Primes, Polynomials, Progressions. B.Sury Indian Statistical Institute Bangalore NISER Bhubaneshwar February 15, PDF Free Download

Indian Statistical Institute Bangalore NISER Bhubaneshwar February 15, 2016

Primes and polynomials It is unknown if there exist infinitely many prime numbers of the form n 2 + 1.

Primes and polynomials It is unknown if there exist infinitely many prime numbers of the form n 2 + 1. As a curiosity, let us note in passing that if there are infinitely many prime numbers of the form n 2 + 1, then the twin prime conjecture will be solved in the ring Z[i] of Gaussian integers.

The question as to which primes p divide integers of the form n 2 + 1 has an easy answer.

The question as to which primes p divide integers of the form n 2 + 1 has an easy answer. In general, consider any irreducible integral polynomial f of degree > 1. Then, it is an easy exercise that: There must be infinitely many primes p such that f does not have roots modulo p. This is easy to prove elementarily but we will give a non-elementary proof which can take us much further. For instance, that approach can answer more general questions like:

If f is an irreducible integral polynomial, is it necessarily irreducible modulo some prime? Modulo infinitely many primes?

If f is an irreducible integral polynomial, is it necessarily irreducible modulo some prime? Modulo infinitely many primes? The answer to this is:

If f is an irreducible integral polynomial, is it necessarily irreducible modulo some prime? Modulo infinitely many primes? The answer to this is: yes!...

If f is an irreducible integral polynomial, is it necessarily irreducible modulo some prime? Modulo infinitely many primes? The answer to this is: yes!... and:

If f is an irreducible integral polynomial, is it necessarily irreducible modulo some prime? Modulo infinitely many primes? The answer to this is: yes!... and: NO!

It is yes if the degree is prime and no if it is composite!

It is yes if the degree is prime and no if it is composite! There exist irreducible, integer polynomials of any composite degree which are reducible modulo every prime p.

It is yes if the degree is prime and no if it is composite! There exist irreducible, integer polynomials of any composite degree which are reducible modulo every prime p. In what follows, we show how such questions are attacked.

Hilbert was the first to find examples in degree 4. One such is the polynomial x 4 10x 2 + 1.

Hilbert was the first to find examples in degree 4. One such is the polynomial x 4 10x 2 + 1. A general example is given by: ( ) ( ) Let p, q be odd primes such that p q = q p = 1 and p 1(8). Then, the polynomial (X 2 p q) 2 4pq is irreducible whereas it is reducible modulo any integer.

To make precise statements about infinite sets of primes, it is useful to have a notion of density. One such notion is:

To make precise statements about infinite sets of primes, it is useful to have a notion of density. One such notion is: A set S of prime numbers is said to have density δ if p S 1/ps all p 1/ps δ as s 1 +. So, any finite set of primes has density 0. To use this in our situation, recall from basic Galois theory that if f is an irreducible, integral polynomial of degree n, its Galois group is a subgroup of the permutation group S n, permuting the roots transitively.

A density theorem due to Frobenius implies:

A density theorem due to Frobenius implies: Let f be a monic integral, irreducible polynomial. The set of prime numbers p such that the polynomial f modulo p decomposes as a product of irreducible polynomials of degrees n 1, n 2,, n r, has a well-defined density. This density equals N/O(Gal(f )) where N is the number of elements σ in Gal(f) which have a cycle pattern n 1, n 2,, n r.

Here s a way to show that for an irreducible integral polynomial f of degree > 1, there are infinitely many primes which do not divide any of the values f (n).

Here s a way to show that for an irreducible integral polynomial f of degree > 1, there are infinitely many primes which do not divide any of the values f (n). If f has roots modulo all but finitely many primes p, the Frobenius theorem shows that each σ in Gal(f ) has a cycle pattern of the form 1, n 2, n 3, So, each element of Gal(f ) fixes a root of f. Since the roots of f are transitively moved around by Gal(f ), this Galois group would be the union of the conjugates of its subgroup H consisting of elements which fix a particular root of f.

An amusing application is the solution of a problem which appeared in an International Mathematical Olympiad:

An amusing application is the solution of a problem which appeared in an International Mathematical Olympiad: If p is a prime number, show that there is another prime number q such that n p p is not a multiple of q for any natural number n.

Strong as the above theorem of Frobenius is, there is an even stronger result due to Chebotarev.

Strong as the above theorem of Frobenius is, there is an even stronger result due to Chebotarev. To state Chebotarev s theorem, we recall one or two facts from basic algebraic number theory.

When k is a finite extension field of Q, we have a nice subring called the ring of integers of k.

When k is a finite extension field of Q, we have a nice subring called the ring of integers of k. It is O k = {x k : x is the root of a monic integral polynomial }.

When k is a finite extension field of Q, we have a nice subring called the ring of integers of k. It is O k = {x k : x is the root of a monic integral polynomial }. For example, if k = Q(i), O k = Z[i]. More generally, if d is a square-free integer, and k = Q( d), then O k = Z[ d] or Z[ 1+ d 2 ] according as to whether d 2, 3 mod 4 or 1 mod 4.

For the important class of number fields k = Q(ζ n ) generated by the n-th roots of unity - the so-called cyclotomic fields - the ring O k = Z[ζ n ] of all integral polynomial expressions in the n-th roots of unity. The rings O k are a lot like the usual integers but there is one very essential difference - in general, the ideals in O k are NOT singly generated (unlike Z). In fact, this fact that, for cyclotomic fields k = Q(ζ p ), the ring O k is not a principal ideal domain when p 23, is the serious reason why Fermat s last theorem has no elementary proof using factorization.

Ordinary prime numbers may not exhibit the same property in a bigger ring.

Ordinary prime numbers may not exhibit the same property in a bigger ring. Look, for instance, at the ring of Gaussian integers Z[i]. The prime number 13 divides the product of 1 + 5i and 1 5i in this ring but 13 does not divide either of these. Why should one care about what happens to ordinary prime numbers in these bigger rings?

Although O k is not a principal ideal domain (PID) in general, it is a so-called Dedekind domain.

Although O k is not a principal ideal domain (PID) in general, it is a so-called Dedekind domain. Without getting into technical details, one can roughly say that the multiplication of numbers is replaced by multiplication of ideals in which case unique factorization into prime ideals holds. A useful fact to keep in mind is that the ideal po k for a prime number p can be a product of at the most d ideals where d is the degree of k.

For instance, the prime number 2 gives in Z[i] the ideal 2Z[i] which becomes the square of a prime ideal generated by 1 + i. An odd prime number p which is 3 modulo 4 gives the ideal pz[i] which remains a prime ideal. An odd prime number p which is 1 modulo 4 gives the ideal pz[i] which is a product of two different prime ideals generated by complex conjugates x + iy and x iy for integers x, y; indeed, x, y are obtained from p = x 2 + y 2. In general, for a quadratic extension k = Q( d), the decomposition of the ideal po k into prime ideals (either two distinct or a single prime ideal or the square of a prime ideal) is an expression of the cases whether d is a square or non-square modulo p or is a multiple of it.

If k = Q(ζ n ), the decomposition of an ideal po k for a prime number p is governed by what is known as the cyclotomic reciprocity law.

If k = Q(ζ n ), the decomposition of an ideal po k for a prime number p is governed by what is known as the cyclotomic reciprocity law. In general, the set of prime numbers p such that po k breaks up into the degree [k : Q] ideals (the maximum possible) determines the abelian extension field k.

Fact 1 (primes splitting in cyclotomic extensions) : If p is a prime and p n and ζ n is a primitive n th root of unity, then p splits completely in Q(ζ n ) if and only if p 1(modn).

Fact 1 (primes splitting in cyclotomic extensions) : If p is a prime and p n and ζ n is a primitive n th root of unity, then p splits completely in Q(ζ n ) if and only if p 1(modn). Fact 2 (primes splitting in a radical extension) : If a N, p is a prime not dividing a, then p splits completely in Q(ζ q, a 1 q ) if and only if p 1(modq) and a is a q th power in Z/pZ. Fact 3 (split primes determine hierarchy) : If K, L are finite extensions of Q and Spl(K) and Spl(L) denote the sets of prime integers which split completely in K and L respectively, then Spl(K) Spl(L), then K L.

Returning to integral polynomials f, the key fact is that given any prime number p not dividing the discriminant of f, there is a conjugacy class defined in Gal(f), called the Frobenius conjugacy class [Frob p ] which comes about as follows. If K is the splitting field of f, the ideal generated by p in the ring O K of integers of K, decomposes uniquely as a product P 1 P 2 P g of distinct prime ideals of O K. These prime ideals are said to lie over p; they are permuted transitively by Gal(f). Each O K /P i is a finite field of p f elements for some r (and fg = [K : Q]).

The stabilizer subgroup G Pi := {σ Gal(f ) : σ(p i ) = P i } which is called the decomposition group at P i maps isomorphically onto the Galois group of the finite field extension (O K /P i )/(Z/pZ). In particular, there is an element Frob Pi of Gal(f ) which generates G Pi. For the different i s they are conjugate giving a conjugacy class [Frob p ] in Gal(f ).

The Frobenius conjugacy class is trivial if and only if the prime p splits completely into degree number of prime ideals.

The Frobenius conjugacy class is trivial if and only if the prime p splits completely into degree number of prime ideals. In general, the primes which split completely in an extension field is an important set as it essentially determines the abelian extensions. We shall soon see how this is of use for us.

In Frobenius s density theorem, one cannot distinguish between two primes p, q which define different conjugacy classes C(x) and C(y) but, which are such that some power of x and y are conjugate.

In Frobenius s density theorem, one cannot distinguish between two primes p, q which define different conjugacy classes C(x) and C(y) but, which are such that some power of x and y are conjugate. For instance, for the polynomial X 10 1, the decomposition type modulo primes congruent to 1, 3, 7, 9 mod 10 are, respectively, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1; 1, 1, 4, 4; 1, 1, 4, 4; 1, 1, 2, 2, 2, 2. Frobenius s theorem cannot distinguish between primes which are 3 mod 10 and those which are 7 mod 10 which define different conjugacy classes in Gal(X 10 1). Thus, it would imply that the number of primes 3 or 7 mod 10 is infinite but does not say whether each congruence class contains infinitely many primes. This requires Chebotarev s theorem.

Chebotarev s density theorem. Let f be a monic, irreducible, integral polynomial. Let C be a conjugacy class of Gal(f ). Then, the set of primes p not dividing disc (f ) for which σ p C, has a well-defined density which equals C G. This theorem implies in particular Dirichlet s theorem that for each a co-prime to n, the set of primes in the congruence class a mod n has density 1 φ(n).

Chebotarev s idea of proving this has been described by two prominent mathematicians as a spark from heaven.

Chebotarev s idea of proving this has been described by two prominent mathematicians as a spark from heaven. In fact, this theorem was proved in 1922 ( while carrying water from the lower part of town to the higher part, or buckets of cabbages to the market, which my mother sold to feed the entire family )! Emil Artin wrote to Helmut Hasse in 1925: Did you read Chebotarev s paper?... If it is correct, then one surely has the general abelian reciprocity laws in one s pocket... Artin found the proof of the general reciprocity law in 1927 using Chebotarev s technique (he had already boldly published the reciprocity law in 1923 but admitted that he had no proof). Nowadays, Artin s reciprocity law is proved in some other way and Chebotarev s theorem is deduced from it!

Before moving on, we show how to answer the earlier, more difficult question we asked: If f is an irreducible integral polynomial, is it necessarily irreducible modulo some prime?

Before moving on, we show how to answer the earlier, more difficult question we asked: If f is an irreducible integral polynomial, is it necessarily irreducible modulo some prime? The answer lies in knowing whether or not Gal(f ) has an element of order n, where deg(f ) = n.

Indeed, if f is irreducible modulo p, the (mutually conjugate) decomposition groups G Pi for prime ideals P i lying over p, are cyclic groups of order equal to n. Therefore, Gal(f ) has elements of order n. Therefore, if Gal(f ) does not contain an element of order n, then f is reducible modulo every prime!

Conversely, if Gal(f ) has an element of order n, the Chebotarev density theorem guarantees the existence of an infinite number of primes p whose Frobenius has order n; that is, the polynomial f is irreducible modulo p for such p. Now, Gal(f ) is a transitive subgroup of S n and, hence, its order is a multiple of n. If n is a prime, then evidently Gal(f ) must contain an element of order p (!) If n is composite, Gal(f ) may or may not contain an element of order n. For Hilbert s example of degree 4, the Galois group is isomorphic to Klein s four group.

Indeed, for every composite n, one can produce an irreducible polynomial over the integers, which is reducible modulo every prime. We don t go into that here but discuss another problem involving primes and integer polynomials.

Kronecker-conjugacy For f Z[X ] and, p prime, consider the value set Val p (f ) = {f (a) mod p ; a Z}.

Kronecker-conjugacy For f Z[X ] and, p prime, consider the value set Val p (f ) = {f (a) mod p ; a Z}. Call f, g Z[X ] to be Kronecker-conjugate if Val p (f ) = Val p (g) for all but finitely many primes p.

To mention in passing: Schur conjectured that the Tchebychev polynomials are essentially the only integer polynomials f which are permutation polynomials modulo p (that is, Val p (f ) = Z p ) for almost all primes p. This was proved by M.Fried using group-theoretic techniques similar to our discussion.

Given f, g Z[X ] which are non-constant, consider a Galois extension K of the field Q(t) of rational functions in a variable t such that K contains a root x of f (x) = t and a root y of g(y) = t. Let G denote Gal(K/Q(t)) and let U, V denote the stabilizers of x, y respectively.

Fried s Theorem. Given f, g Z[X ] which are non-constant, consider t, K, G, x, y, U, V as above. Then, f, g are Kronecker-conjugate if and only if gug 1 = gvg 1. g G g G

Idea of Proof Suppose that f, g are Kronecker-conjugate. Write f = ux n + be of degree n. As Kronecker-conjugacy is preserved when we replace f(x) and g(x) by u n 1 f (X /u) and u n 1 g(x ) respectively, we may assume that f is monic. Now, for any integer a, the hypothesis gives that f (X ) g(a) (mod p) has a root for almost all primes p.

Hilbert s irreducibility theorem tells us that the Galois groups Gal(f (X ) g(y) over K(y) and Gal(f (X ) g(a)) over K are isomorphic as permutation groups for infinitely many a. Recall that g(y) = t. Thus every element of the Galois group Gal(f (X ) t) over K(y) fixes at least one root. This Galois group is just the induced action of V on the roots of f (X ) t (but V need not act faithfully). Hence every element in V fixes a root. But these roots are the conjugates of x whose stabilizer is U. So every element of V lies in some conjugate of U. By symmetry, the result follows.

Conjecture. Over a field of characteristic 0, two polynomials f and g are Kronecker-conjugate if and only if, the permutation representations IndU G 1 and Ind V G 1 are equivalent (that is, U and V are Gassmann equivalent). Actually, this question is not purely group-theoretic because there are examples of abstract finite groups G and subgroups U, V such that g G gug 1 = g G gvg 1 but U, V are not Gassmann equivalent.

Rational values What about two nonconstant polynomials f, g with integer (or rational) coefficients such that the value sets of f, g at all rational numbers are the same?

Rational values What about two nonconstant polynomials f, g with integer (or rational) coefficients such that the value sets of f, g at all rational numbers are the same? Evidently, f (x) = g(ax + b) for some rational a, b with a 0 is a possibility. Are there others? More generally, if the value set of f at rationals is contained in the corresponding value set of g. If h is some rational polynomial so that f (x) = g(h(x)), this holds; are there other situations?

The answer to the question is yes and this can be proved in an elementary manner.

The answer to the question is yes and this can be proved in an elementary manner. Let us indicate a powerful method which can prove many such results.

The answer to the question is yes and this can be proved in an elementary manner. Let us indicate a powerful method which can prove many such results. The idea is that a rational polynomial in two (or more variables) which is irreducible admits infinitely many specializations of all but one variable to rationals so that the resulting polynomials in one variable are irreducible.

We are given two nonconstant polynomials f, g with rational coefficients such that f (Q) g(q).

We are given two nonconstant polynomials f, g with rational coefficients such that f (Q) g(q). Write the two variable polynomial f (X ) g(y ) = h 1 (X, Y )h 2 (X, Y ) h d (X, Y ) where h i are irreducible. By Hilbert s irreducibility theorem, there are infinitely many rational x 0 such that h i (x 0, Y ) are irreducible in Y. On the other hand, for each such x 0, f (x 0 ) f (Q) g(q); that is, there is a corresponding y 0 such that f (x 0 ) = g(y 0 ).

Therefore, h 1 must have degree 1 in Y ; that is, h 1 (X, Y ) = p(x ) + q(x )Y.

Therefore, h 1 must have degree 1 in Y ; that is, h 1 (X, Y ) = p(x ) + q(x )Y. Hence f (X ) g(y ) = (p(x ) + q(x )Y )h 2 (X, Y ) h d (X, Y ); so, f (X ) = g( p(x )/q(x )).

Therefore, h 1 must have degree 1 in Y ; that is, h 1 (X, Y ) = p(x ) + q(x )Y. Hence f (X ) g(y ) = (p(x ) + q(x )Y )h 2 (X, Y ) h d (X, Y ); so, f (X ) = g( p(x )/q(x )). This is easily seen to imply that p(x )/q(x ) is actually a polynomial in X ; this proves f (X ) = g(h(x )).

The support problem Let x, y be positive integers such that each prime divisor of x n 1 also divides y n 1 for every n. What can we say about the relation between x and y?

The support problem Let x, y be positive integers such that each prime divisor of x n 1 also divides y n 1 for every n. What can we say about the relation between x and y? For instance, if y is a power of x, the above property holds. We may ask whether the converse also holds. This problem came to be known as the support problem.

This is proved using Kummer theory.

This is proved using Kummer theory. Kummer theory is a correspondence between abelian extensions of a field K and subgroups of the n-th powers of K.

This is proved using Kummer theory. Kummer theory is a correspondence between abelian extensions of a field K and subgroups of the n-th powers of K. For our purposes in this application, we may think of the following concrete statement as Kummer theory: If K contains the n-th roots of unity, then abelian extensions L of K whose Galois groups have exponent n correspond bijectively to subgroups Ω of K containing (K ) n via L K (L ) n and its inverse map Ω K(Ω 1/n ).

The support problem is:

The support problem is: Theorem: Let x, y Q. If, for all but finitely many primes p and, for all n N, one has the implication p (x n 1) = p (y n 1), then y is a power of x. Here p divides a rational number x = a b, (a, b) = 1, means that p a.

This is in reality a local-global theorem. To explain:

This is in reality a local-global theorem. To explain: For a prime p not dividing the numerator and denominators of x and y, look at the order of x mod p; viz.,

This is in reality a local-global theorem. To explain: For a prime p not dividing the numerator and denominators of x and y, look at the order of x mod p; viz., The smallest n such that p (x n 1).

This is in reality a local-global theorem. To explain: For a prime p not dividing the numerator and denominators of x and y, look at the order of x mod p; viz., The smallest n such that p (x n 1). We have p (y n 1) so that the order of y mod p divides n, the order of x. A simple exercise in the cyclic group Z p shows that y must be a power of x mod p. Therefore, the support theorem says that if, modulo all but finitely many primes p, y is a power of x modulo p, then y is actually a power of x; this is what Kummer theory accomplishes.

Recall the three facts we mentioned earlier which will now be useful in solving the support problem. Fact 1 (primes splitting in cyclotomic extensions) : If p is a prime and p n and ζ n is a primitive n th root of unity, then p splits completely in Q(ζ n ) if and only if p 1(modn). Fact 2 (primes splitting in a radical extension) : If a N, p is a prime not dividing a, then p splits completely in Q(ζ q, a 1 q ) if and only if p 1(modq) and a is a q th power in Z/pZ. Fact 3 (split primes determine hierarchy) : If K, L are finite extensions of Q and Spl(K) and Spl(L) denote the sets of prime integers which split completely in K and L respectively, then Spl(K) Spl(L), then K L.

Assume x, y are as in the theorem, that is, for all n and all but finitely many primes p, one has p (x n 1) = p (y n 1).

Assume x, y are as in the theorem, that is, for all n and all but finitely many primes p, one has p (x n 1) = p (y n 1). We us Kummer s theorem which, for our purposes, gives using the facts above:

Assume x, y are as in the theorem, that is, for all n and all but finitely many primes p, one has p (x n 1) = p (y n 1). We us Kummer s theorem which, for our purposes, gives using the facts above: Let q be an odd prime power. Then the natural map Q /(Q ) q Q(ζ q ) /(Q(ζ q ) ) q is injective.

Look at our x, y in the support theorem. We can show:

Look at our x, y in the support theorem. We can show: Let q be a power of a prime l and let ζ q be a primitive q th root of unity. Then Q(ζ q, x 1 q ) Q(ζ q, y 1 q ).

Look at our x, y in the support theorem. We can show: Let q be a power of a prime l and let ζ q be a primitive q th root of unity. Then Q(ζ q, x 1 q ) Q(ζ q, y 1 q ). This is proved using a generalization of the so-called Hilbert Theorem 90 slightly.

Basically, the hypothesis of the support theorem shows that y = x d in the group Q(ζ q ) /(Q(ζ q ) ) q. So, by the last quoted result based on Kummer, we have y = x d in Q /(Q ) q also.

Basically, the hypothesis of the support theorem shows that y = x d in the group Q(ζ q ) /(Q(ζ q ) ) q. So, by the last quoted result based on Kummer, we have y = x d in Q /(Q ) q also. It is a simple matter to show that this is a power over integers also; I don t say any more about it.

Fermat s letter on June 15th, 1641 In a letter written by Fermat to Mersenne on 15th June 1641, he made the following 4 conjectures. For positive integers a, b, let us denote the sequence {a n + b n } n=1 by S a,b and we write p S a,b if p divides at least one member of S a,b. Fermat stated : Conjecture (Fermat, 1641) 1) If p S 3,1, then p 1(mod 12). 2) If p S 3,1, then p +1(mod 12). 3) If p S 5,1, then p 1(mod 10). 4) If p S 5,1, then p +1(mod 10). The quadratic reciprocity law implies that Conjecture 1 holds true. However, the remaining three conjectures are all false. We can even show that there is a positive density of primes for which each of the conclusions of these three conjectures is false.

One can determine explicitly in terms of a, b, c, d the natural density of primes in the residue class c mod d which divide S a,b. For instance, denoting by δ a,b (c, d) the density of primes in the residue class c mod d which divides S a,b, our theorem has the following bearing on the above conjectures : Corollary δ 3,1 (1, 12) = 1 6, δ 3,1(5, 12) = 1 4, δ 3,1(7, 12) = 1 4 and δ 3,1(11, 12) = 0. Furthermore, we have δ 5,1 (1, 10) = 1 12, δ 5,1(3, 10) = 1 4, δ 5,1(7, 10) = 1 4 and δ 5,1(9, 10) = 1 12.

The ideas to prove come from the approach to the so-called Artin conjecture. To motivate it:

The ideas to prove come from the approach to the so-called Artin conjecture. To motivate it: Gauss considers in articles 315-317 of his Disquisitiones Arithmeticae (1801) the decimal expansion of numbers of the form 1 p with p prime. For example, 1/7 = 0.142857, 1/11 = 0.09. Note that the decimal expansion of 1/p is periodic for p 2, 5 and the period is p 1 if and only if 10 is a primitive root modulo p (that is, p 1 is the smallest k such that 10 k 1 is a multiple of p).

Start with two given integers a, b and consider the set of all prime numbers which divide a n + b n for some n.

Start with two given integers a, b and consider the set of all prime numbers which divide a n + b n for some n. For instance, it turns out that the two-thirds of the set of all primes divide a number of the form 10 n + 1. This means : Two-thirds of the prime numbers p, the decimal expansion of 1/p has even period!

Artin s primitive root conjecture (1927) Let g Q \ { 1, 0, 1}. For a prime p not dividing neither the numerator nor the denominator, one can look at the subgroup of F p generated by g and call g a primitive root modulo p if this is the whole group. That is, the order of g modulo p is p 1. Denote by P(g) the set of prime numbers for which g is a primitive root. Qualitative form of Artin s primitive root conjecture: The set P(g) is infinite if g is not a square of a rational number. Quantitative form: Let h be the largest integer such that g = g0 h with g 0 Q. We have, as x tends to infinity, P(g)(x) := 1 p P(g),p x = 1 (1 p(p 1) ) (1 1 p 1 ) x + o(x/ log x) log x p h p h

Recall the Prime Number Theorem in the form π(x) := #{p x} Li(x) as x tends to infinity (i.e. the quotient of the r.h.s. and l.h.s. tends to 1 as x tends to infinity), where The logarithmic integral Li(x) is defined as x 2 dt/ log t. The theorem suggests that the probability that a number n is prime is of the order 1/ log n. The prime number theorem is useful especially when used with an error term. For instance, we shall prove statements like the following: The number of prime numbers p x dividing some term of the sequence a n + b n equals δ(a, b)li(x) + O(E(x)) where δ(a, b) is a positive real number and E(x)/Li(x) 0 as x.

The starting point to analyze Artin s primitive root conjecture is the observation :

The starting point to analyze Artin s primitive root conjecture is the observation : p P(g) g p 1 q 1( mod p) for every prime q dividing p 1. (1)

= Obvious. = Suppose p P(g). Then g p 1 k 1( mod p) for some k p 1, k > 1. But this implies that g p 1 d 1( mod p) for some prime divisor d of k. This is a contradiction.

= Obvious. = Suppose p P(g). Then g p 1 k 1( mod p) for some k p 1, k > 1. But this implies that g p 1 d 1( mod p) for some prime divisor d of k. This is a contradiction. Thus, associated with a prime p we have conditions for various q. Interchanging the roles of p and q, that is for a fixed q we can consider the set of primes p such that p 1( mod q) and g p 1 q 1( mod p).

Fix any prime q and let us try to compute the density of primes p such that both p 1( mod q) and g p 1 q 1( mod p).

Fix any prime q and let us try to compute the density of primes p such that both p 1( mod q) and g p 1 q 1( mod p). The main observation is that the above two conditions can be interpreted as the prime p splitting completely in a certain field extension and, Chebotarev density theorem determines the density of such primes in terms of the field degree.

About the other condition g p 1 q 1( mod p), using Fermat s little theorem, we infer (in case p g) that g p 1 q x q 1( mod p). is a solution of

About the other condition g p 1 q 1( mod p), using Fermat s little theorem, we infer (in case p g) that g p 1 q x q 1( mod p). is a solution of We expect there to be q solutions and we want a solution to be 1 modulo q. Thus we expect to be successful with probability 1 q, except when q h. Then g p 1 q = g h p 1 q 0 1( mod p), trivially. If we assume that these events are independent then the probability 1 1 that both events occur is q(q 1) if q h and q 1 otherwise. The above events should not occur for any q in order to ensure that p P(g). This suggests a natural density of q h ( 1 1 q(q 1) ) q h ( 1 1 ) q 1 for such primes and hence we expect the analytic formulation of Artin s conjecture to hold true.

Artin conjecture as splitting conditions In Artin s conjecture, a number r divides the index [F p :< g mod p >] if, and only if, r (p 1) and the order of g divides (p 1)/r.

Artin conjecture as splitting conditions In Artin s conjecture, a number r divides the index [F p :< g mod p >] if, and only if, r (p 1) and the order of g divides (p 1)/r. This is just the condition that p splits completely in the splitting field Q(ζ r, g 1/r ) of the polynomial X r g over Q. In particular, if v 2 (p 1) = v 2 (r), then this is equivalent to g having odd order modulo p. This, in turn, is characterized by the property that p 1 + 2 k mod 2 k+1 and splits completely in the field extension K k := Q(ζ 2 k, g 1/2k ).

Primes dividing a n + b n as splitting conditions Returning to our problem of primes dividing sequences, we interpret the statement that a prime p divides some term of {a n + b n } as statements about splitting of primes in field extensions. Leaving out primes dividing b, a prime p divides some a n + b n if and only if, g := a/b has odd order in F p. We already know this as splitting conditions in the fields K k := Q(ζ 2 k, g 1/2k ). If we have additional conditions on the prime (like p c mod d), we need to count the primes splitting completely in fields obtained by intersecting fields like K k s above with Q(ζ d ) etc. and with a given Frobenius conjugacy class.

Theorem Let a and b be positive integers with a b. Let N a,b (x) count the number of primes p x that divide S a,b. Put r = a/b. Let λ be the largest integer such that r = u 2λ, with u a rational number. Let L = Q( u). We have ( ) x(log log x) 4 N a,b (x) = δ(r)li(x) + O log 3, x where the implied constant may depend on a and b, and δ(r) is a positive rational number that is given in Table 0.

Table 0: The value of δ(r) L λ δ(r) L Q( 2) λ 0 2 1 λ /3 L = Q( 2) λ = 0 17/24 L = Q( 2) λ = 1 5/12 L = Q( 2) λ 2 2 λ /3

Let a, b, c, d be integers with (c, d) = 1 and assume that a b. We can use the ideas used in the approach to Artin s conjecture to determine the density of prime numbers which divide some number of the form a n + b n and are in the residue class c mod d. The data are tabulated in a convenient form so that given particular values a, b, c, d it is easy to find which table and which row of it to look at to find the corresponding value.

Compute r 0, h from a/b = r0 h, where r 0 is not a proper power of a rational number. Then, write λ = v 2 (h). Find δ, d in d = 2 δ d, with δ = v 2 (d). Get γ = v 2 (c 1), where it is understood that γ is larger than any number when c = 1. Write the discriminant D(r 0 ) of the quadratic field Q( r 0 ) as D(r 0 ) = 2 δ 0 D. Writing r 0 = u/v, write t = r 0 if u, v are odd and, write t = k i=1 ( 1 p i )p i if uv = 2 k i=1 p i.

Table 1 : Q( r 0 ) Q( 2), D d λ δ φ(d)δ a,b (c, d) < δ γ 1 2λ+1 δ 3 > 0, min(λ, γ) 2 δ λ 3 2 0 1 λ 3 γ > γ 0 < γ > γ 1 2 λ γ

Table 2 : Q( r 0 ) Q( 2), D d, δ 0 δ λ δ ( D(r 0) c ) φ(d)δ a,b (c, d) δ 1 > 0, γ 1 2 δ 1 λ 3 2 δ 1 λ 1 0 1 2 λ 1 3 2 λ < δ 1 γ 1 1 2λ+2 δ 3 1 1 δ > γ 0 γ 1 > γ 1 1 2 λ+1 γ 1 1 γ > λ 0

Table 3 : Q( r 0 ) Q( 2), D d and δ 0 > δ λ δ ( D(t) c ) φ(d)δ a,b (c, d) < δ 1 γ 1 1 2λ+1 δ 3 + 2λ+2+δ 2δ 0 3 < δ 1 γ 1 1 2λ+1 δ 3 2λ+2+δ 2δ 0 3 = δ 1 γ 1 2 3 + 22δ+1 2δ 0 3 = δ 1 γ 1 2 3 22δ+1 2δ 0 3 γ 1 > γ 1 2 λ γ γ > λ 0 δ > γ 0 2 δ 0 2 > 0, min(γ, λ) 1 δ λ 3 + 2λ+2+δ 2δ 0 3 2 δ 0 2 > 0, min(γ, λ) 1 δ λ 3 2λ+2+δ 2δ 0 3 δ 0 1 > 0, γ 1 2 δ 1 λ 3 δ 0 1 > 0, γ 1 2 δ λ 1 2 δ 0 2 0 1 1 λ 3 + 2λ+3 2δ 0 3 2 δ 0 2 0 1 1 λ 3 2λ+3 2δ 0 3 δ 0 1 0 1 2 λ 3 δ 0 1 0 1 2 λ

Primes, Polynomials, Progressions. B.Sury Indian Statistical Institute Bangalore NISER Bhubaneshwar February 15, 2016