Primitive roots in algebraic number fields

Transcription

1 Primitive roots in algebraic number fields Research Thesis Submitted in partial fulfillment of the requirements for the Degree of Doctor of Philosophy Joseph Cohen Submitted to the Senate of the Technion Israel Institute of Technology Elul 5764 Haifa September 004 1

2 The research was carried out in the Faculty of Mathematics under supervision of Professor Jack Sonn and Professor Zeev Rudnick. The generous financial help of the Technion is gratefully acknowledged.

3 Contents 1. Abbreviations and Notations 5. Introduction 6 3. The work of Gupta-Murty and of Heath-Brown 7 4. Artin s conjecture in a real quadratic field for units Notation and Preliminaries Proof of Theorem 4. - the sieve part Proof of Theorem 4.- The algebraic part 0 5. Artin s conjecture for nonunits Notation and Preliminaries Proof of Theorem the sieve part Proof of Theorem The algebraic part Gcd of eponent functions Application with sieve method Proof of The result 45 References 49 3

4 Abstract We consider an analogue of Artin s primitive root conjecture for units in real quadratic fields. Given such a nontrivial unit, for a rational prime p which is inert in the field. The maimal order of this unit modulo p is p + 1. An etension of Artin s conjecture is that there are infinitely many such inert primes for which this order is maimal. This is known at present only under the Generalized Riemann Hypothesis. Unconditionally, we show that for any choice of 7 units in different real quadratic fields satisfying a certain simple restriction, there is at least one which satisfies the above version of Artin s conjecture. Likewise, we consider an analogue of Artin s primitive root conjecture for nonunits in real quadratic fields. Given such an element, for a rational prime p which is inert in the field the maimal order of the unit modulo p is p 1. As before, the etension of Artin s conjecture is that there are infinitely many such inert primes for which this order is maimal. We show that out of any choice of 85 algebraic numbers satisfying a certain simple restriction, there is at least one which satisfies the above version of Artin s conjecture. Gupta and Murty s method to attack the former problems raises question regarding gcd (a n 1, b n 1) where a, b are multiplicatively independent rational positive integers. It is known that there are infinitely many integers n with big gcd (a n 1, b n 1). We show the same property for gcd (a n + 1, b n + 1). By the principal method which we use for primitive roots we can obtain another result. Let l r (p) (l nr (p) respectively) denote the smallest prime which is quadratic residue (non-residue respectively) mod p, and z() any unbounded increasing real function. We show the known results of Erdös and Elliot with an elementary method, and generalize this result in the sense that the l r (p) s (l nr (p) s) can be chosen from a specific infinite set which fulfills a certain condition. Then, we present an interesting application of the proof technique of the result which relates to the former problem. 4

5 1. Abbreviations and Notations Z - The ring of integers. Z/p - The ring of integers modulo prime p F p - Multiplicative group of the field of p elements < a > - Subgroup of F p which generate by a gcd(, ) - Greatest common divisor of two integers ord(q) - Order of element in F p ) - Legendre symbol ( q p Ω(n) - Number of prime factors of n (with multiplicity) f g (or g f or f = O(g))) - The inequality f() cg(). f() = o(g()) - Means that f() g() 0 - The discriminant of the real quadratic field K = Q( ) - A real quadratic field O K - Integer ring of K Π(y; m, s) - Number of primes p y such that p s (mod m) Li(y) - The term y dt log t E(y; m, s) - The term Π(y; m, s) - Li(y) ϕ(m) E(y; m, s) E(, m) = ma 1 y ma (s,m)=1 ϕ(m) - Euler totient C ɛ (p) -Kernel of a map (O K /(p)) (Z/p) N - The norm map ν - Function which gives, the number of prime factors of an integer µ - Möbius function S(A, z, v) {a a A, (a, p) = 1} p<z, p v ord(a, N) - Order of the matri A mod N. ord(a, b; N) - Order of (a, b) mod N P a (z) - The product of all odd primes up to z and a(mod 4) P(z) = P 1 (z)p 3 (z). P 3 - Integer with at most three prime divisors l r (p) (l nr (p)) - The smallest prime which is a quadratic residue (non-residue) mod p 5

6 . Introduction A natural question to ask is if there are many primes for which is a primitive root, that is if the subgroup of the multiplicative group F p of the field of p elements generated by is the whole group. Is there a finite number of such groups F p? Does the same apply for any integer a? In 197 Emil Artin [] made the following conjecture: Conjecture.1. Let a 1 be an integer which is not a perfect square. Then there are infinitely many primes p such that < a >= F p. In addition, for > 0, the number of primes p with this property is asymptotic as to A(a) log where A(a) is a constant which depends on a. In 1967 Hooley [13] proved Artin s conjecture with the asymptotic formula under the Generalized Riemann Hypothesis. In 1983 Gupta and Murty [9] proved that there are 13 specific integers such that at least one of them fulfills the Artin conjecture. From the proof we can deduce that Artin s conjecture is true for almost all integers. R. Murty, K. Murty and Gupta [10] showed that we can reduce the specific set of integers from 13 to 7. Improving the analytic part of Gupta and Murty enabled to give the best result till now: Theorem.. ([1]) Let q, r and s any three primes. Then at least one of them is a primitive root mod p for infinitely many primes p. We note that theorem 5.1 holds for any three non-zero integers, q, r and s which are multiplicatively independent where q, r, s, 3qr, 3qs, 3rs and qrs are not a square. (we say that r integers a 1,..., a r are multiplicatively independent if for any integers n 1,..., n r, a n 1 1 anr r = 1 n 1 =... = n r = 0). In this work we present an analog of Artin s conjecture in a quadratic field and we will prove a result similar to the one just shown (we will 6

7 show that a set which contains a specific number of elements or more always contains a primitive root) for units (in chapter ) and for any algebraic number which is not unit (in chapter 3). Another interesting problem which was raised from the Gupta and Murty work is to find an infinite sequence of integers n such that gcd(a n 1, b n 1) = 1. Ailon and Rudnick [1] have conjectured that the answer is true. From this, a natural question to ask is if there eists an infinite sequence of integers n such that the gcd(a n 1, b n 1) is big. Let a, b be multiplicatively independent. Y. Bugeaud, P. Corvaja and U. Zannier [4] proved that for all ɛ > 0, gcd(a n 1, b n 1) < c(ɛ)ep(ɛn). L. Adleman, C. Pomerance and R. Rumely [3] proved that there are infinitely many integers n such that gcd(a n 1, b n 1) > ep(ep(c log n/ log log n)) for any integers a, b. In chapter 6 we prove the same result as APR found result for gcd(a n + 1, b n + 1). In addition we note that this is also true for gcd(a n + 1, b n 1). In the last chapter we show an interesting use by elementary sieve method (the method which we use for the primitive roots result) to solve two problems from number theory one of them related to the above mentioned problem. 3. The work of Gupta-Murty and of Heath-Brown Since our work is based on the idea of Gupta and Murty with the advanced version as in the paper of Heath-Brown it will be natural to present their work. We start with following trivial idea: since the number of elements in F p is p 1, if we show for all integer d, and infinitely many primes p d p 1, d p 1 a d 1 mod p we will have proven the conjecture. So our first goal is to find infinitely many primes p with a small number of prime divisors of p 1. Heath-Brown proved the following lemma. 7

8 Lemma 3.1. Let q, r and s be any three primes. There eist K = k, k = 1,, 3 such that for any sufficiently large R + we have two numbers ɛ, δ (0, 1/4) and c = c(ɛ, δ) > 0 so that there are at least c log primes p which satisfy: Either p 1 p 1 is prime or K K p 1 < p 1/ δ. Furthermore, p satisfies = p 1p for p 1, p primes > p 1/4+ɛ and ( q p ) = (r p ) = (s p ) = 1. Now we prove theorem 5.1 from this lemma. Assume for simplicity that K = and that we have infinitely many primes p as in the Lemma 3.1 such that p 1 = l where l is a prime. Take one of the three primes in the lemma, say, q. If the order of q equals l we get a contradiction to the fact that ( q ) = 1 by the theorem on cyclic groups p (the order of the squares subgroup of F p 1 p is = l). If ord(q) = l we are done. If not, the only possibility left is ord(q) = but this does not occur for sufficiently large primes p and hence q is a primitive root. Assume now that there eist c primes p as in Lemma 3.1 log such that p 1 = p 1 p. As before the order of q, r and s can be (if they are not primitive), p 1 or p. As before there is only a small number of cases where ord(q) =. Assume that ord(q) = p 1. For this case we need some observation. Let n be a natural number and Ω(n) denote the number of prime factors of n (with multiplicity) and write f g (or g f or f = O(g)), where g is a positive function, if there eists a constant c > 0 such that f() cg(). Then Observation 3.. (1) For any natural number n, Ω(n) log n. () Given an integer a, The number of primes p such that ord(a) < y (mod p) is O(y ). To see (1), use n = q α 1 1 qr αr α1 αr α α r = Ω(n). To see (), use Ω(a m 1) a log(a m 1) a m a y. m<y m<y m<y Now, if ord(q) = p 1 < 1/ δ (mod p), then by observation 3. this occurs for at most ( 1/ δ ) = 1 δ primes. Since 1 δ 0 as there are a negligible number of primes c/log p such that ord(q) = p 1 (mod p). This fact is also true for r and s. 8

9 Now, assume that q and r and s have order p. Since F p is a cyclic group, ord(< q, r, s >) = p < 3/4 ɛ. By Lemma in [9] the number of primes p such that ord(< q, r, s >) < y is O(y 4/3 ). So the number of primes p such that ord(< q, r, s >) = p < 3/4 ɛ is O( 1 4ɛ/3 ) and as before, is negligible in comparison to c. log 4. Artin s conjecture in a real quadratic field for units Let d 1 be a square-free natural number and let = d if d 1(mod 4) and = 4d otherwise. Let K = Q( ) be a real quadratic field and denote the integer ring of K by O K. The principal ideals po K that are generated by a rational prime p, take one of the following forms (1) po K = P (inert); () po K = P 1 P, P 1 P (splits); (3) po K = P (ramified) where P and P i are prime ideals in O K. We note that the option (3) occurs only in a finite number of cases and so does not interest us. Now, the norm map gives a homomorphism N : O K Z (O K /(p)) (Z/p) For any unit ɛ with N(ɛ) = 1 the kernel of this map contains the residue class ɛ modulo p. Denote this kernel by C ɛ (p). By Lemma 19 in [15] (appendi B) ord(c ɛ (p)) = { p 1, p splits p + 1, p inert Assuming GRH, Cooke and Weinberger ([8]) and Lenstra ([16]) showed that given a real quadratic field K, there are infinitely many split primes for which the fundamental unit of the field has maimal order (namely p 1) in C ɛ (p). Using the strong analytic theorem of Heath-Brown [13], Narkiewicz [19] proved the following unconditional theorem: Theorem 4.1. Let ɛ 1, ɛ, ɛ 3 be units in the integer rings O K1, O K, O K3 of K 1 = Q( 1 ), K = Q( ), K 3 = Q( 3 ), respectively, which 9

10 are not roots of unity. There is an inde j, 1 j 3, such that for infinitely many split primes p, c j ɛ j, c j = ±1, has order p 1 (mod (p)). For inert primes, one wants similar results. Under GRH, an analogue of [8] [16] was only proven recently by Roskam ([0]). We want to etend the result of Narkiewicz for inert primes. In this case the order of C ɛ (p) (mod p) is p + 1. So we cannot use the result of Heath-Brown on the divisors of p 1. We shall use a simpler method to get infinitely many primes p such that p+1 = P 3 (we write P 3 for an integer with at most three prime factors) but with almost same magnitude of the prime divisors. With this result we obtain: Theorem 4.. Let ɛ 1,...,ɛ 7 be units (assume that they have norm +1) in the rings of integers O 1,..., O 7 of Q( 1 ),...,Q( 7 ), respectively, which are not roots of unity, with 1,..., 7 multiplicatively independent and distinct from 3. Assume that all the numbers ( 1) a 1 3 a 7 i=1 b i i, a i, b i {0, 1}, are not perfect squares if 7 b i is odd. Then there eists an inde 1 j 7, such that for infinitely many inert primes p, the unit c j ɛ j, (c j = ±1), has order p + 1 modulo po j. Corollary 4.3. Let ɛ 1,...,ɛ 7 be units (assume that they have norm +1) in the rings of integers O 1,..., O 7 of Q( 1 ),...,Q( 7 ), respectively, which are not roots of unity, with 1,..., 7 primes distinct from 3. Then there eists an inde 1 j 7, such that for infinitely many inert primes p, the unit c j ɛ j, (c j = ±1) has order p+1 modulo po j Notation and Preliminaries. Now before we prove the theorem about the prime divisors of p + 1 (as in Lemma 3.1 for p 1) we need to decide on some notation. Let Π(y; m, s) denote the number of primes p such that p s (mod m) where m and s are some integers, and where Li(y) = y E(y; m, s) := Π(y; m, s) Li(y) ϕ(m) dt. Also set log t E(; m) := ma ma E(y; m, s). 1 y (s,m)=1 10 i=1

11 Define A = {p + 1 p, p u (mod v)} where u, v are some integers such that (u, v) = 1, u 1( mod ), 8 v, ( u+1, v) = 1 and take X = Li(). ϕ(v) For a square-free integer d, (d, v) = 1, let A d := {a A : a 0 mod d)} = {p + 1 : p, p u mod v, p 1 mod d} By the Chinese remainder theorem there eists an l such that A d = #{p + 1 p, p l (mod dv)}. By the definition of E(; dv, l), A d = Li 1 +E(; dv, l) = ϕ(dv) ϕ(d) Define ω(d) := d ϕ(d) and Li ϕ(v) R d := A d ω(d) X = E(; dv, l) d X +E(; dv, l) = +E(; dv, l) ϕ(d) Finally, we define two arithmetical functions for a square-free d = p 1 p k. µ(d) = ( 1) k and ν(d) = k (where µ(1) = 1 and ν(1) = 0). Now we want to prove two lemmas. Lemma 4.4. For any prime q, which is relatively prime to v we have: (4.1) 0 1 q c 1 where c 1 > 1 is some suitable constant. (4.) w q<z log q q 1 log z w where O does not depend on z or w. = O(1) ( w z) (4.3) <q<z q v (1 1 q 1 ) 1 log z. 11

12 Proof. Since q > it is clear that (7.3) holds. log q As for the second equation, = q 1 w q<z w q<z 1 ) = log q + log q = log z + O(1) q 1 q q(q 1) w w q<z w q<z O(1)). Hence we get (7.4). Finally, (1 ω(q) Since and Since <q<z we have <q<z q v q ) = <q<z q v = ep(log = ep( log q q (1 1 q 1 ) <q<z <q<z <q<z (1 1 q 1 )) log (1 1 q 1 )) q = q 1 ( p< ep( ( 1 q 1 1 (q 1) )) <q<z 1 q 1 = 1 q + 1 q(q 1) 1 q + 1 (q 1) 1 (q 1) converges, we get <q<z ep( (1 1 q 1 ) ep( <q<z <q<z 1 q log log z <q<z 1 q ). 1 1 ) ep( log log z) = q log z w q<z log q q (1+ log p p = log + (1 1 q 1 ) Lemma 4.5. For any natural square-free number d, (d, v) = 1, given an A > 0 there eist constants c ( 1) and c 3 ( 1) such that (4.4) µ (d)3 ν(d) X R d c 3, (X ) log A X d< X 1 (log ) c 1

13 Proof. Denote by S Rd the term which we need to estimate: S Rd = µ (d)3 ν(d) R d. d< X 1 (log ) c By the definitions of R d and E(; dv) S Rd µ (d)3 ν(d) E(; dv). d< X 1 (log ) c Since E(; dv) if d, we get that dv v S Rd µ (d)3 ν(d) E(; dv) 1 ( dv )1. Hence, S Rd 1 d< X 1 (log ) c By Cauchy s inequality, We have, S Rd 1 ( d< X 1 (log ) c S Rd 1 ( d<x 1 d< X 1 (log ) c µ (d)3 ν(d) E(; dv) 1. d 1 µ (d)3 ν(d) ) 1 ( d d< X 1 (log ) c µ (d)3 ν(d) ) 1 ( d dv< vx 1 (log ) c For sufficiently large we obtain S Rd 1 µ (d)3 ν(d) ( ) 1 ( d d< 1 dv< 1 (log ) c E(; dv) ) 1. E(; dv) ) 1. E(; dv) ) 1. With Bombieri-Vinogradov theorem ([5]) (given any positive constant e 1, there eist a positive constant e such that E(; d) = O( log e 1 )) for the last sum and since 13 d<w µ (d)9 ν(d) d d< 1 log e (log w + 1) 9 (see

14 [11], p.115, equation (6.7)) we find that for given constant B there eist c such that S Rd log B So, for given A there eist c such that where depends on v and c. S Rd X log A X 4.. Proof of Theorem 4. - the sieve part. In this section we will show that for a sufficiently small 0 < δ < 1/4 there eists some constant c(δ) > 0 (which depends on δ) such that for at least c(δ) log p+1 primes p, p u (mod v), = P 3 where q p+1 q > 1/4 δ. Later we will sharpen this result further Use of the lower bound linear sieve. In the following subsection we will show, using the linear sieve, that for a sufficiently small 0 < δ < 1/4 there eists some constant c 1 (δ) > 0 (which depends on δ) such that for at least c 1 (δ) primes p, p u (mod v), p+1 log has at most four prime divisors all of them greater than 1 4 δ. Define S(A, z, v) = #{a a A, (a, p) = 1} and let f denote p<z p v the lower bound function for the linear sieve which is defined as f(t) = e γ t 1 log(t 1) for t 4, where γ is Euler constant. Then (see [11, Theorem 8.4, page 36]): Lemma 4.6. Assume (7.3), (7.4) and (5.4). Then for X 1/8 < z < X 1/4 we have (4.5) S(A, z, v) X q<z q v (1 ω(q) log ){f( q logz ) + O( 1 log )} where the O-term does not depend on X or on z. Note 4.7. Obviously z influences the number of primes which divide the elements of A and their magnitude. Heath-Brown used a stronger version of this lemma which gives z = 1/4+ɛ 0 where ɛ 0 is a specific small real number. 14

15 By Lemmas 5.3 and 4.5, (7.3), (7.4) and (5.4) hold. Hence we can use Lemma 5.5 with z = X 1 4 δ. S(A, X 1 4 δ, v) X (1 1 q 1 ){f(1 log log ) + O( δ log )}. q<x 1 4 δ q v By Lemma 5.3 (5.3) we have S(A, X 1 4 δ, v) X log f( 1 4 δ 1 4δ ). But for t 4, f(t) = e γ t 1 log(t 1), and so, S(A, X 1 4 δ, v) X log 1/4 δ eγ ( 1 4δ ) log 1 + 4δ 1 4δ 1 + 4δ log log 1 4δ = log log(1 + 8δ 1 4δ ). Since log(1 + s)/s 1 as s 0 and for, 0 < δ < 1/4, 1 4δ are bounded, we have: Lemma 4.8. S(A, X 1 4 δ, v) δ log where the implied constant in does not depend on δ. Note 4.9. By definition of S(A, X 1 4 δ, v), for all sufficiently small 0 < δ < 1/4, there are δ primes p, such that any prime divisor log of p+1 (for a p in our sequence) is greater than X 1 4 δ or divides v. Since by our assumption ( u+1 Li(), v) = 1 where p u (mod v) and X = we ϕ(v) obtain that all odd prime divisor of p+1 are greater than 1 4 δ. Hence there are at most four prime divisors of p+1 which are greater than 1 4 δ. In the net subsection we will show that there are only a small number of primes p such that p+1 has eactly four prime divisors all of which are greater than 1 4 δ First use of the Selberg upper bound sieve. In order to prove that there are only a small number of primes p such that eactly four primes divide p+1 we need to use Selberg s upper bound sieve (see [11, theorem 3.1]): 15

16 Proposition Let a, b be integers satisfying ab 0, gcd(a, b) = 1, ab Then as we have uniformly in a, b that {p : p, ap + b = prime} 8 1 p>(1 p 1 log (p 1) ) p log {1 + O(log log )} <p ab From this proposition we derive the following: Lemma For any 0 < δ < 1/4, there eists c (δ) log (c (δ) > 0) primes p such that p+1 has at most three prime divisors all of which are greater than 1/4 δ. Proof. Assume that p+1 = p 1 p p 3 p 4, p where the p i are primes greater than 1/4 δ. Instead of counting the elements in this set we can count the products of primes p 1 p p 3 p 4 such that p 1 p p 3 p 4 1 = p where the p i s are primes greater than 1/4 δ. To count the latter set we use Proposition We take a = p 1 p p 3, b = 1 and Y = +1 p 1 p p 3 (since p 1 p p 3 p 4 1 p 4 +1 p 1 p p 3 ). By the Proposition 4.10, S p4 = #{p 4 Y : ap 4 + b = prime} = #{p p 1 p p 3 : p 1 p p 3 p 4 1 = prime} Since the p i s are big primes, the term one. Then S p4 + 1 p 1 p p 3 log +1 p 1 p p 3 p p 1 p p 3 p + 1 p 1 p p 3 log +1 p 1 p p 3 16 p 1 p p p 1 p p 3 p p 1 p is approimately

17 From the fact that for all i = 1,, 3, p i < 1/4+3δ we have for a sufficiently small δ S p4 1 p 1 p p 3 log 1 p 1 p p 3 ( 1/4+3δ ) 3 log ( 1 1 1/4 9δ 1/4 9δ ) p 1 p p 3 log Now we shall sum-up the last term over all possibilities for p 1, p, p 3. This number is bounded by Sp = 4 log p p 1 p 1 p p p 3 3 where the sum is over 1/4 δ < p i < 1/4+3δ, i = 1,, 3. 1 Observation 4.1. We have = log α +o(1). p β By observation 4.1, β <p< α Sp 3 1/4 + 3δ 4 log 1/4 δ log log3 (1 + 16δ 1 4δ ) log Since log(1 + s) = O(s) for 0 < s < 1 and for, 0 < δ < 1/4, 1 4δ is bounded, we have Sp 4 δ 3 log where does not depend on δ. Hence, Sp 4 is a small number in comparison to S(A, X 1 4 δ, v) δ. log Second use of Selberg s upper bound sieve. Up till now we know that for any sufficiently small number δ > 0, there are c (δ) primes log p such that p+1 has at most three prime divisors all of which are greater than 1/4 δ. In this section we want to prove the eistence of p+1 c 3 (δ) primes p, such that = P log 3 and if p+1 is a product of eactly three primes q 3 q q 1 then q 1 > 1/4 δ, q > 1/4+δ, q 3 > 1/3+δ. First we prove the claim about q (by the previous subsections it is clear that q 1 > 1/4 δ ). Assume that q 1 and q take values between 1 4 δ and 1 4 +δ. Instead of counting the number of primes p such that p+1 = q 1 q q 3 where q 1 and q are between 1 4 δ and 1 4 +δ, we shall count the products 17

18 q 1 q q 3 such that q 1 q q 3 1 = p where q 1 and q are between 1 4 δ and 1 4 +δ. To count this set we use Proposition Define a = q 1 q, b = 1 and Y = +1 q 1 q (since q 1 q q 3 1 q 3 +1 q 1 q ). By Proposition 4.10 S q3 = #{q 3 Y : aq 3 + b = prime} = #{q : q 1 q q 3 1 = prime} q 1 q + 1 p 1 q 1 q log +1 p q 1 q p q 1 q p As in the previous subsection, since the q i s are big primes the term is approimately one, so p q 1 q p p 1 p S q3 + 1 q 1 q log +1 q 1 q. Now we sum-up the last term over all possibilities for q 1, q. This number is bounded by, (see the previous subsection) Sq 3 = 1 1 log 1 + 8δ q log log 1 4δ. q δ q δ 4 1 δ q 1 4 +δ Since log 1+8δ = O(δ ), S 1 4δ q 3 = O(δ ). Hence for any δ sufficiently small we get a small number of primes p such that log p+1 = q 1 q q 3 where q 1 and q are between 1 4 δ and 1 4 +δ. Thus for most such p, we have q > 1/4+δ. Finally we prove the claim about q 3. Assume that p+1 = q 1 q q 3, q 3 q q 1 then we have that q 3 ( p+1 3. The following lemma sharpens )1 this result. Lemma For any 0 < δ < 1/4 there are at most O(δ ) log primes p for which ( p+1 )1 3 q 3 p 1 3 +δ where O does not depend on δ. Proof. Note that if p+1 then q log 3 ( p+1 )1 3 ( log ) δ for (δ) (the number of primes p for which p+1, is o( ) log log by the prime number theorem and so may be ignored). 18

19 Assume now that p+1 = q 1 q q 3 with 1/3 δ q 3 1/3+δ and 1/4+δ q 5/1+δ+δ (this is the maimum range which q can be in). Using Proposition 4.10, we take a = q q 3, b = 1, Y = +1 q q 3, and so S q1 = #{q 1 Y : aq 1 + b = prime} = #{q : q q 3 1 = prime} q q p 1 q q 3 log +1 p. q q 3 p q q 3 p Since 3/4+δ+δ is the maimum of q q 3 (q 1 > 1/4 δ ) we obtain S q1 q q 3 log 3/4+δ q q 3 log Now we sum-up the last term over all possibilities for q, q 3. this number is bounded by, (see the proof of Lemma 4.11) S q 1 = log 1 4 +δ q 5 1 +δ+δ 1 1 q q δ q δ 5/1 + δ + δ 1/3 + δ log log log 1/4 + δ 1/3 δ log log(1 + 6δ 1 3δ ) = O(δ log ) and for a sufficiently small δ we can ignore this number. By the same method (see Lemma 3 in [13]) there are only O(δ ) log primes p such that p+1 = r 1 r where r i s are primes, i = 1,, r r 1, p 1/ δ r 1 ( p+1 )1/. If we summarize this section we conclude that for any sufficiently small 0 < δ < 1/4 there are at least c 3 (δ) c 3(δ) > 0, primes log, p, p u (mod v) such that we can factor p+1 in at least one of the following ways: (1) p+1 is a prime number. () p+1 = r 1 (p)r (p) where r 1 (p), r (p) are some prime numbers, p 1/4 δ < r 1 (p) < p 1/ δ, p 1/+δ < r (p) < p 3/4+δ. 19

20 (3) p+1 = q 1 (p)q (p)q 3 (p) where q 1 (p) q (p) q 3 (p) are some prime numbers, q 1 (p) > p 1/4 δ, q (p) > p 1/4+δ, q 3 (p) > p 1/3+δ Proof of Theorem 4.- The algebraic part Construction of the arithmetic sequence. In this section we want to construct integers u and v, (u, v) = 1 such that for all primes p such that p u (mod v), the discriminants 1,..., 7, i 3 of Q( 1 ),..., Q( 7 ), respectively, satisfy ( 1 p ) = ( p ) =... = ( 7 p ) = 1. This means that p is inert simultaneously in all of the fields. In addition we want to insure that p+1 will be an odd integer and so we take u 1 (mod 4) where 8 v. Finally, to get ( p+1, v) = 1 we shall construct u and v so that ( u+1, v) = 1 (since after sieving the small factors of p+1 we may be left with small factors which divide v, see previous section). In order to fulfill these demands, we will first show that there eist infinitely many primes p with the following simultaneous conditions (4.6) ( 1 p ) = (3 p ) = 1 and ( 1 p ) = ( p ) =... = ( 7 p ) = 1 This condition is equivalent to the condition: B(p) = (1 + ( 1 p ))(1 + (3 p ))(1 ( 1 p )) (1 ( 7 )) 0. p Since the Legendre symbol is a multiplicative function, we obtain, B(p) = (1+( 1 p ))(1+(3 p ))(1 Σ( i p )+Σ( i j )... ( 1 7 )) p p Let S be the set of all integers of the form n = ( 1) a a b i i, a i, b i {0, 1}. Then (4.7) B(p) = ( 1) b b 7 ( n), b p i {0, 1} p Z n S By the assumption in the theorem (see the introduction) each n S is not a square when 7 b i is odd. i=1 0 p Z i=1

21 This assumption, together with the fact that for n not a perfect square (by reciprocity law for Legendre symbol), ( n ) = o(π(z)) as Z p p Z p Z implies that B(p) is asymptotic to at least π(z) (since all the negative summands contribute o(π(z)) and at least the natural number 1 contributes π(z)). This shows that the simultaneous conditions have infinitely many solutions p. We fi some particular p 0 satisfying the condition. We define u = p 0 and for each odd prime l, such that l 1 7 we define u l = p 0 if l p and u l = 4p 0 otherwise. Claim l u l + 1 Proof. If u l = p 0 then by the assumption l p 0 +1, l u l +1. If u l = 4p 0, assume, by reductio ad absurdum, that l u l + 1. Hence l 4p Because u l = 4p 0 and l p 0 + 1, we obtain that l 3p 0. On the other hand, by our condition, ( 3 p 0 ) = 1 so ( p 0 ) = 1 (p (mod 4)). Hence p 0 1 (mod 3). Since l p and p 0 1 (mod 3) we conclude that l 3. Using the assumption that l p we deduce that l p 0 (if l = p 0 then l p 0 + 1). Hence l 3p 0, a contradiction. Let v = and u be the common solution of u u (mod 8) and all the congruences u u l (mod l). Such a solution eists, by the Chinese Remainder Theorem. Since l u+1 for every odd prime l v and the fact that u 1 (mod 4) (by the construction u u (mod 8) where u = p 0 1 (mod 4)) we conclude that ( u+1, v) = 1. Finally, if p u (mod v) then p p 0 (mod 8) and p p 0 or 4p 0 (mod l) for all odd primes l v. So, ( 1 ) = p ( 1 p 0 ) = 1, and similarly for all i s. This completes the construction of u and v. Note that by the construction of the integers u and v we have that (u, v) = 1. (Take l an odd prime number, l v = and assume that l u. Since u u l (mod l), l u l, hence l p 0 or 4p 0 ; in other words l = p 0. But p (p 0 fulfills the simultaneous condition (5.6)) and l 1 7 ). 1

22 4.3.. The last step of the proof. For the last step of the proof we need to use Lemma 4 from Narkiewicz [18], which generalized Lemma in [9]; Lemma If a 1,...a k are multiplicatively independent integers of an algebraic number-field K, G the subgroup of K generated by a 1,...a k, and for any prime ideal P not dividing a 1, a k we denote by G P the reduction of G (mod P), then for all positive y one can have #G P < y for at most O(y 1+1 k) prime ideals P, with the implied constant being dependent on the a i s and K. Now, as we saw at the end of section 3, for any sufficiently small 0 < δ < 1/4 there is some constant c 3 (δ) > 0 such that for c 3 (δ) log primes p, p u (mod v) at least one of the following occur: (1) p+1 is a prime number. () p+1 = r 1 (p)r (p) where r 1 (p), r (p) are primes so that, p 1/4 δ < r 1 (p) < p 1/ δ, p 1/+δ < r (p) < p 3/4+δ. (3) p+1 = q 1 (p)q (p)q 3 (p) where q 1 (p) q (p) q 3 (p) are some prime numbers, q 1 (p) > p 1/4 δ, q (p) > p 1/4+δ, q 3 (p) > p 1/3+δ It is clear by the construction of u and v that p 1 (mod 4). Because #C ɛ (p) = p+1 when p is inert in Q( ) the unit 1 is a non-square in the group C ɛ (p). Hence for any unit ɛ, we can choose constant c = ±1 such that cɛ is a non-square in C ɛ (p). Similarly, since cɛ is a non-square and the inde of the group of squares is, the order of cɛ is even. Now we look at our cases: (1) In this case, by the above note, cɛ, if not primitive, has order But the number of p s with this property is O(1) (by Lemma 5.7). () Let c 1 ɛ 1,...,c 4 ɛ 4, be units in the orders O 1,...,O 4 of Q( 1 ),...,Q( 4 ), i 3, i = 1,, 3, 4, respectively. We will show that one of them is primitive for infinitely many primes. If ord(c i ɛ i ) mod(p) = r 1 (p) < 1/ δ for some i = 1,, 3, 4 by Lemma 5.7 this occurs for at most O( 1/ δ ) = 1 δ

23 primes p and this is a negligible number compared to c 3 (δ). log Assume ord(c i ɛ i ) mod(p) = r (p), i = 1,, 3, 4. Consider the ring of integers O M of the compositum field M of Q( i ) i = 1,..., 4. Proposition For any prime ideal P (p) = po M : (O M /P) (O 1 /po 1 )... (O 4 /po 4 ). Proof. Since p is inert in each Q( i ), the order of O i /po i is p, i.e., [O i /po i : Z/pZ] =. Since all these residue fields are finite fields and two finite fields with the same number of elements are isomorphic, it is enough to show that, f = [O M /P : Z/pZ] =. Consider the Galois group G = Gal[M/Q] and define two subgroups of G, the decomposition group D and the inertia group E: and D = D(P (p)) = {σ G σ(p) = P } E = E(P (p)) = {σ G σ(α) α (mod P), α Z}. Now, consider the Galois group Ḡ, Ḡ = Gal[O M /P / Z/pZ] By [17, beginning of Chapter 4], D/E Ḡ By theorem 8 in in [17] (since (p) is inert in all the fields Q( i ), (p) is unramified in all the fields Q( i ). Hence (p) is also unramified in O M, i.e., e = the eponent of P in the decomposition of (p), is equal to 1) E = e = 1 D = f 3

24 Immediately we conclude that, D Ḡ Since Ḡ is a cyclic group, we get that D is a cyclic group of order f. But D is a subgroup of G and G = C... C where C is a group of order. So, f. Since [O i /po i : Z/pZ] = we also see that f. Hence f =. Because the compositum of normal etensions is normal, this claim is true for all the prime ideals P in the decomposition of (p) (they have the same e and the same f) By Proposition 4.16 < c 1 ɛ 1, c ɛ, c 3 ɛ 3, c 4 ɛ 4 > = r (p) p 3/4+δ in (O M /P) for (p) P (this is a cyclic group). By Lemma 5.7 the number of p that have this order is at most O( 3/4+δ ) 5 4 = O( 15/16+5δ/4 ) but we can choose δ to be as small as needed. Hence the number of primes p, p such that c i ɛ i i = 1,..., 4 have order r (p) is a small in comparison with c 3 (δ) log. (3) Denote by S (1,n) the set {c 1 ɛ 1,..., c n ɛ n } where c 1 ɛ 1,..., c n ɛ n are multiplicatively independent units and take four units c i1 ɛ i1,..., c i4 ɛ i4 from S (1,n) and assume that q 1 (p), which is greater than 1/4 δ, divides [C ɛ (p) : c ik ɛ ik ] for k = 1,..., 4. For any unit ɛ let be ɛ its image in C ɛ (p). Then c ik ɛ ik = p+1 [C ɛ(p): c ik ɛ p 3/4+δ 3/4+δ for k = 1,..., 7. Since q ik ] 1 (p) divides [C ɛ (p) : c ik ɛ ik ] for k = 1,..., 4 and C ɛ (p) is a cyclic group, c i1 ɛ i1,..., c i4 ɛ i4 mod p 3/4+δ. By Lemma 5.7 it occurs in at most O(( 3/4+δ ) 5/4 ) = O( 15/16+5δ/4 ) primes p which is a negligible number relatively to c 3 (δ) for sufficiently small δ. log So, for at most three units from c i1 ɛ i1,..., c i4 ɛ i4, q 1 (p) divides [C ɛ (p) : c ik ɛ ik ] for k = 1,..., 4. In other words for at most three from c i1 ɛ i1,..., c i4 ɛ i4 q 1 (p) does not divide c ik ɛ ik. Hence 4

25 for at least one unit, say c n ɛ n, q 1 (p) divide c n ɛ n. Denote by S (1,n 1) the set {c 1 ɛ 1,..., c n 1 ɛ n 1 }. By repeating the former process for S (1,n 1) we obtain that for at least one integer, say c n 1 ɛ n 1, q 1 (p) divide c n 1 ɛ n 1. We continue this process till we obtain the set S (4,n) = {c 4 ɛ 4,..., c n ɛ n } where c 4 ɛ 4,..., c n ɛ n are multiplicatively independent units such that q 1 (p) divides c t ɛ t for t = 4,..., n. Now, we are repeating the process which we use to q 1 (p) and S (1,n) for q (p) > p 1/4+δ and the set S (4,n) = {c 4 ɛ 4,..., c n ɛ n }. Take three units c i1 ɛ i1,..., c i3 ɛ i3 from S (4,n) and assume that q (p), which is greater than 1/4+δ, divides [C ɛ (p) : c ik ɛ ik ] for k = 1,, 3. p+1 Then c ik ɛ ik = [C ɛ(p): c ik ɛ p 3/4 δ 3/4 δ for k = ik ] 1,, 3. Since q (p) divides [C ɛ (p) : c ik ɛ ik ] for k = 1,, 3 and C ɛ (p) is a cyclic group, c i1 ɛ i1,..., c i3 ɛ i3 mod p 3/4+δ. By Lemma 5.7 it occurs in at most O(( 3/4 δ ) 4/3 ) = O( 1 8δ/3 ) primes p which is a negligible number relatively to c 3 (δ) log for sufficiently small δ. So, for at most two units from c i1 ɛ i1,..., c i3 ɛ i3, q (p) divides [C ɛ (p) : c ik ɛ ik ] for k = 1,, 3. In other words for at most two from c i1 ɛ i1,..., c i3 ɛ i3, q (p) does not divide c ik ɛ ik. Hence for at least one integer, say c n ɛ n, q (p) divide c n ɛ n. Denote by S (4,n 1) the set c 4 ɛ 4,..., c n 1 ɛ n 1. By repeating the former process for S n 1 we obtain that for at least one unit, say ɛ n 1, q (p) divide ɛ n 1. We continue this process till we obtain the set S (6,n) = {c 6 ɛ 6,..., c n ɛ n } where c 6 ɛ 6,..., c n ɛ n are multiplicatively independent units such 5

26 that q (p) divides c t ɛ t, t = 6,..., n and in fact q 1 (p)q (p) divides c t ɛ t for t = 6,..., n. Finally, We repeating this process for q 3 (p) > p 1/3+δ and the set S (6,n) = {c 6 ɛ 6,..., c n ɛ n }. Take two units c i1 ɛ i1, c i ɛ i from S (6,n) and assume that q 3 (p), which is greater than 1/3+δ, divides [C ɛ (p) : c ik ɛ ik ] for k = 1,. p+1 Then c ik ɛ ik = [C ɛ(p): c ik ɛ ik ] p/3 δ /3 δ, k = 1,. Since q 3 (p) divides [C ɛ (p) : c ik ɛ ik ] for k = 1, and C ɛ (p) is a cyclic group, c i1 ɛ i1, c i ɛ i mod p /3 δ. By Lemma 5.7 it occurs in at most O(( /3 δ ) 3/ ) = O( 1 3δ / ) primes p which is a negligible number relatively to c 3 (δ) for sufficiently small log δ. So, for at most one unit from c i1 ɛ i1, c i ɛ i, q 3 (p) divides [C ɛ (p) : c ik ɛ ik ] for k = 1,. In other words for at most one from c i1 ɛ i1, c i ɛ i q 3 (p) does not divide c ik ɛ ik. Hence for at least one integer, say c n ɛ n, q 3 (p) divide c n ɛ n. Denote by S (6,n 1) the set {c 6 ɛ 6,..., c n 1 ɛ n 1 }. By repeating the former process for S n 1 we obtain that for at least one unit, say ɛ n 1, q 3 (p) divide ɛ n 1. We continue this process till we obtain the set S (7,n) = {c 7 ɛ 7,..., c n ɛ n } where c 7 ɛ 7,..., c n ɛ n are multiplicatively independent units such that q 3 (p) divides c t ɛ t for t = 7,..., n and in fact q 1 (p)q (p)q 3 (p) divides c t ɛ t for t = 7,..., n. Hence if we take n = 7 multiplicatively independent integers we obtain that one of them have at least the order p+1. Since we choose the c j s such that the c jɛ js are quadratics non-residue we obtain that one of them have at least the order p + 1 which completes the proof of the Theorem 4. 6

27 Note that Theorem 4. implies Corollary Artin s conjecture for nonunits In this chapter we want to etend the result of the previous chapter for any algebraic integer modulo inert primes p. We will prove Theorem 5.1. Let K = Q( ) be a quadratic field and {α i } 85 i=1 be a set of 85 integers of K such that (1) N(α i ) = α i σ(α i ), the norms of the α is, are multiplicatively independent. () 5N(α i ), N(α i ) are not perfect squares. (3) M(α i ) = σ(α i )/α i are multiplicatively independent. Then at least one of the 85 integers has at least order p 1 4 mod p for infinitely many inert primes p in K. Note that in the case of split primes, Narkiewicz ([18]) proved a much stronger result. Since in our case the order is p 1, which is not linear, their divisors are too big and we can not use the method of [1]. But since p 1 = (p 1)(p+1) can be factored into two linear factors, with the following remark we can still use the method of [1] Remark 5.. Consider an algebraic number α in K = Q( ). Let p α be an inert prime in K. Since: (1) M(α) α p 1 (mod (p)) () N(α) α p+1 (mod (p)) 7

28 We have: ord(m(α)) ord(α) (mod (p)) and ord(n(α)) ord(α) (mod (p)) In addition: ord(m(α)) p + 1 and ord(n(α)) p 1 But, ( p 1 p+1, p + 1) = 1 or (p 1, ) = 1 So, ord(m(α))ord(n(α)) ord(α) (mod (p)) Let e 1 and e be some integers. If we prove that, M(α) and N(α) have simultaneously at least orders p+1 e 1 and p 1 e, respectively, then we will obtain that α have at least order p 1 e 1 e. With this way we reduce the problem to a linear problem. As in the former chapter we need to decide on some notation for the lemma from sieve methods Notation and Preliminaries. Let π(y; m, s) denote the number of primes p y such that p s (mod m) where m and s are some integers, and E(y; m, s) := π(y; m, s) Li(y) ϕ(m) where Li(y) = y dt. Also set log t E(; m) := ma 1 y ma E(y; m, s). (s,m)=1 Define A = {p 1 p, p u (mod v)} where u, v are some given integers such that (u, v) = 1 and take X = Li() ϕ(v). 8

29 For a square-free integer d, (d, v) = 1, denote A d := {a A : a 0 mod d} = {p 1 : p, p u mod v, p 1 0 mod d} = = d m=1 m 1 0 mod d d m=1 m 1 0 mod d (m,d)=1 {p p, p u mod v, p m {p p, p u mod v, p m mod d} mod d} By the Chinese Remainder Theorem, for each m there eists an integer l m such that A d = d m=1 m 1 0 mod d (m,d)=1 {p p, p l m (mod dv)} ; Since {p p, p l m (mod dv)} is asymptotically independent of m, there eists some integer l such that where ρ(d) = A d = π(; dv, l) d m=1 m 1 0 mod d (m,d)=1 1. d m=1 m 1 0 mod d (m,d)=1 1 = π(; dv, l)ρ(d) We note that for any prime q, ρ(q) = and hence for any square-free d, ρ(d) = ν(d) where ν(d) denotes the number of prime divisors of d. By the definition of E(; dv, l), A d = ρ(d) Li ϕ(d) ϕ(v) + ρ(d)e(; dv, l)) = ν(d) ϕ(d) X + ν(d) E(; dv, l) 9

30 For any prime q define ω(q) := q, ω(d) = ω(q) = ν(d) d and ϕ(q) ϕ(d) q d R d := A d ω(d) d X = ν(d) E(; dv, l) Finally, we define the Möbius function, µ(1) = 1 and for a square-free d = p 1 p k, µ(d) = ( 1) k. Now we want to prove two lemmas. Lemma 5.3. For any prime q > 3 which is relatively prime to v we have: (5.1) 0 q 1 1. (5.) w q<z q 1 log q log z w = O(1) ( w z) where O does not depend on z or w. (5.3) <q<z q v (1 q 1 ) 1 log z. Proof. Since q > 3, it is clear that (7.3) holds. As for the second equation, log q = q 1 w q<z w q<z 1 ) = log q + log q = log z + O(1) q 1 q q(q 1) w w q<z log + O(1)). w q<z 30 log q q q = q 1 w q<z ( log p = p p< log q q (1+

31 Hence we get (7.4). Finally, (1 q 1 ) <q<z q v Since and Since <q<z we have <q<z (1 q 1 ) = ep(log = ep( <q<z <q<z (1 q 1 )) log (1 q 1 )) ep( ( q 1 4 (q 1) )) <q<z q 1 = q + q(q 1) q + (q 1) converges, we get (q 1) <q<z ep( <q<z (1 q 1 ) ep( <q<z q log log z <q<z q ) 1 ) ep( log log z) = q log z Lemma 5.4. For any square-free natural number d, (d, v) = 1, and a real number A > 0, there eist constants c ( 1) and c 3 ( 1) such that (5.4) µ (d)3 ν(d) X R d c 3, (X ) log A X d< X 1 (log ) c Proof. (See Lemma 4.5) Denote by S Rd the term which we need to estimate. S Rd = µ (d)3 ν(d) R d d< X 1 (log ) c 31

32 By the definitions of R d and E(; dv) S Rd d< X 1 (log ) c µ (d)6 ν(d) E(; dv). Since E(; dv) if d, we get that dv v S Rd 1 By Cauchy s Inequality, d< X 1 (log ) c µ (d)6 ν(d) E(; dv) 1. d 1 S Rd 1 ( d<x 1 µ (d)6 ν(d) ) 1 ( d dv< vx 1 (log ) c E(; dv) ) 1. For sufficiently large we obtain S Rd 1 ( d< 1 µ (d)6 ν(d) ) 1 ( d dv< 1 (log ) c E(; dv) ) 1. With Bombieri-Vinogradov Theorem ([5]) (given any positive constant e 1, there eists a positive constant e such that E(; d) = O( log e 1 )) for the last sum and the inequality d<w d< 1 log e µ (d)36 ν(d) d (log w + 1) 36 (see [11], p.115, equation (6.7)) we find that for given constant B there eists c such that S Rd log B. So, for given A there eists c such that S Rd X log A X where depends on v and c. 3

33 5.. Proof of Theorem the sieve part. In this section we use the Selberg lower bound sieve and show that there is some small real number δ 1 and some constant c(δ 1 ) > 0 (which depends on δ 1 ) such that for at least c(δ 1 ) primes p, p u (mod v), if log 3 q p 1 then either q > 1/8+δ 1 or q v. Now, define S(A, z, v) = {a a A, (a, p) = 1} and define a p<z p v function g by g(t 0 ) = 1, t 0 = 4.4 and g(t) < 1 for t > t 0. Then (see [11, Theorem 7.4, page 19]): Lemma 5.5. We have (5.5) S(A, z, v) X q<z q v (1 ω(q) log X ){1 g( q where the O-term does not depend on X or on z. q<x 1 8 +δ 0 q v log z ) + O((loglog3X)8 )} logx By Lemmas 5.3 and 4.5, (7.3), (7.4) and (5.4) hold. Hence we can use Lemma 5.5 with z = X 1 8 +δ 0 S(A, X 1 8 +δ 0, v) X (1 q 1 ){1 g(1 log X )+O( (loglog3x)8 )}. log X 1 8 +δ 0 logx By Lemma 5.3 (5.3) we have for δ 0 sufficiently small S(A, X 1 8 +δ 0, v) X log X log 3 Thus for such δ 0 we obtain that there is a constant c(δ 0 ) > 0 (which depends on δ 0 ) such that for at least c(δ 0 ) primes p, p log 3 u (mod v) if q p 1 then either q > 1/8+δ 0 or q v. Hence we obtain that for all 0 < δ 1 < δ 0 there is a constant c(δ 1 ) > 0 (which depends on δ 1 ) such that for at least c(δ 1 ) primes p, p u (mod v), if log 3 q p 1 then either q > 1/8+δ 1 or q v Proof of Theorem The algebraic part Construction of the arithmetic sequence. Let K = Q( ) be any quadratic field, O the integers ring of K, α O any algebraic integer and a = N(α). In this section we want to construct integers u 33

34 and v, (u, v) = 1 such that for all primes p such that p u (mod v), the discriminant of Q( ) and a satisfy ( p ) = (a p ) = 1. This means that p is inert and a is not a quadratic residue (mod p). In addition we want to obtain by the construction that ( p 1, v) = 1 4 (since after sieving the small factors of p 1 we may be left with small 4 factors which divide v, see previous section). In order to fulfill these demands, we will first show that there eist infinitely many primes p satisfying the following simultaneous conditions (5.6) ( 1 p ) = (5 p ) = (a p ) = ( p ) = 1 This condition is equivalent to the condition: B(p) = (1 ( 1 p ))(1 (5 p ))(1 (a p ))(1 ( p )) 0 Since the Legendre symbol is a multiplicative function, we obtain, (1 ( 1 p ))(1 (a p ) ( p ) + (a p ) (5 p ) + (5a p ) + (5 p ) (5a p )) Let S be the set of all integers of the form n = ( 1) b 0 5 b 1 a b b 3, b i {0, 1}. Then (5.7) B(p) = ( 1) b 0+b 1 +b +b 3 ( n ), b p i {0, 1} p Z n S p Z By the assumption in the theorem each n S is not a perfect square when 3 b i is odd. i=0 p Z This assumption, together with the fact that for n not a perfect square (by reciprocity law for Legendre symbol) ( n ) = o(π(z)) as Z p p Z implies that B(p) is asymptotic to at least π(z) (since all the negative summands contribute o(π(z)) and at least the natural number 1 contributes π(z)). This shows that the simultaneous conditions have infinitely many solutions p. 34

35 We fi some particular p 0 satisfying the condition (5.6) and for each odd prime l 3, such that l 4a we define u l = p 0 if l p 0 1, and u l = 9p 0 otherwise. Claim 5.6. l u l 1 Proof. If u l = p 0 then by the assumption l p 0 1, so l u l 1. If u l = 9p 0, assume, by reductio ad absurdum, that l u l 1. Hence l 81p 0 1. Since, l p 0 1, we obtain that l 80p 0. On the other hand, by our condition, ( 5 p 0 ) = 1 so ( p 0 ) = 1 (p (mod 4)). Hence p 0 or 3 (mod 5). Since l p 0 1 and p 0 or 3 (mod 5) we conclude that l 5. Using the assumption that l p 0 1 we deduce that l p 0 (if l = p 0 then l p 0 1). Hence (l, 3) l 80p 0, which is a contradiction. In addition let u = p 0 if 8 p 0 1 and u = p 0 8 if 16 p 0 1. Likewise we take u 3 = p 0 if 3 p 0 1 and u 3 = p 0 3 if 9 p 0 1. Let v = 4a and u be the common solution of u u (mod 16), u u 3 (mod 9) and all the congruences u u l (mod l). Such a solution eists, by the Chinese Remainder Theorem. Since l u 1 for every odd prime l, 3, l v, and by the construction ( u 1, 6) = 1 we conclude that 1, v) = 1. 4 (u 4 Finally, if p u (mod v), then p p 0 (mod 4) and p p 0 or 4p 0 (mod l) for all odd primes l v. So, ( p ) = ( p 0 ) = 1, and similarly for a. This completes the construction of u and v. Note that by the construction of the integers u and v we have that (u, v) = 1. (take l an odd prime number, l v = 4a and assume that l u. Since u u l (mod l), l u l. Hence l p 0 or 9p 0 (in this case l, 3). In other words l = p 0. But p 0 4a (p 0 fulfills the simultaneous condition (5.6)) and l 4a ) The last step of the proof. As we saw at the previous subsections, for at least c(δ 1 ) log 3 primes p, p u (mod v), if q p 1 35

36 then q > 1/8+δ 1 or q v. Since ( p 1 4, v) = 1, if q p 1 4 then q > 1/8+δ 1. Since by the construction of u and v, p 1 (mod 4) and ( p 1, v) = 1, 4 we have that p 1 and p+1 are odd. In addition if p 1 (mod 3) then 4 ( p 1 1, v) = 1 and if p 1 (mod 3) then (p+1 6, v) = 1 If we conclude the result about p 1 and p + 1 we have for at least p 1 c 1 (δ 1 ) primes p, p u (mod v), if q log 3 d or q p+1 d + then q > 1/8+δ 1, where d = 4 or 1 and d + = 6 or, respectively. For the last step of the proof we need to use a version of Lemma 4 from Narkiewicz [18], which generalized Lemma in [9]. Lemma 5.7. If a 1,...a k are multiplicatively independent algebraic numbers of an algebraic number-field K, G the subgroup of K generated by a 1,...a k, and for any prime ideal P not dividing a 1, a k we denote by G P the reduction of G (mod P), then for all positive y one can have G P < y for at most O(y 1+1 k) prime ideals P, with the implied constant being dependent on the a i s and K. Proof. According to [18], For any real number T denote by M = M(T) the set of all k-element sequences (r 1,..., r k ) of non-negative integers satisfying r 1 + r r k T It is easy to see that for T tending to infinity M(T) = (c + o(1))t k with suitable positive constant c = c k. If now P is a prime ideal for which G P < y, then select T to be the smallest rational integer with ct k > y and let a i = b i c i for i = 1,..., k. There eist two distinct sequences Z = (z i ), W = (w i ) in M(T) for which P bz 1 1 bz k k c z 1 1 c z k k bw 1 1 b w k k b w 1 1 c w k Hence for sufficient large P (since the a is are multiplicatively independent). 36 k

37 P bz 1 1 b z k k bw 1 1 b w k k c [z 1,w 1 ] 1 c [z k,w k ] k Thus for sufficient large P = D 0 ν P (Πa z i w i i 1) 0 where ν P denotes the P-adic valuation and it follows that for fied z 1 w 1,..., z k w k we obtain log (ma j ā j T ) T possibilities for P. Finally we obtain {P G P < y} T 1+k y 1+ 1 k Look at the p 1 case (the case of p + 1 is similar). We have for at least c 1 (δ 1 ) log 3 primes p, p u (mod v) such that if q p 1 d then q > 1/8+δ 1 where d = 4 or 1. Let p 1 = d q 1 (p)q (p) q m (p), q m (p) > q m 1 (p) >... > q 1 (p), m 7, and let a be some integer where ā its image in F p. Denote by S n the set S n = {a 1,..., a n } where a 1,..., a n are multiplicatively independent integers and take seven integers a i1,..., a i7 from S n and assume that at least one prime, say q 1 (p), which is greater than 1/8+δ 1, divides [F p : ā i k ] for k = 1,..., 7. p 1 Then ā ik = [F p: ā ik p 7/8 δ 1 7/8 δ 1, k = 1,..., 7. Since ] q 1 (p) divides F p : ā i k for k = 1,..., 7 and F p is a cyclic group, a i1,..., a i7 mod p 7/8 δ 1. By Lemma 5.7 it occurs in at most O(( 7/8 δ 1 ) 8/7 ) = O( 1 8/7δ 1 ) primes p which is a negligible number relatively to c 1 (δ 1 ) for sufficiently small δ log 3 1. So, for at most si integers from a i1,..., a i7, q 1 (p) divides [F p : a ik ] for k = 1,..., 7. In other words for at most si from a i1,..., a i7 q 1 (p) does not divide ā ik. Hence for at least one integer, say a n, q 1 (p) divide ā n. 37

38 Denote by S n 1 the set {a 1,..., a n 1 }. By repeating the former process for S n 1 we obtain that for at least one integer, say a n 1, q 1 (p) divide ā n 1. We continue this process till we obtain the set T 1 = {a 7,..., a n } where a 7,..., a n are multiplicatively independent integers such that q 1 (p) divides ā t for t = 7,..., n. By repeating this process for q (p) we obtain for the set T = {a 13,..., a n } that q (p) divides ā t for t = 13,..., n. Again, by repeating this process for q m (p) we obtain for the set T m = {a 6m+1,..., a n } that q m (p) divides ā t for t = 6m + 1,..., n. Since the maimum value of m is 7, if we take n = 6m + 1 = 43 multiplicatively independent integers we obtain that one of them have at least the order p 1 d Let us look on the algebraic number M(α). By the same method we obtain that one of 43 M(α) have at least the order p+1 d +. one of = 85 has, at least, the order p 1. This completes the proof of 4 Theorem Gcd of eponent functions As we mentioned, Gupta and Murty s method to attack the former problems raises question regarding gcd(a n 1, b n 1) where a, b are multiplicatively independent rational positive integers. We prove a result on the divisors of gcd (a n + 1, b n + 1) which is the same to APR result [3] for gcd (a n 1, b n 1). Theorem 6.1. Let a and b be two multiplicatively independent integers. Then there eist infinitely many integers n such that (a n + 1, b n + 1) > ep(ep(c log n/ log log n)) where c is some positive constant. 38

39 Since the proof is based on the work of APR we now give the principal proof of this result. The proof is based on the following simple fact: Let a be any integer. By the Little Fermat Theorem, if (a, p) = 1, where p is a prime number, then (6.1) a p 1 1 (mod p) (or p (a p 1 1)). It is clear that (6.1) is true for all integer n with p 1 n. So, for all integers n with p 1 n the following holds (6.) a n 1 (mod p) (or p a n 1). Hence, for finding big divisor of a n 1 it is enough to find a lot of primes p where p 1 n. Let K be a square-free positive integer (which will be a product of a lot of small primes). We want to count the number of integers m and primes p such that (6.3) m(p 1) 0 (mod K) Let A be the number of solutions of the congruence (6.3) where m and p. To estimate A, for all d K define the integer A d to be the number of solutions of (6.3) where d p 1 and (m, K) = K/d. Now, we estimate the integer A d By APR: {p p 1 square free and d p 1} > 10d log 39