pair of integers u, v 6= ±1 and some divisor d | p − 1, and the generalization to number ﬁelds, and abelian varieties are studied in several papers as [4], [1], et cetera, and counterexamples are produced in [4] and [20].

Deﬁnition 1.1. A k-tuple u1, u2, . . . , uk 6= ±1 of rational numbers is called an admissible k-tuple if the product is multiplicatively independent over the rational numbers:

e1 e2 ek u1 u2 · · · uk 6= 1, (3) for any list of integer exponents e1, e2,...,ek ∈ Z×. Relatively prime k-tuples, and relatively prime and squarefree k-tuples are automatically multiplicatively independent over the rational numbers. However, squarefree k-tuples are not necessarily multiplicatively independent over the rational numbers, for example, 3, 5, 15. Any k-tuple of integers that satisﬁes the Lang-Waldschmid conjecture is an admissible k-tuple. Speciﬁcally,

k e1 e2 ek C(ε) E u1 u2 · · · uk − 1 ≥ 1+ε , (4) |u1u2 · · · uk · e1e2 · · · aek|

where C(ε) > 0 is a constant, E = max{|ei|}, and ε > 0 is a small number, confer [31, Conjecture 2.5] for details.

Deﬁnition 1.2. Fix an admissible k-tuple u1, u2, . . . , uk 6= ±1 of rational numbers. The elements u1, u2, . . . , uk ∈ Fp are said to be a simultaneous k-tuple of equal multiplicative orders modulo a prime p ≥ 2, if ordp u1 | p − 1, and

ordp u1 = ordp u2 = · · · = ordp uk, (5) inﬁnitely often as p →∞.

Deﬁnition 1.3. Fix an admissible k-tuple u1, u2, . . . , uk 6= ±1 of rational numbers. The elements u1, u2, . . . , uk ∈ Fp are said to be a simultaneous k-tuple of decreasing multiplicative orders modulo a prime p, if ordp ui | p − 1, and

ordp u1 > ordp u2 > · · · > ordp uk, (6) inﬁnitely often as p →∞. Simultaneous Elements Of Prescribed Multiplicative Orders 3

Deﬁnition 1.4. Fix an admissible k-tuple u1, u2, . . . , uk 6= ±1 of rational numbers, and ﬁx an index integers k-tuple d1, d2,...,dk ≥ 1. The elements u1, u2, . . . , uk ∈ Fp are called a simultaneous k-tuple of prescribed multiplicative orders modulo a prime p, if

ordp u1 = (p − 1)/d1, ordp u2 = (p − 1)/d2, · · · ordp uk = (p − 1)/dk, (7) inﬁnitely often as p →∞.

Conditional on the GRH and or the k-tuple primes conjecture, several results for the existence and the densities of simultaneous rationals k-tuples have been proved in the literature, see [26], [8], and [14]. This note studies the unconditional asymptotic formulas for the number of primes with simultaneous elements of prescribed multiplicative orders.

Theorem 1.1. Fix a pair of relatively prime squarefree u 6= ±1, and v 6= ±1 rational numbers , and ﬁx a pair of integers d ≥ 1, and e ≥ 1. If x ≥ 1 is a suﬃciently large number, and the indices d, e ≪ (log x)B, with B ≥ 0, then, the number of primes p ∈ [x, 2x] with simultaneous elements u and v of prescribed multiplicative orders ordp u = (p − 1)/d and ordp v = (p − 1)/e modulo p ≥ 2, has the asymptotic lower bound x R (x, u, v) ≫ (8) 2 (log x)4B+1(log log x)2 as x →∞, unconditionally.

In general, the number of primes p ∈ [x, 2x] such that a ﬁxed admissible k-tuples of rational numbers u1, u2, . . . , uk 6= ±1 has simultaneous prescribed multiplicative orders

ordp u1 = (p − 1)/d1, ordp u2 = (p − 1)/d2, ..., ordp uk = (p − 1)/dk, (9)

B where di ≪ (log x) are ﬁxed indices, and B ≥ 0, has the lower bound x R (x, u, v) ≫ (10) k (log x)2Bk+1(log log x)k as x →∞, unconditionally.

The maximal number of simultaneous primitive roots is bounded by k = O (log p), see [3, Section 14]. The same upper bound is expected to hold for any combination of simultaneous k-tuple prescribed of multiplicative orders, but it has not been verified yet. The average multiplicative order of a fixed rational number u 6= ±1 has the asymptotic lower bound 1 x 3/2 T (x)= ord u ≫ ec(log log log x) , (11) u x n log x n x gcd(Xu,n≤ )=1 where c > 0 is a constant, the fine details are given in [17]. Other information on the average multiplicative orders in finite cyclic groups are given in [18], and the literature.

A special case illustrates the existence of simultaneously prescribed primitive roots and quadratic residues (or quadratic nonresidues) in the prime finite field Fp for infinitely many primes p.

Corollary 1.1. Let u ≥ 3 be ﬁxed a squarefree odd integer and let v = 2. If x ≥ 1 is a large number, then elements u, 2 ∈ Fp have multiplicative orders ordp u = p − 1 Simultaneous Elements Of Prescribed Multiplicative Orders 4

and ordp 2 = (p − 1)/2, simultaneously. Moreover, the number of such primes has the asymptotic lower bound x R (x, u, 2) ≫ (12) 2 (log x)(log log x)2 as x →∞, unconditionally. The unconditional number of primes with simultaneous admissible triple of rational numbers a, b, c ∈ Q − {−1, 0, 1} of equal multiplicative orders

ordp a = ordp b = ordp c, (13) B such that ordp a = (p − 1)/d, with d ≪ (log x) , and B ≥ 0, has almost identical analysis as the proof of Theorem 1.1. In fact, any permutation of equality or inequality between the multiplicative orders can be produced by selecting any small ﬁxed indices d, e, f ≥ 1 to prescribe the multiplicative orders ordp a = (p − 1)/d, ordp b = (p − 1)/e, and ordp c = (p − 1)/f respectively.

Some of the foundation works on the calculations of the implied constants in (3) appear in [26], [8], [14], et alii. The proof of Theorem 1.1 appears in Section 11, this result is completely unconditional. The other sections cover foundational and auxiliary materials.

2 Divisors Free Characteristic Function

Deﬁnition 2.1. The multiplicative order of an element in the cyclic group Fp× is deﬁned k by ordp(v) = min{k : v ≡ 1 mod p}. Primitive elements in this cyclic group have order p − 1=#G.

Definition 2.2. Let p ≥ 2 be a prime, and let Fp be the prime field of characteristic p. If d | p − 1 is a small divisor, the multiplicative subgroup of d-powers, and the subgroup d d d index are defined by Fp = {α : α ∈ Fp×}, and [Fp× : Fp]= d respectively.

Definition 2.3. Fix an integer d ≥ 1, and a rational number u ∈ Q×. An element u ∈ Fp has index d = indp u modulo a prime p if and only if indp u = (p−1)/ordp u. In particular, if d | p − 1, then the prime finite field Fp contains ϕ(d) elements of index d ≥ 1.

Each element u ∈ Fp of index d is a d-power, but not conversely. A new divisor-free representation of the characteristic function of primitive element is introduced here. This representation can overcomes some of the limitations of the standard divisor-dependent representation of the characteristic function ϕ(p − 1) µ(d) Ψ(u) = χ(u) (14) p − 1 ϕ(d) d p 1 ord(χ)=d X| − X 1 if ord (u)= p − 1, = p 0 if ord (u) 6= p − 1, p of primitive roots. The works in [5], and [32] attribute this formula to Vinogradov. The proof and other details on this representation of the characteristic function of primitive roots are given in [7, p. 863], [19, p. 258], [23, p. 18]. Equation (14) detects the multiplicative order ordp(u) of the element u ∈ Fp by means of the divisors of the totient p − 1. In contrast, the divisors-free representation of the characteristic function in (15) detects the multiplicative order ordp(u) ≥ 1 of the element u ∈ Fp by means of the solutions of the n equation τ − u =0 in Fp, where u, τ are constants, and 1 ≤ n

Lemma 2.1. Let p ≥ 2 be a prime, and let τ be a primitive root mod p. Let ψ(z) = i2πz/p e 6= 1 be a nonprincipal additive character of order ord ψ = p. If u ∈ Fp is a nonzero element, then, 1 Ψ(u) = ψ ((τ n − u)k) (15) p gcd(n,p 1)=1 0 k p 1 X− ≤X≤ − 1 if ord (u)= p − 1, = p 0 if ord (u) 6= p − 1. p Proof. As the index n ≥ 1 ranges over the integers relatively prime to p − 1, the element n τ ∈ Fp ranges over the primitive roots mod p. Ergo, the equation τ n − u = 0 (16) has a solution if and only if the ﬁxed element u ∈ Fp is a primitive root. Next, replace ψ(z)= ei2πkz/p to obtain

1 i2π(τ n u)k/p Ψ(u) = e − (17) p gcd(n,p 1)=1 0 k p 1 X− ≤X≤ − 1 if ord (u)= p − 1, = p 0 if ord (u) 6= p − 1. p k N This follows from the geometric series identity 0 k N 1 w = (w − 1)/(w − 1) with w 6= 1, applied to the inner sum. ≤ ≤ − P Let d | p−1. A new representation of the indicator function for d-power v ∈ Fp or elements of order ordp(v) = (p − 1)/d is consider below. Lemma 2.2. Let p ≥ 2 be a prime, and let τ be a primitive root mod p. Let ψ(z) = i2πz/p e 6= 1 be a nonprincipal additive character of order ord ψ = p. If u ∈ Fp is a d-power, then, 1 Ψ(u, d) = ψ (τ dn − u)k (18) p gcd(n,(p 1)/d)=1 0 k (p 1)/d X− ≤ ≤X− 1 if ord (u) = (p − 1)/d, = p 0 if ord (u) 6= (p − 1)/d. p Proof. Similar to the proof of Lemma 2.1, mutatis mutandis.

3 Finite Summation Kernels

Let f : C −→ C be a function, and let q ∈ N be a large integer. The finite Fourier transform 1 fˆ(t)= eiπst/q (19) q 0 s q 1 ≤X≤ − and its inverse are used here to derive a summation kernel function, which is almost identical to the Dirichlet kernel. Definition 3.1. Let p and q be primes, and let ω = ei2π/q, and ζ = ei2π/p be roots of unity. The finite summation kernel is defined by the finite Fourier transform identity

1 t(n s) K(f(n)) = ω − f(s)= f(n). (20) q 0 t q 1, 0 s p 1 ≤X≤ − ≤X≤ − Simultaneous Elements Of Prescribed Multiplicative Orders 6

This simple identity is very eﬀective in computing upper bounds of some exponential sums f(n)= K(f(n)), (21) n x n x X≤ X≤ where x ≤ p

Proof. (i) Use the geometric series to compute this simple exponential sum as

ωt − ωtp ωtn = . 1 − ωt n p 1 ≤X− (ii) Observe that the parameters q = p+o(p) is prime, ω = ei2π/q, the integers t ∈ [1,p−1], and d ≤ p − 1

Lemma 3.2. Let p ≥ 2 and q = p + o(p) be large primes, and let ω = ei2π/q be a qth root of unity. Then, ωrt − ωdrt((p 1)/r(d+1)) (i) ωtn = µ(d) − , 1 − ωrdt gcd(n,(p 1)/d)=1 r (p 1)/d X− | X− 4q log log p (ii) ωtn ≤ , πt gcd(n,p 1)=1 X− where µ(k) is the Mobius function, for any ﬁxed pair d | p − 1 and t ∈ [1,p − 1].

Proof. (i) Use the inclusion exclusion principle to rewrite the exponential sum as ωtn = ωtn µ(d) gcd(n,p 1)=1 n p 1 d p 1 ≤ − X− X X|d −n | = µ(d) ωtn d p 1 n p 1 X| − ≤Xd n− | = µ(d) ωdtm (23) d p 1 m (p 1)/d X| − ≤X− ωdt − ωdt((p 1)/d+1) = µ(d) − . 1 − ωdt d p 1 X| − Simultaneous Elements Of Prescribed Multiplicative Orders 7

(ii) Observe that the parameters q = p+o(p) is prime, ω = ei2π/q, the integers t ∈ [1,p−1], and d ≤ p − 1

for 1 ≤ d ≤ p − 1. Finally, the upper bound is

ωdt − ωdt((p 1)/d+1) 2q 1 µ(d) − ≤ (25) 1 − ωdt πt d d p 1 d p 1 X| − X| − 4q log log p ≤ . πt 1 The last inequality uses the elementary estimate d n d− ≤ 2 log log n. | P 4 Gaussian Sums, And Weil Sums

Theorem 4.1. (Gauss sums) Let p ≥ 2 and q ≥ 2 be large primes. Let τ be a primitive t root modulo p. If χ(t) = ei2πt/q and ψ(t) = ei2πτ /p are a pair of characters, then, the Gaussian sum has the upper bound

χ(t)ψ(t) ≤ 2q1/2 log q. (26)

1 t q 1 ≤X≤ −

Theorem 4.2. (Weil sums) Let p ≥ 2 and q ≥ 2 be large primes, and let f(t) be a powerfree polynomial of degree deg f = d ≥ 1. If χ(t)= ei2πt/q and ψ(t)= ei2πf(t)/p are a pair of characters. Then, the Weil sum has the upper bound

χ(t)ψ(f(t)) ≤ 2dq1/2 log q. (27)

1 t q 1 ≤X≤ −

Theorem 4.3. Let p ≥ 2 and q ≥ 2 be large primes. Let τ be a primitive root modulo p, and let κ = τ d be an element of large multiplicative order. If χ(t) = ei2πt/q and ψ(t)= ei2πt/p are a pair of characters. Then, the exponential sum has the upper bound

χ(t)ψ(τ dt) ≤ 2dq1/2 log q. (28)

1 t q 1 ≤X≤ −

Proof. Use the change of variable z = τ t to rewrite the exponential sum as a Weil sum with a polynomial f(z)= zd of degree d.

5 Incomplete And Complete Exponential Sums

Two applications of the generalizes the sum of resolvents method used in [22], and [28], to estimate exponential sums are illustrated here. The ﬁrst application is a nonlinear counterpart of the Polya-Vinogradov inequality χ(n) ≤ 2p1/2 log p (29) n x X≤ for nonprincipal character χ 6= 1 modulo p. Simultaneous Elements Of Prescribed Multiplicative Orders 8

Theorem 5.1. Let p ≥ 2 be a large prime, and let κ ∈ Fp be an element of large multiplicative order ordp(κ) | p − 1. Then, for any ﬁxed integer a ∈ [1,p − 1], and x ≤ p − 1,

n ei2πaκ /p ≪ p1/2 log3 p. (30) n x X≤ dn Proof. Let q = p+o(p) be a large prime, and write f(n)= ei2πaτ /p, where τ is a primitive root modulo p, and κ = τ d has large multiplicative order modulo p modulo p. Applying the ﬁnite summation kernel in Deﬁnition 3.1, yields

dn R(d, x) = ei2πaτ /p (31) n x X≤ 1 t(n s) i2πaτ ds/p = ω − e . q n x 0 t q 1, 1 s p 1 X≤ ≤X≤ − ≤X≤ − Use the Weil sum upper bound, see Theorem 4.3, to show that the value t = 0 contributes

1 ds x ds ei2πaτ /p = ei2πaτ /p (32) q q n x, 1 s p 1 1 s p 1 X≤ ≤X≤ − ≤X≤ − x d = ei2πaz /p q 1 z p 1 x ≤X≤ − ≤ · (2 log p)p1/2 q 4x log p ≤ , p1/2 where 1/q ≤ 2/p. Replacing (32) into (31), and rearranging it, yield

dn R(d, x) = ei2πaτ /p (33) n x X≤ 1 t(n s) i2πaτ ds/p 4x log p = ω − e + O q p1/2 n x, 1 t q 1, 1 s p 1 X≤ ≤X≤ − ≤X≤ −

1 ts i2πaτ ds/p tn 4x log p = ω− e ω + O . q     p1/2 1 t q 1 1 s p 1 n x ≤X≤ − ≤X≤ − X≤     Taking absolute value, and applying Lemma 3.1 to the inner sum, and Theorem 4.3 to the middle sum, yield

dn |R(d, x)| = ei2πaτ /p (34)

n x X≤

1 ts i2πaτ ds/p tn 4x log p ≤ ω− e · ω + O q p1/2 1 t q 1 1 s p 1 n x ≤X≤ − ≤X≤ − X≤ 1 2q 4 x log p ≪ 2p1/2 log2 p · + q πt p1/2 1 t q 1 ≤X≤ − 4x log p ≪ p1/2 log3 p + . p1/2 Simultaneous Elements Of Prescribed Multiplicative Orders 9

The last summation in (34) uses the estimate 1 ≪ log q ≪ log p (35) t 1 t q 1 ≤X≤ − since q = p + o(p).

This result is nontrivial for x ≥ p1/2+δ, and elements of large multiplicative orders 1/2+δ ordp κ ≥ p , where δ > 0. A similar upper bound for composite moduli p = m is also proved in [22, Equation (2.29)].

The second application is a complete exponential sum version of the previous one, but restricted to relatively prime arguments.

Theorem 5.2. Let p ≥ 2 be a large prime, let τ be a primitive root modulo p, and let κ = τ d be an element of large multiplicative order modulo p. Then,

i2πaτ dn/p 1 ε e ≪ p − (36) gcd(n,(p 1)/d)=1 X− for any ﬁxed integer a ∈ [1,p − 1], and any arbitrary small number ε ∈ (0, 1/2).

dn Proof. Let q = p+o(p) be a large prime, and write f(n)= ei2πaτ /p, where τ is a primitive root modulo p. Start with the representation

dn ds i2πaτ 1 t(n s) i2πaτ e p = ω − e p , (37) q gcd(n,(p 1)/d)=1 gcd(n,(p 1)/d)=1 0 t q 1, 1 s p 1 X− X− ≤X≤ − ≤X≤ − see Deﬁnition 3.1. Use the inclusion exclusion principle to rewrite the exponential sum as

dn ds i2πaτ 1 t(n s) i2πaτ e p = ω − e p µ(r). (38) q gcd(n,(p 1)/d)=1 n (p 1)/d 0 t q 1, 1 s p 1 r (p 1)/d ≤ ≤ − ≤ ≤ − X− ≤ X− X X | Xr−n | The value t = 0 contributes

1 i2πaτds T (d, p) = e p µ(r) 0 q n (p 1)/d 1 s p 1 r (p 1)/d ≤ ≤ − ≤ X− X | Xr−n | 1 i2πaτds ≤ e p 1 (39) q r (p 1)/d 1 s p 1 m (p 1)/rd | X− ≤X≤ − ≤ X−

1 p − 1 i2πaτds 1 ≤ e p q d r 1 s p 1 r (p 1)/d ≤X≤ − | X− 1/2 2 2p log p ≤ , d where the middle sum is a Weil sum, see Theorem 4.3, and (p − 1)/q ≤ 1. Replacing (39) Simultaneous Elements Of Prescribed Multiplicative Orders 10 into (38), and rearranging it, yield

i2πaτdn ρ(d, p) = e p (40) gcd(n,(p 1)/d)=1 X− ds 1 t(n s) i2πaτ = ω − e p µ(d) q n (p 1)/d 1 t q 1, 1 s p 1 r (p 1)/d ≤ X− ≤X≤ − ≤X≤ − | Xr−n | 2p1/2 log2 p +O d !

ds 1 ts i2πaτ tn = ω− e p  µ(d) ω  q   1 t q 1 1 s p 1 r (p 1)/d n (p 1)/d, ≤X≤ − ≤X≤ −  | X− ≤ X−     r n   |  2p1/2 log2 p +O . d ! Taking absolute value, and applying Lemma 3.2 to the inner sum, and Theorem 4.3 to the middle sum, yield

i2πaτdn |ρ(d, p)| = e p (41)

gcd(n,(p 1)/d)=1 X−

1 ts i2πaτ ds/p tn ≤ ω− e · µ(d) ω q 1 t q 1 1 s p 1 r (p 1)/d n (p 1)/d, ≤X≤ − ≤X≤ − | X− ≤ X− r n | 2p1/2 log2 p +O d ! 1 4q log log p 2p1/2 log2 p ≪ 2p1/2 log2 p · + O q πt d 1 t q 1 ! ≤X≤ − ≪ p1/2 log3 p. The last summation in (41) uses the estimate 1 ≪ log q ≪ log p (42) t 1 t q 1 ≤X≤ − 1/2 3 1 ε since q = p + o(p). This is restated in the simpler notation p log p ≤ p − for any arbitrary small number ε ∈ (0, 1/2), as x →∞. The upper bound given in Theorem 5.2 for maximal ε < 1/2 seems to be optimum. A diﬀerent proof, which has a weaker upper bound is included here as a reference for a second independent proof. Theorem 5.3. ([10, Theorem 6]) Let p ≥ 2 be a large prime, and let τ be a primitive root modulo p. Then, i2πaτ n/p 1 ε e ≪ p − (43) gcd(n,p 1)=1 X− for any integer a ∈ [1,p − 1], and any arbitrary small number ε> 0 is a small number. Simultaneous Elements Of Prescribed Multiplicative Orders 11

Other related results are given in [2], [9], [11], and [12, Theorem 1].

6 Explicit Exponential Sums

An explicit version of Theorem 5.2 and Theorem 5.3 is computed below. Theorem 6.1. Let m ≥ 1 be an integer, and let Q ≥ 1 be the period of the element w ∈ Z/mZ. If the number P

i2πawn/m 1 ε e ≤ c0P − . (44) 1 n P ≤X≤ where a 6= 0, ε = c1 log P/log m< 1 is a small number, and c0, c1 > 0 are constants. Proof. A discussion of this exponential sum and a proof appears in [16, p. 8].

The complete exponential sum n ei2πaw /m (45) 1 n Q ≤X≤ is known to be a very small number or to vanish. An upper bound for a related and diﬀerent exponential sum will be used in the analysis of the orders of elements in ﬁnite rings. Theorem 6.2. Let m ≥ 1 be an integer, and let Q ≥ 1 be the period of the element w ∈ Z/mZ. If the number P

i2πawn/m ε 1 2ε e ≤ c0m P − . (46) 1 n P gcd(≤Xn,ϕ≤(m)) where a 6= 0, 2ε = c1 log P/log m< 1 is a small number, and c0, c1 > 0 are constants. Proof. Let P = Q − 1, and rewrite the exponential sum in the form

n n ei2πaw /m = ei2πaw /m µ(d) (47) 1 n P 1 n P d n gcd(≤Xn,ϕ≤(m) ≤X≤ dXϕ|(m) | n = µ(d) ei2πaw /m. d ϕ(m) 1 n P |X ≤Xd n≤ | Taking absolute value, and applying Theorem 6.1 to the inner exponential sum return

n n ei2πaw /m ≤ 1 ei2πaw /m (48)

where c0 = c4, c1, c2, c3 > 0 are constants. Plugging the trivial upper bounds P ≤ m, and ϕ(m) ≤ m complete the veriﬁcation of the inequality

i2πawn/m 1 2ε ε 1 ε e ≤ c P − ϕ(m) ≤ c m − , (49) 4 4 1 n P ≤X≤ gcd(n,ϕ(m))

for suﬃciently large m.

7 Equivalent Exponential Sums

This section demonstrate that the exponential sums

dn dn ei2πaτ /p and ei2πτ /p, (50) gcd(n,(p 1)/d)=1 gcd(n,(p 1)/d)=1 X− X− where d | p − 1, and a 6= 0, are asymptotically equivalent. This result expresses this exponential sum as a sum of simpler exponential sum and an error term. The proof is entirely based on established results and elementary techniques. Theorem 7.1. Let p ≥ 2 be a large primes, and let d | p − 1 be a small divisor. If τ be a primitive root modulo p, then,

dn dn ei2πaτ /p = ei2πτ /p + O(p1/2 log4 p), (51) gcd(n,(p 1)/d)=1 gcd(n,(p 1)/d)=1 X− X− for any integer a ∈ [1,p − 1]. Proof. For any integer a ≥ 1, the exponential sum has the representation

i2πaτdn ρ(a, d, p) = e p (52) gcd(n,(p 1)/d)=1 X−

ds 1 ts i2πaτ tn = ω− e p  µ(r) ω  q   1 t q 1 1 s p 1 r (p 1)/d n (p 1)/d, ≤X≤ − ≤X≤ −  | X− ≤ X−     r n   |  2p1/2 log2 p +O , d ! confer equations (37) to (40) for details. And, for a = 1,

i2πτdn ρ(1,d,p) = e p (53) gcd(n,(p 1)/d)=1 X−

ds 1 ts i2πτ tn = ω− e p  µ(r) ω  q   1 t q 1 1 s p 1 r (p 1)/d n (p 1)/d, ≤X≤ − ≤X≤ −  | X− ≤ X−     r n   |  2p1/2 log2 p +O , d ! Simultaneous Elements Of Prescribed Multiplicative Orders 13 respectively, see equations (37) to (40). Diﬀerencing (52) and (53) produces

dn dn D(d, p) = ei2πaτ /p − ei2πτ /p (54) gcd(n,(p 1)/d)=1 gcd(n,(p 1)/d)=1 X− X− ds ds 1 ts i2πaτ ts i2πτ = ω− e p − ω− e p q   0 t q 1 1 s p 1 1 s p 1 ≤X≤ − ≤X≤ − ≤X≤ −  

×  µ(r) ωtn . r (p 1)/d n (p 1)/d,  | X− ≤ X−   r n   |  By Lemma 3.2, the relatively prime summation kernel is bounded by

µ(r) ωtn = ωtn

r (p 1)/d n (p 1)/d, gcd(n,(p 1)/d)=1 | X− ≤ X− X− r n | 4q log log p ≤ , (55) πt and by Theorem 4.3, the diﬀerence of two Weil sums (or Gauss sums) is bounded by

i2πaτds i2πτds δ(d, p) = ω tse p − ω tse p − − 1 s p 1 1 s p 1 ≤X≤ − ≤X≤ −

= χ(s)ψ (s) − χ(s)ψ (s) a 1 1 s p 1 1 s p 1 ≤X≤ − ≤X≤ − 1/2 2 ≤ 4p log p, (56)

iπst/p i2πaτ ds/p where χ(s)= e , and ψa(s)= e . Taking absolute value in (54) and replacing (55), and (56), return

n n |D(d, p)| = ei2πbτ /p − ei2πτ /p (57)

gcd(n,p 1)=1 gcd(n,p 1)=1 X− X− 1 4q log log p ≤ 4p1/2 log2 p · q t 0 t q 1 ≤X≤ − ≤ 16p1/2(log2 p)(log q)(log log p) ≤ 16p1/2 log4 p, where q = p + o(p).

8 Double Exponential Sums

Lemma 8.1. Given a small number ε> 0. Let p be a large prime number, and let d | p−1 be a small divisor. If τ is a primitive root modulo p, then,

i2πa(τdn−u) 1 ε e p ≪ p − . (58) 0

Proof. To compute an upper bound, rearrange the double ﬁnite sum as follows.

i2πa(τdn−u) i2πau/p i2πaτ dn/p e p = e− e . (59) 0

i2πau/p i2πaτ dn/p T (p) = e− e (60) 0

i2πau/p i2πτ dn/p 1/2 4 = e− e + O p log p .   0

dn |T (p)| ≤ e i2πau/p ei2πτ /p + O p1/2 log4 p − 0

i2πau/p i2πτ dn/p 1/2 4 ≤ e− e + O p log p   0

The main term of two or more simultaneous elements requires the average order of a product of totient functions. A lower bound for the product two totients will be computed here. Lemma 9.1. If x ≥ 1 is a large number, and d, e ≪ (log x)B, with B ≥ 0, then ϕ((p − 1)/d) ϕ((p − 1)/e) x · ≫ . (62) p p (log x)4B+1(log log x)2 x p 2x ≤X≤ Proof. The totient function has the lower bound ϕ(n)/n ≫ 1/log log n, see [27, Theorem 15]. Replacing this estimate yields ϕ((p − 1)/d) ϕ((p − 1)/e) M(x, u, v) = · (63) p p x p 2x ≤X≤ ϕ((p − 1)/d) ϕ((p − 1)/e) 1 2 = · 1 − p − 1 p − 1 p x p 2x ≤X≤ 1 1 1 1 ≫ · d log log p e log log p x p 2x p 1≤X mod≤ de ≡ 1 1 ≫ 1, (log x)2B (log log x)2 x p 2x p 1≤X mod≤ de ≡ Simultaneous Elements Of Prescribed Multiplicative Orders 15 since d, e ≪ (log x)B, with B ≥ 0. Applying the prime number theorem on arithmetic progression over the short interval [x, 2x], yields 1 1 M(x, u, v) = 1 (64) (log x)2B (log log x)2 x p 2x p 1≤X mod≤ de ≡ x x c√log x ≫ · + O xe− (log x)2B(log log x)2 ϕ(de) log x x ≫ , (log x)4B+1(log log x)2 where ϕ(de) ≤ de ≤ (log x)2B , and c> 0 is an absolute constant.

This analysis is eﬀective and unconditional for the prescribed indices d, e ≪ logB x, where B ≥ 0 is a constant, as limited by the current version of the prime number theorem on arithmetic progressions, confer [6, Theorem 3.10].

The exact asymptotic for the average order of a product k totient functions over the primes is proved in [30], and related discussions are given in [25, p. 16]. The generalization to number ﬁelds appears in [13]. However, the exact asymptotic for the average order of a product of k totient functions over the primes in arithmetic progressions seems to be unknown. For example, for equal prescribed multiplicative orders (p−1)/d, it should have the form

k ϕ((p − 1)/d) ? x c√log x = A + O xe− , (65) p − 1 1 ϕ(q) log x x p 2x p ≤aXmod≤ q ≡ where q = dk ≤ (log x)Bk, B ≥ 0 constant, 1 ≤ a 0 is a constant, and c > 0 is an absolute constant. But, for distinct prescribed multiplicative orders (p − 1)/di, it should have the form

ϕ((p − 1)/di) ? x c√log x = A + O xe− , (66) p − 1 k ϕ(q) log x x p 2x 1 i k p ≤aXmod≤ q ≤Y≤ ≡ Bk where q = d1d2 · · · dk ≤ (log x) , and Ak = Ak(a, q) > 0, which is slightly more complex.

10 The Error Terms

Lemma 10.1. Assume ordp u 6= (p − 1)/d. If x is a large number, then

1 i2πa(τdn−u) 1 i2πb(τem−v) E (x)= e p · e p = 0. 1 p p x p 2x 0 a

Proof. By hypothesis, ordp u 6= (p − 1)/d, so the ﬁrst ﬁnite sum

i2πa(τdn−u) e p = 0 (67) 0 a

Lemma 10.2. Assume ordp v 6= (p − 1)/e. If x is a large number, then

1 i2πb(τem−v) 1 i2πb(τem−v) E (x)= e p · e p = 0. 2 p p x p 2x 0

Proof. By hypothesis, ordp v 6= (p − 1)/e, so the second ﬁnite sum

i2πb(τem−v) e p = 0 (68) 0 b

Lemma 10.3. If x is a large number, then

i2πa(τdn−u) i2πb(τem−v) 1 1 1 2ε E (x)= e p · e p ≪ x − . 3 p p x p 2x 0

Proof. To compute an upper bound, deﬁne the exponential sum

i2πa(τdn−u) T (d, p)= e p . (69) 0

Now, apply Lemma 8.1 to each factor T (d, p) and T (e, p) to obtain the followings.

1 1 E (x) = · T (d, p) · · T (e, p) (70) 3 p p x p 2x ≤X≤ 1 1 ε 1 1 ε ≪ · p − · · p − . p p x p 2x ≤X≤ Take an upper bound, and apply the prime number theorem: 1 E (x) ≪ 1 (71) 3 x2ε x p 2x ≤X≤ 1 2ε ≪ x − , as x →∞.

11 Simultaneous Prescribed Multiplicative Orders

The multiplicative order of an element u ∈ Fp in ﬁnite ﬁeld is the smallest integer n ≥ 1 for n which u =1 in Fp, see Section 2 for additional details. The simpler case of simultaneous prescribed multiplicative orders of two admissible rational numbers is investigated in this section. The analysis of simultaneous and prescribed multiplicative orders for k-tuple of admissible rational numbers are similar to this analysis, but have bulky and cumbersome notation. Simultaneous Elements Of Prescribed Multiplicative Orders 17

Definition 11.1. Fix a pair of rational numbers u, v 6= ±1 such that uavb 6= ±1 for any a, b ∈ Z. The elements u, v ∈ Fp are said to be simultaneous of equal orders ordp u | p − 1, and ordp v | p − 1 respectively, if ordp u = ordp v infinitely often as p →∞. Otherwise, the elements are said to be simultaneous of unequal orders, if ordp u 6= ordp v, infinitely often as p →∞.

Given a pair of small integer indices d ≥ 1 and e ≥ 1, the number of primes x ≤ p ≤ 2x, which have simultaneous elements u> 1 and v> 1 of prescribed orders ordp u | (p − 1)/d and ordp v | (p − 1)/e modulo p ≥ 2, respectively, is deﬁned by

R(x, u, v) = # { x ≤ p ≤ 2x : p is prime and (72)

ordp(u) = (p − 1)/d, ordp(v) = (p − 1)/e } ,

The small integers indices d ≥ 1 and e ≥ 1 prescribed the multiplicative orders of the ﬁxed pair u> 1 and v> 1

Proof. (Theorem 1.1) Substitute the indicator function Ψp(u, d) for elements u of order ordp u = (p − 1)/d modulo p, and the indicator function Ψp(u, e) for elements v of order ordp v = (p−1)/e modulo p, see Lemma 2.2, to construct the associated counting function for the number of such primes p ∈ [x, 2x]:

R(x, u, v) = Ψp(u, d) · Ψp(v, e) (73) x p 2x ≤X≤ 1 i2πa(τ dn u)/p 1 i2πb(τ em v)/p = e − · e − p p x p 2x 0 a

= M(x, u, v) + E1(x)+ E2(x)+ E3(x).

The main term M(x, u, v) is determined by (a, b) = (0, 0), the error terms E1(x), E2(x), and E3(x) are determined by (a, b) = (0, b 6= 0), (a, b) = (a 6= 0, 0), and (a, b) 6= (0, 0), respectively.

Summing the main term M(x, u, v), estimated in Lemma 9.1, and the error terms E1(x), E2(x), and E3(x) estimated in Lemma 10.1, Lemma 10.2, and Lemma 10.3 respectively, yield

R(x, u, v) = M(x, u, v) + E1(x)+ E2(x)+ E3(x) (74) x 1 2ε ≫ +0+0+ x − (log x)4B+1(log log x)2 x ≫ , (log x)4B+1(log log x)2 where ε> 0 is a small number, as x →∞.

The current analysis is unconditional for any indices product de ≪ (log x)2B , where B ≥ 0 is a constant. Assuming the RH, it appears that this analysis can handled any indices 1/2 δ product de as large as de ≤ p − , where δ > 0, but no eﬀort was made to verify this observation, confer the proof for the lower bound of the main term in Section 9 for some information. Simultaneous Elements Of Prescribed Multiplicative Orders 18

12 Probabilistic Results For Simultaneous Orders

Theorem 12.1. For any pair of random relatively prime integers a > 1 and b > 1, and any suﬃciently large prime p, the ratio ord (a) p 6= 1 (75) ordp(b) is true with probability 1+ o(1) > 1/2. Proof. Fix a large prime p. Two random relatively prime integers a and b have the same multiplicative order if and only if ordp(a) = ordp(b) = (p − 1)/d0 for at least one divisor d0 | p − 1. Let α2 > 0 denotes the probability that ord (a) p = 1 (76) ordp(b) is true. Otherwise, α2 = 0, and the claim in (75) is trivially true. Assuming statistical pseudo independence, this event occurs with probability 1 1 ϕ((p − 1)/d ) ϕ((p − 1)/d ) c · < α = c 0 · 0 (77) 0 p − 1 p − 1 2 0 p − 1 p − 1 ϕ((p − 1)/d) 2 < c d p − 1 d p 1 X| − ϕ((p − 1)/d) ≤ p − 1 d p 1 X| − ϕ(p − 1) 1 = ≤ , p − 1 2 where ci = ci(a, b) ≤ 1 is a statistical independence correction factor. The lower bound 1/(p − 1)2 in (77) follows from the hypothesis a> 1 and b> 1. Hence, random relatively prime integers a and b of multiplicative order ordp(a) 6= ordp(b), satisfy the ratio ord (a) p 6= 1 (78) ordp(b) with probability 1 1 − α > . (79) 2 2

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