Primes, Polynomials, Progressions · Let f be a monic integral, irreducible polynomial. The set of...

Primes, Polynomials, Progressions

B.SuryIndian Statistical Institute Bangalore

NISER BhubaneshwarFebruary 15, 2016

B.Sury Primes, Polynomials, Progressions

Primes and polynomials

It is unknown if there exist infinitely many prime numbers of theform n2 + 1.

As a curiosity, let us note in passing that if there are infinitelymany prime numbers of the form n2 + 1, then the twin primeconjecture will be solved in the ring Z[i ] of Gaussian integers.

Indeed, there are infinitely many twin primes in − 1, in + 1 in Z[i ].

The question as to which primes p divide integers of the formn2 + 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 haveroots modulo p.This is easy to prove elementarily but we will give anon-elementary proof which can take us much further. Forinstance, that approach can answer more general questions like:

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 haveroots modulo p.

This is easy to prove elementarily but we will give anon-elementary proof which can take us much further. Forinstance, that approach can answer more general questions like:

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 haveroots modulo p.This is easy to prove elementarily but we will give anon-elementary proof which can take us much further. Forinstance, that approach can answer more general questions like:

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

The answer to this is:

yes!...

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

There exist irreducible, integer polynomials of any compositedegree which are reducible modulo every prime p.In what follows, we show how such questions are attacked.

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

In what follows, we show how such questions are attacked.

It is yes if the degree is prime and no if it is composite!There exist irreducible, integer polynomials of any compositedegree 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 thepolynomial x4 − 10x2 + 1.

A general example is given by:

Let p, q be odd primes such that

)= 1 and p ≡ 1(8).

Then, the polynomial (X 2 − p − q)2 − 4pq is irreducible whereas itis reducible modulo any integer.

Hilbert was the first to find examples in degree 4. One such is thepolynomial x4 − 10x2 + 1.A general example is given by:

Let p, q be odd primes such that

)= 1 and p ≡ 1(8).

Then, the polynomial (X 2 − p − q)2 − 4pq is irreducible whereas itis reducible modulo any integer.

To make precise statements about infinite sets of primes, it isuseful 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/p

s → 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 iff is an irreducible, integral polynomial of degree n, its Galois groupis a subgroup of the permutation group Sn, permuting the rootstransitively.

To make precise statements about infinite sets of primes, it isuseful 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/p

s → 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 iff is an irreducible, integral polynomial of degree n, its Galois groupis a subgroup of the permutation group Sn, permuting the rootstransitively.

To make precise statements about infinite sets of primes, it isuseful 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/p

s → 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 iff is an irreducible, integral polynomial of degree n, its Galois groupis a subgroup of the permutation group Sn, permuting the rootstransitively.

A density theorem due to Frobenius implies:

Let f be a monic integral, irreducible polynomial. The set of primenumbers p such that the polynomial f modulo p decomposes as aproduct of irreducible polynomials of degrees n1, n2, · · · , nr , has awell-defined density. This density equals N/O(Gal(f )) where N isthe number of elements σ in Gal(f) which have a cycle patternn1, n2, · · · , nr .

A density theorem due to Frobenius implies:Let f be a monic integral, irreducible polynomial. The set of primenumbers p such that the polynomial f modulo p decomposes as aproduct of irreducible polynomials of degrees n1, n2, · · · , nr , has awell-defined density. This density equals N/O(Gal(f )) where N isthe number of elements σ in Gal(f) which have a cycle patternn1, n2, · · · , nr .

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

If f has roots modulo all but finitely many primes p, the Frobeniustheorem shows that each σ in Gal(f ) has a cycle pattern of theform 1, n2, n3, · · ·So, each element of Gal(f ) fixes a root of f .Since the roots of f are transitively moved around by Gal(f ), thisGalois group would be the union of the conjugates of its subgroupH consisting of elements which fix a particular root of f .It is an elementary exercise that a finite group cannot be the unionof conjugates of a proper subgroup. Thus, in our case H = Gal(f ).This means that Gal(f ) fixes each root and is, therefore, identity.

Here’s a way to show that for an irreducible integral polynomial fof degree > 1, there are infinitely many primes which do not divideany of the values f (n).If f has roots modulo all but finitely many primes p, the Frobeniustheorem shows that each σ in Gal(f ) has a cycle pattern of theform 1, n2, n3, · · ·So, each element of Gal(f ) fixes a root of f .

Since the roots of f are transitively moved around by Gal(f ), thisGalois group would be the union of the conjugates of its subgroupH consisting of elements which fix a particular root of f .It is an elementary exercise that a finite group cannot be the unionof conjugates of a proper subgroup. Thus, in our case H = Gal(f ).This means that Gal(f ) fixes each root and is, therefore, identity.

Here’s a way to show that for an irreducible integral polynomial fof degree > 1, there are infinitely many primes which do not divideany of the values f (n).If f has roots modulo all but finitely many primes p, the Frobeniustheorem shows that each σ in Gal(f ) has a cycle pattern of theform 1, n2, n3, · · ·So, each element of Gal(f ) fixes a root of f .Since the roots of f are transitively moved around by Gal(f ), thisGalois group would be the union of the conjugates of its subgroupH consisting of elements which fix a particular root of f .

It is an elementary exercise that a finite group cannot be the unionof conjugates of a proper subgroup. Thus, in our case H = Gal(f ).This means that Gal(f ) fixes each root and is, therefore, identity.

Here’s a way to show that for an irreducible integral polynomial fof degree > 1, there are infinitely many primes which do not divideany of the values f (n).If f has roots modulo all but finitely many primes p, the Frobeniustheorem shows that each σ in Gal(f ) has a cycle pattern of theform 1, n2, n3, · · ·So, each element of Gal(f ) fixes a root of f .Since the roots of f are transitively moved around by Gal(f ), thisGalois group would be the union of the conjugates of its subgroupH consisting of elements which fix a particular root of f .It is an elementary exercise that a finite group cannot be the unionof conjugates of a proper subgroup. Thus, in our case H = Gal(f ).This means that Gal(f ) fixes each root and is, therefore, identity.

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

If p is a prime number, show that there is another prime number qsuch that np − p is not a multiple of q for any natural number n.

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

If p is a prime number, show that there is another prime number qsuch that np − p is not a multiple of q for any natural number n.

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

To state Chebotarev’s theorem, we recall one or two facts frombasic algebraic number theory.

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

To state Chebotarev’s theorem, we recall one or two facts frombasic algebraic number theory.

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

It is Ok = x ∈ k : x is the root of a monic integral polynomial .For example, if k = Q(i), Ok = Z[i ].More generally, if d is a square-free integer, and k = Q(

√d), then

Ok = Z[√

d ] or Z[1+√d

2 ] according as to whether d ≡ 2, 3 mod 4or 1 mod 4.

When k is a finite extension field of Q, we have a nice subringcalled the ring of integers of k .It is Ok = x ∈ k : x is the root of a monic integral polynomial .

For example, if k = Q(i), Ok = Z[i ].More generally, if d is a square-free integer, and k = Q(

√d), then

Ok = Z[√

d ] or Z[1+√d

When k is a finite extension field of Q, we have a nice subringcalled the ring of integers of k .It is Ok = x ∈ k : x is the root of a monic integral polynomial .For example, if k = Q(i), Ok = Z[i ].

More generally, if d is a square-free integer, and k = Q(√

d), then

Ok = Z[√

d ] or Z[1+√d

When k is a finite extension field of Q, we have a nice subringcalled the ring of integers of k .It is Ok = x ∈ k : x is the root of a monic integral polynomial .For example, if k = Q(i), Ok = Z[i ].More generally, if d is a square-free integer, and k = Q(

√d), then

Ok = Z[√

d ] or Z[1+√d

For the important class of number fields k = Q(ζn) generated bythe n-th roots of unity - the so-called cyclotomic fields - the ringOk = Z[ζn] of all integral polynomial expressions in the n-th rootsof unity.

The rings Ok are a lot like the usual integers but there is one veryessential difference - in general, the ideals in Ok are NOT singlygenerated (unlike Z).In fact, this fact that, for cyclotomic fields k = Q(ζp), the ring Ok

is not a principal ideal domain when p ≥ 23, is the serious reasonwhy Fermat’s last theorem has no elementary proof usingfactorization.

For the important class of number fields k = Q(ζn) generated bythe n-th roots of unity - the so-called cyclotomic fields - the ringOk = Z[ζn] of all integral polynomial expressions in the n-th rootsof unity.The rings Ok are a lot like the usual integers but there is one veryessential difference - in general, the ideals in Ok are NOT singlygenerated (unlike Z).

In fact, this fact that, for cyclotomic fields k = Q(ζp), the ring Ok

For the important class of number fields k = Q(ζn) generated bythe n-th roots of unity - the so-called cyclotomic fields - the ringOk = Z[ζn] of all integral polynomial expressions in the n-th rootsof unity.The rings Ok are a lot like the usual integers but there is one veryessential difference - in general, the ideals in Ok are NOT singlygenerated (unlike Z).In fact, this fact that, for cyclotomic fields k = Q(ζp), the ring Ok

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

Look, for instance, at the ring of Gaussian integers Z[i ]. The primenumber 13 “divides” the product of 1 + 5i and 1− 5i in this ringbut 13 does not divide either of these.Why should one care about what happens to ordinary primenumbers in these bigger rings?We’ll see that in facts 1,2,3 below.

Ordinary prime numbers may not exhibit the same property in abigger ring.Look, for instance, at the ring of Gaussian integers Z[i ]. The primenumber 13 “divides” the product of 1 + 5i and 1− 5i in this ringbut 13 does not divide either of these.

Why should one care about what happens to ordinary primenumbers in these bigger rings?We’ll see that in facts 1,2,3 below.

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

We’ll see that in facts 1,2,3 below.

Ordinary prime numbers may not exhibit the same property in abigger ring.Look, for instance, at the ring of Gaussian integers Z[i ]. The primenumber 13 “divides” the product of 1 + 5i and 1− 5i in this ringbut 13 does not divide either of these.Why should one care about what happens to ordinary primenumbers in these bigger rings?We’ll see that in facts 1,2,3 below.

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

Without getting into technical details, one can roughly say thatthe multiplication of numbers is replaced by multiplication of idealsin which case unique factorization into prime “ideals” holds.A useful fact to keep in mind is that the ideal pOk for a primenumber p can be a product of at the most d ideals where d is thedegree of k.

Although Ok is not a principal ideal domain (PID) in general, it isa so-called Dedekind domain.Without getting into technical details, one can roughly say thatthe multiplication of numbers is replaced by multiplication of idealsin which case unique factorization into prime “ideals” holds.

A useful fact to keep in mind is that the ideal pOk for a primenumber p can be a product of at the most d ideals where d is thedegree of k.

Although Ok is not a principal ideal domain (PID) in general, it isa so-called Dedekind domain.Without getting into technical details, one can roughly say thatthe multiplication of numbers is replaced by multiplication of idealsin which case unique factorization into prime “ideals” holds.A useful fact to keep in mind is that the ideal pOk for a primenumber p can be a product of at the most d ideals where d is thedegree 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 bycomplex conjugates x + iy and x − iy for integers x , y ; indeed, x , yare obtained from p = x2 + y2.

In general, for a quadratic extension k = Q(√

d), thedecomposition of the ideal pOk into prime ideals (either twodistinct or a single prime ideal or the square of a prime ideal) is anexpression of the cases whether d is a square or non-square modulop or is a multiple of it.

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 bycomplex conjugates x + iy and x − iy for integers x , y ; indeed, x , yare obtained from p = x2 + y2.

In general, for a quadratic extension k = Q(√

d), thedecomposition of the ideal pOk into prime ideals (either twodistinct or a single prime ideal or the square of a prime ideal) is anexpression of the cases whether d is a square or non-square modulop or is a multiple of it.

If k = Q(ζn), the decomposition of an ideal pOk for a primenumber p is governed by what is known as the cyclotomicreciprocity law.

In general, the set of prime numbers p such that pOk breaks upinto the degree [k : Q] ideals (the maximum possible) determinesthe abelian extension field k.More precisely,

If k = Q(ζn), the decomposition of an ideal pOk for a primenumber p is governed by what is known as the cyclotomicreciprocity law.In general, the set of prime numbers p such that pOk breaks upinto the degree [k : Q] ideals (the maximum possible) determinesthe abelian extension field k.

More precisely,

If k = Q(ζn), the decomposition of an ideal pOk for a primenumber p is governed by what is known as the cyclotomicreciprocity law.In general, the set of prime numbers p such that pOk breaks upinto the degree [k : Q] ideals (the maximum possible) determinesthe abelian extension field k.More precisely,

Fact 1 (primes splitting in cyclotomic extensions) :If p is a prime and p 6| n and ζn is a primitive nth root of unity, thenp splits completely in Q(ζn) 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(ζq, a1q ) if and only if p ≡ 1(modq) and a is a qth power in Z/pZ.

Fact 3 (split primes determine hierarchy) :If K , L are finite extensions of Q and Spl(K ) and Spl(L) denotethe sets of prime integers which split completely in K and Lrespectively, then Spl(K ) ⊆ Spl(L), then K ⊇ L.

• Returning to integral polynomials f , the key fact is that givenany prime number p not dividing the discriminant of f , there is aconjugacy class defined in Gal(f), called the Frobenius conjugacyclass [Frobp] which comes about as follows.

• If K is the splitting field of f , the ideal generated by p in the ringOK of integers of K , decomposes uniquely as a productP1P2 · · ·Pg of distinct prime ideals of OK .

• These prime ideals are said to lie over p; they are permutedtransitively by Gal(f).

• Each OK/Pi is a finite field of pf elements for some r (andfg = [K : Q]).

• The stabilizer subgroup GPi:= σ ∈ Gal(f ) : σ(Pi ) = Pi which

is called the decomposition group at Pi maps isomorphically ontothe Galois group of the finite field extension (OK/Pi )/(Z/pZ ).In particular, there is an element FrobPi

of Gal(f ) which generatesGPi

.For the different i ’s they are conjugate giving a conjugacy class[Frobp] in Gal(f ).

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

In general, the primes which split completely in an extension field isan important set as it essentially determines the abelian extensions.We shall soon see how this is of use for us.For instance, the prime numbers which split completely in the fieldcorresponding to f (x) = xn − 1 are the primes p ≡ 1 mod n.

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

For instance, the prime numbers which split completely in the fieldcorresponding to f (x) = xn − 1 are the primes p ≡ 1 mod n.

The Frobenius conjugacy class is trivial if and only if the prime psplits completely into degree number of prime ideals.In general, the primes which split completely in an extension field isan important set as it essentially determines the abelian extensions.We shall soon see how this is of use for us.For instance, the prime numbers which split completely in the fieldcorresponding to f (x) = xn − 1 are the primes p ≡ 1 mod n.

In Frobenius’s density theorem, one cannot distinguish between twoprimes 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 typemodulo 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 3mod 10 and those which are 7 mod 10 which define differentconjugacy classes in Gal(X 10 − 1). Thus, it would imply that thenumber of primes ≡ 3 or 7 mod 10 is infinite but does not saywhether each congruence class contains infinitely many primes.This requires Chebotarev’s theorem.

In Frobenius’s density theorem, one cannot distinguish between twoprimes 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 typemodulo 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 3mod 10 and those which are 7 mod 10 which define differentconjugacy classes in Gal(X 10 − 1). Thus, it would imply that thenumber of primes ≡ 3 or 7 mod 10 is infinite but does not saywhether each congruence class contains infinitely many primes.This requires Chebotarev’s theorem.

In Frobenius’s density theorem, one cannot distinguish between twoprimes 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 typemodulo 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 3mod 10 and those which are 7 mod 10 which define differentconjugacy classes in Gal(X 10 − 1). Thus, it would imply that thenumber of primes ≡ 3 or 7 mod 10 is infinite but does not saywhether 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 aconjugacy class of Gal(f ). Then, the set of primes p not dividingdisc (f ) for which σp ∈ C , has a well-defined density which equals|C ||G | .

This theorem implies in particular Dirichlet’s theorem that for eacha co-prime to n, the set of primes in the congruence class a mod nhas density 1

φ(n) .

Chebotarev’s density theorem.Let f be a monic, irreducible, integral polynomial. Let C be aconjugacy class of Gal(f ). Then, the set of primes p not dividingdisc (f ) for which σp ∈ C , has a well-defined density which equals|C ||G | .This theorem implies in particular Dirichlet’s theorem that for eacha co-prime to n, the set of primes in the congruence class a mod nhas density 1

φ(n) .

Chebotarev’s idea of proving this has been described by twoprominent mathematicians as “ a spark from heaven”.

In fact, this theorem was proved in 1922 (“while carrying waterfrom the lower part of town to the higher part, or buckets ofcabbages to the market, which my mother sold to feed the entirefamily”)!Emil Artin wrote to Helmut Hasse in 1925:“Did you read Chebotarev’s paper? ... If it is correct, then onesurely has the general abelian reciprocity laws in one’s pocket...”Artin found the proof of the general reciprocity law in 1927 usingChebotarev’s technique (he had already boldly published thereciprocity law in 1923 but admitted that he had no proof).Nowadays, Artin’s reciprocity law is proved in some other way andChebotarev’s theorem is deduced from it !

Chebotarev’s idea of proving this has been described by twoprominent mathematicians as “ a spark from heaven”.In fact, this theorem was proved in 1922 (“while carrying waterfrom the lower part of town to the higher part, or buckets ofcabbages to the market, which my mother sold to feed the entirefamily”)!

Emil Artin wrote to Helmut Hasse in 1925:“Did you read Chebotarev’s paper? ... If it is correct, then onesurely has the general abelian reciprocity laws in one’s pocket...”Artin found the proof of the general reciprocity law in 1927 usingChebotarev’s technique (he had already boldly published thereciprocity law in 1923 but admitted that he had no proof).Nowadays, Artin’s reciprocity law is proved in some other way andChebotarev’s theorem is deduced from it !

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

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

The answer lies in knowing whether or not Gal(f ) has an elementof order n, where deg(f ) = n.

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

Indeed, if f is irreducible modulo p, the (mutually conjugate)decomposition groups GPi

for prime ideals Pi lying over p, arecyclic 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 fis reducible modulo every prime !

Indeed, if f is irreducible modulo p, the (mutually conjugate)decomposition groups GPi

for prime ideals Pi lying over p, arecyclic 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 fis reducible modulo every prime !

Conversely, if Gal(f ) has an element of order n, the Chebotarevdensity theorem guarantees the existence of an infinite number ofprimes p whose Frobenius has order n; that is, the polynomial f isirreducible modulo p for such p.

Now, Gal(f ) is a transitive subgroup of Sn and, hence, its order isa multiple of n.If n is a prime, then evidently Gal(f ) must contain an element oforder p (!)If n is composite, Gal(f ) may or may not contain an element oforder n. For Hilbert’s example of degree 4, the Galois group isisomorphic to Klein’s four group.

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

Indeed, for every composite n, one can produce an irreduciblepolynomial over the integers, which is reducible modulo everyprime. We don’t go into that here but discuss another probleminvolving primes and integer polynomials.

Kronecker-conjugacy

For f ∈ Z[X ] and, p prime, consider the value set

Valp(f ) = f (a) mod p ; a ∈ Z.

Call f , g ∈ Z[X ] to be Kronecker-conjugate if Valp(f ) = Valp(g)for all but finitely many primes p.There is a group-theoretic criterion for checkingKronecker-conjugacy.

Kronecker-conjugacy

Call f , g ∈ Z[X ] to be Kronecker-conjugate if Valp(f ) = Valp(g)for all but finitely many primes p.

There is a group-theoretic criterion for checkingKronecker-conjugacy.

Kronecker-conjugacy

Call f , g ∈ Z[X ] to be Kronecker-conjugate if Valp(f ) = Valp(g)for all but finitely many primes p.There is a group-theoretic criterion for checkingKronecker-conjugacy.

To mention in passing:Schur conjectured that the Tchebychev polynomials are essentiallythe only integer polynomials f which are permutation polynomialsmodulo p (that is, Valp(f ) = Zp) for almost all primes p. This wasproved by M.Fried using group-theoretic techniques similar to ourdiscussion.

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

Fried’s Theorem.Given f , g ∈ Z[X ] which are non-constant, considert,K ,G , x , y ,U,V as above. Then, f , g are Kronecker-conjugate ifand only if ⋃

g∈GgUg−1 =

⋃g∈G

gVg−1.

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) andg(X) by un−1f (X/u) and un−1g(X ) respectively, we may assumethat 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 groupsGal(f (X )− g(y) over K (y) and Gal(f (X )− g(a)) over K areisomorphic as permutation groups for infinitely many a.Recall that g(y) = t. Thus every element of the Galois groupGal(f (X )− t) over K (y) fixes at least one root.This Galois group is just the induced action of V on the roots off (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 areKronecker-conjugate if and only if, the permutation representationsIndG

U 1 and IndGV 1 are equivalent (that is, U and V are Gassmann

equivalent).

Actually, this question is not purely group-theoretic because thereare examples of abstract finite groups G and subgroups U,V suchthat

⋃g∈G gUg−1 =

⋃g∈G gVg−1 but U,V are not Gassmann

equivalent.

Conjecture.Over a field of characteristic 0, two polynomials f and g areKronecker-conjugate if and only if, the permutation representationsIndG

U 1 and IndGV 1 are equivalent (that is, U and V are Gassmann

equivalent).Actually, this question is not purely group-theoretic because thereare examples of abstract finite groups G and subgroups U,V suchthat

⋃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 (orrational) coefficients such that the value sets of f , g at all rationalnumbers are the same?

Evidently, f (x) = g(ax + b) for some rational a, b with a 6= 0 is apossibility. Are there others?

More generally, if the value set of f at rationals is contained in thecorresponding value set of g .

If h is some rational polynomial so that f (x) = g(h(x)), this holds;are there other situations?

Think of the special case g(x) = xn; we are asking if a polynomialtakes only n-th powers as values, is it necessarily the n-th power ofa polynomial?

Rational values

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

Let us indicate a powerful method which can prove many suchresults.

The idea is that a rational polynomial in two (or more variables)which is irreducible admits infinitely many specializations of all butone variable to rationals so that the resulting polynomials in onevariable are irreducible.

This goes under the name of Hilbert’s irreducibility theorem. Letus use this to answer our question.

We are given two nonconstant polynomials f , g with rationalcoefficients such that f (Q) ⊆ g(Q).

Write the two variable polynomial

f (X )− g(Y ) = h1(X ,Y )h2(X ,Y ) · · · hd(X ,Y )

where hi are irreducible.

By Hilbert’s irreducibility theorem, there are infinitely manyrational x0 such that hi (x0,Y ) are irreducible in Y .

On the other hand, for each such x0, f (x0) ∈ f (Q) ⊆ g(Q); thatis, there is a corresponding y0 such that f (x0) = g(y0).

Hence, one hi (say h1) has the property that hi (x0,Y ) has a zero(in Y ) for infinitely many values of x0 (while being irreducible!).

f (X )− g(Y ) = h1(X ,Y )h2(X ,Y ) · · · hd(X ,Y )

Therefore, h1 must have degree 1 in Y ; that is,

h1(X ,Y ) = p(X ) + q(X )Y .

f (X )− g(Y ) = (p(X ) + q(X )Y )h2(X ,Y ) · · · hd(X ,Y );

so, f (X ) = g(−p(X )/q(X )).

This is easily seen to imply that p(X )/q(X ) is actually apolynomial in X ; this proves f (X ) = g(h(X )).

h1(X ,Y ) = p(X ) + q(X )Y .

so, f (X ) = g(−p(X )/q(X )).

h1(X ,Y ) = p(X ) + q(X )Y .

so, f (X ) = g(−p(X )/q(X )).

The support problem

Let x , y be positive integers such that each prime divisor of xn − 1also divides yn − 1 for every n. What can we say about the relationbetween 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 cameto be known as the ‘support problem’.One calls the ‘support’ of a natural number a > 1 to be the set ofits prime divisors.

The support problem

Let x , y be positive integers such that each prime divisor of xn − 1also divides yn − 1 for every n. What can we say about the relationbetween 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 cameto be known as the ‘support problem’.One calls the ‘support’ of a natural number a > 1 to be the set ofits prime divisors.

The support problem

Let x , y be positive integers such that each prime divisor of xn − 1also divides yn − 1 for every n. What can we say about the relationbetween 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 cameto be known as the ‘support problem’.

One calls the ‘support’ of a natural number a > 1 to be the set ofits prime divisors.

The support problem

Let x , y be positive integers such that each prime divisor of xn − 1also divides yn − 1 for every n. What can we say about the relationbetween 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 cameto be known as the ‘support problem’.One calls the ‘support’ of a natural number a > 1 to be the set ofits prime divisors.

This is proved using Kummer theory.

Kummer theory is a correspondence between abelian extensions ofa field K and subgroups of the n-th powers of K ∗.For our purposes in this application, we may think of the followingconcrete statement as Kummer theory:

If K contains the n-th roots of unity, then abelian extensions L ofK whose Galois groups have exponent n correspond bijectively tosubgroups Ω of K ∗ containing (K ∗)n via L 7→ K ∗ ∩ (L∗)n and itsinverse map Ω 7→ K (Ω1/n).

Kummer theory is a correspondence between abelian extensions ofa field K and subgroups of the n-th powers of K ∗.

For our purposes in this application, we may think of the followingconcrete statement as Kummer theory:

Kummer theory is a correspondence between abelian extensions ofa field K and subgroups of the n-th powers of K ∗.For our purposes in this application, we may think of the followingconcrete statement as Kummer theory:

Kummer theory is a correspondence between abelian extensions ofa field K and subgroups of the n-th powers of K ∗.For our purposes in this application, we may think of the followingconcrete statement as Kummer theory:If K contains the n-th roots of unity, then abelian extensions L ofK whose Galois groups have exponent n correspond bijectively tosubgroups Ω of K ∗ containing (K ∗)n via L 7→ K ∗ ∩ (L∗)n and itsinverse map Ω 7→ K (Ω1/n).

The support problem is:

Theorem:Let x , y ∈ Q∗. If, for all but finitely many primes p and, for alln ∈ N, one has the implication

p | (xn − 1) =⇒ p | (yn − 1),

then y is a power of x.Here p divides a rational number x = a

b , (a, b) = 1, means thatp | a.

The support problem is:Theorem:Let x , y ∈ Q∗. If, for all but finitely many primes p and, for alln ∈ N, one has the implication

p | (xn − 1) =⇒ p | (yn − 1),

then y is a power of x.Here p divides a rational number x = a

b , (a, b) = 1, means thatp | a.

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

For a prime p not dividing the numerator and denominators of xand y , look at the order of x mod p; viz.,The smallest n such that p|(xn − 1).We have p|(yn − 1) so that the order of y mod p divides n, theorder of x .A simple exercise in the cyclic group Z∗p shows that y must be apower of x mod p.Therefore, the support theorem says that if, modulo all but finitelymany primes p, y is a power of x modulo p, then y is actually apower of x ; this is what Kummer theory accomplishes.

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

The smallest n such that p|(xn − 1).We have p|(yn − 1) so that the order of y mod p divides n, theorder of x .A simple exercise in the cyclic group Z∗p shows that y must be apower of x mod p.Therefore, the support theorem says that if, modulo all but finitelymany primes p, y is a power of x modulo p, then y is actually apower of x ; this is what Kummer theory accomplishes.

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

We have p|(yn − 1) so that the order of y mod p divides n, theorder of x .A simple exercise in the cyclic group Z∗p shows that y must be apower of x mod p.Therefore, the support theorem says that if, modulo all but finitelymany primes p, y is a power of x modulo p, then y is actually apower of x ; this is what Kummer theory accomplishes.

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

A simple exercise in the cyclic group Z∗p shows that y must be apower of x mod p.Therefore, the support theorem says that if, modulo all but finitelymany primes p, y is a power of x modulo p, then y is actually apower of x ; this is what Kummer theory accomplishes.

This is in reality a local-global theorem. To explain:For a prime p not dividing the numerator and denominators of xand y , look at the order of x mod p; viz.,The smallest n such that p|(xn − 1).We have p|(yn − 1) so that the order of y mod p divides n, theorder of x .A simple exercise in the cyclic group Z∗p shows that y must be apower of x mod p.

Therefore, the support theorem says that if, modulo all but finitelymany primes p, y is a power of x modulo p, then y is actually apower of x ; this is what Kummer theory accomplishes.

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

Recall the three facts we mentioned earlier which will now beuseful in solving the support problem.

Assume x , y are as in the theorem, that is, for all n and all butfinitely many primes p, one has

p | (xn − 1) =⇒ p | (yn − 1).

We us Kummer’s theorem which, for our purposes, gives using thefacts above:Let q be an odd prime power. Then the natural mapQ∗/(Q∗)q → Q(ζq)∗/(Q(ζq)∗)q is injective.

p | (xn − 1) =⇒ p | (yn − 1).

We us Kummer’s theorem which, for our purposes, gives using thefacts above:

Let q be an odd prime power. Then the natural mapQ∗/(Q∗)q → Q(ζq)∗/(Q(ζq)∗)q is injective.

p | (xn − 1) =⇒ p | (yn − 1).

We us Kummer’s theorem which, for our purposes, gives using thefacts above:Let q be an odd prime power. Then the natural mapQ∗/(Q∗)q → Q(ζq)∗/(Q(ζq)∗)q is injective.

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

Let q be a power of a prime l and let ζq be a primitive qth root of

unity. Then Q(ζq, x1q ) ⊇ Q(ζq, y

This is proved using a generalization of the so-called HilbertTheorem 90 slightly.

Look at our x , y in the support theorem. We can show:Let q be a power of a prime l and let ζq be a primitive qth root of

Basically, the hypothesis of the support theorem shows thaty = xd in the group Q(ζq)∗/(Q(ζq)∗)q. So, by the last quotedresult based on Kummer, we have y = xd in Q∗/(Q∗)q also.

It is a simple matter to show that this is a power over integersalso; I don’t say any more about it.

Basically, the hypothesis of the support theorem shows thaty = xd in the group Q(ζq)∗/(Q(ζq)∗)q. So, by the last quotedresult based on Kummer, we have y = xd in Q∗/(Q∗)q also.It is a simple matter to show that this is a power over integersalso; 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, hemade the following 4 conjectures. For positive integers a, b, let usdenote the sequence an + bn∞n=1 by Sa,b and we write p|Sa,b if pdivides at least one member of Sa,b. Fermat stated :

Conjecture

(Fermat, 1641)1) If p|S3,1, then p 6≡ −1(mod 12).2) If p|S3,1, then p 6≡ +1(mod 12).3) If p|S5,1, then p 6≡ −1(mod 10).4) If p|S5,1, then p 6≡ +1(mod 10).

The quadratic reciprocity law implies that Conjecture 1 holds true.However, the remaining three conjectures are all false. We caneven show that there is a positive density of primes for which eachof the conclusions of these three conjectures is false.

One can determine explicitly in terms of a, b, c , d the naturaldensity of primes in the residue class c mod d which divide Sa,b.For instance, denoting by δa,b(c , d) the density of primes in theresidue class c mod d which divides Sa,b, our theorem has thefollowing bearing on the above conjectures :

Corollary

δ3,1(1, 12) =1

6, δ3,1(5, 12) =

4, δ3,1(7, 12) =

4and δ3,1(11, 12) = 0.

Furthermore, we have

δ5,1(1, 10) =1

12, δ5,1(3, 10) =

4, δ5,1(7, 10) =

4and δ5,1(9, 10) =

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

Gauss considers in articles 315-317 of his DisquisitionesArithmeticae (1801) the decimal expansion of numbers of the form1p 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 6= 2, 5 andthe 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 10k − 1 is a multiple ofp).Artin conjectured that any integer a 6= 0, 1,−1 is a primitive rootmodulo infinitely many primes.

The ideas to prove come from the approach to the so-called Artinconjecture. To motivate it:Gauss considers in articles 315-317 of his DisquisitionesArithmeticae (1801) the decimal expansion of numbers of the form1p 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 6= 2, 5 andthe 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 10k − 1 is a multiple ofp).Artin conjectured that any integer a 6= 0, 1,−1 is a primitive rootmodulo infinitely many primes.

For example, 1/7 = 0.142857, 1/11 = 0.09.Note that the decimal expansion of 1/p is periodic for p 6= 2, 5 andthe 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 10k − 1 is a multiple ofp).

Artin conjectured that any integer a 6= 0, 1,−1 is a primitive rootmodulo infinitely many primes.

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

For instance, it turns out that the two-thirds of the set of allprimes divide a number of the form 10n + 1. This means :Two-thirds of the prime numbers p, the decimal expansion of 1/phas even period!Similarly, 17/24-th of the proportion of prime numbers divide anumber of the form 2n + 1.

Start with two given integers a, b and consider the set of all primenumbers which divide an + bn for some n.For instance, it turns out that the two-thirds of the set of allprimes divide a number of the form 10n + 1. This means :

Two-thirds of the prime numbers p, the decimal expansion of 1/phas even period!Similarly, 17/24-th of the proportion of prime numbers divide anumber of the form 2n + 1.

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

Similarly, 17/24-th of the proportion of prime numbers divide anumber of the form 2n + 1.

Start with two given integers a, b and consider the set of all primenumbers which divide an + bn for some n.For instance, it turns out that the two-thirds of the set of allprimes divide a number of the form 10n + 1. This means :Two-thirds of the prime numbers p, the decimal expansion of 1/phas even period!Similarly, 17/24-th of the proportion of prime numbers divide anumber of the form 2n + 1.

Artin’s primitive root conjecture (1927)

Let g ∈ Q \ −1, 0, 1. For a prime p not dividing neither thenumerator nor the denominator, one can look at the subgroup ofF∗p generated by g and call g a primitive root modulo p if this isthe 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 aprimitive 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 = gh

0 with g0 ∈ Q. Wehave, as x tends to infinity,

P(g)(x) :=∑

p∈P(g),p≤x

=∏p-h

(1− 1

p(p − 1))∏p|h

(1− 1

p − 1)

log x+ o(x/ log x)

Artin’s primitive root conjecture (1927)

Let g ∈ Q \ −1, 0, 1. For a prime p not dividing neither thenumerator nor the denominator, one can look at the subgroup ofF∗p generated by g and call g a primitive root modulo p if this isthe 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 aprimitive 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 = gh

0 with g0 ∈ Q. Wehave, as x tends to infinity,

P(g)(x) :=∑

p∈P(g),p≤x

=∏p-h

(1− 1

p(p − 1))∏p|h

(1− 1

p − 1)

log x+ o(x/ log x)

Recall the Prime Number Theorem in the formπ(x) := #p ≤ x ∼ Li(x) as x tends to infinity (i.e. the quotientof the r.h.s. and l.h.s. tends to 1 as x tends to infinity), where Thelogarithmic integral Li(x) is defined as

∫ x2 dt/ log t.

The theorem suggests that the ‘probability that a number n isprime’ is of the order 1/ log n.

The prime number theorem is useful especially when used with anerror term.For instance, we shall prove statements like the following:The number of prime numbers p ≤ x dividing some term of thesequence an + bn equals δ(a, b)Li(x) + O(E (x)) where δ(a, b) is apositive real number and E (x)/Li(x)→ 0 as x →∞.

The starting point to analyze Artin’s primitive root conjecture isthe observation :

p ∈ P(g) ⇐⇒ gp−1q 6≡ 1( mod p) for every prime q dividing p−1.

The starting point to analyze Artin’s primitive root conjecture isthe observation :

p ∈ P(g) ⇐⇒ gp−1q 6≡ 1( mod p) for every prime q dividing p−1.

”=⇒” Obvious.”⇐=” Suppose p 6∈ P(g).

Then gp−1k ≡ 1( mod p) for some k |p − 1, k > 1.

But this implies that gp−1d ≡ 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 canconsider the set of primes p such that p ≡ 1( mod q) and

gp−1q 6≡ 1( mod p).

”=⇒” Obvious.”⇐=” Suppose p 6∈ P(g).

Then gp−1k ≡ 1( mod p) for some k |p − 1, k > 1.

But this implies that gp−1d ≡ 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 canconsider the set of primes p such that p ≡ 1( mod q) and

gp−1q 6≡ 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 gp−1q ≡ 1( mod p).

The main observation is that the above two conditions can beinterpreted as the prime p splitting completely in a certain fieldextension and, Chebotarev density theorem determines the densityof such primes in terms of the field degree.

For instance, p ≡ 1( mod q) is true for primes p with density1/ϕ(q) = 1/(q − 1).

About the other condition gp−1q ≡ 1( mod p), using Fermat’s

little theorem, we infer (in case p - g) that gp−1q is a solution of

xq ≡ 1( mod p).

We expect there to be q solutions and we want a solution to be 1modulo q.Thus we expect to be successful with probability 1

q , except when

q|h. Then gp−1q = g

h p−1q

0 ≡ 1( mod p), trivially.

If we assume that these events are independent then the probabilitythat both events occur is 1

q(q−1) if q - h and 1q−1 otherwise.

The above events should not occur for any q in order to ensurethat p ∈ P(g).This suggests a natural density of∏

(1− 1

q(q − 1)

)∏q|h

(1− 1

q − 1

)for such primes and hence we expect the analytic formulation ofArtin’s conjecture to hold true.

xq ≡ 1( mod p).

q , except when

h p−1q

(1− 1

q(q − 1)

)∏q|h

(1− 1

q − 1

xq ≡ 1( mod p).

q , except when

h p−1q

(1− 1

q(q − 1)

)∏q|h

(1− 1

q − 1

xq ≡ 1( mod p).

q , except when

h p−1q

(1− 1

q(q − 1)

)∏q|h

(1− 1

q − 1

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 gdivides (p − 1)/r .

This is just the condition that p splits completely in the splittingfield Q(ζr , g

1/r ) of the polynomial X r − g over Q.In particular, if v2(p − 1) = v2(r), then this is equivalent to ghaving odd order modulo p.

This, in turn, is characterized by the property that p ≡ 1 + 2k mod2k+1 and splits completely in the field extensionKk := Q(ζ2k , g

1/2k ).

So, g is a primitive root modulo p if and only if p does not splitcompletely in any of the fields Kk for k > 1.

1/2k ).

Primes dividing an + bn as splitting conditions

Returning to our problem of primes dividing sequences, we interpretthe statement that a prime p divides some term of an + bn asstatements about splitting of primes in field extensions.

Leaving out primes dividing b, a prime p divides some an + bn ifand only if, g := a/b has odd order in F∗p.We already know this as splitting conditions in the fieldsKk := Q(ζ2k , 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 fieldsobtained by intersecting fields like Kk ’s above with Q(ζd) etc. andwith a given Frobenius conjugacy class.

Returning to our problem of primes dividing sequences, we interpretthe statement that a prime p divides some term of an + bn asstatements about splitting of primes in field extensions.Leaving out primes dividing b, a prime p divides some an + bn ifand only if, g := a/b has odd order in F∗p.We already know this as splitting conditions in the fieldsKk := Q(ζ2k , 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 fieldsobtained by intersecting fields like Kk ’s above with Q(ζd) etc. andwith a given Frobenius conjugacy class.

Returning to our problem of primes dividing sequences, we interpretthe statement that a prime p divides some term of an + bn asstatements about splitting of primes in field extensions.Leaving out primes dividing b, a prime p divides some an + bn ifand only if, g := a/b has odd order in F∗p.We already know this as splitting conditions in the fieldsKk := Q(ζ2k , 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 fieldsobtained by intersecting fields like Kk ’s above with Q(ζd) etc. andwith a given Frobenius conjugacy class.

Theorem

Let a and b be positive integers with a 6= b. Let Na,b(x) count thenumber of primes p ≤ x that divide Sa,b. Put r = a/b. Let λ be

the largest integer such that r = u2λ , with u a rational number.Let L = Q(

√u). We have

Na,b(x) = δ(r)Li(x) + O

(x(log log x)4

log3 x

where the implied constant may depend on a and b, and δ(r) is apositive rational number that is given in Table 0.

Table 0: The value of δ(r)

L λ δ(r)

L 6= Q(√

2) λ ≥ 0 21−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 6= b.We can use the ideas used in the approach to Artin’s conjecture todetermine the density of prime numbers which divide some numberof the form an + bn and are in the residue class c mod d .

The data are tabulated in a convenient form so that givenparticular values a, b, c , d it is easy to find which table and whichrow of it to look at to find the corresponding value.

Let a, b, c , d be integers with (c , d) = 1 and assume that a 6= b.We can use the ideas used in the approach to Artin’s conjecture todetermine the density of prime numbers which divide some numberof the form an + bn and are in the residue class c mod d .

The data are tabulated in a convenient form so that givenparticular values a, b, c , d it is easy to find which table and whichrow of it to look at to find the corresponding value.

Compute r0, h from a/b = rh0 , where r0 is not a proper power of arational number.Then, write λ = v2(h).Find δ, d ′ in d = 2δd ′, with δ = v2(d).Get γ = v2(c − 1), where it is understood that γ is larger than anynumber when c = 1.Write the discriminant D(r0) of the quadratic field Q(

√r0) as

D(r0) = 2δ0D ′.Writing r0 = u/v , write t = −r0 if u, v are odd and,write t =

∏ki=1(−1pi )pi if uv = 2

∏ki=1 pi .

Table 1 : Q(√

r0) 6= Q(√

2),D ′ - d ′

λ δ φ(d)δa,b(c , d)

< δ ≤ γ 1− 2λ+1−δ

∗ > 0,≤ min(λ, γ) 2δ−λ

∗ 0 21−λ

≥ γ > γ 0

< γ > γ 1− 2λ−γ

Table 2 : Q(√

r0) 6= Q(√

2),D ′|d ′, δ0 ≤ δ

λ δ (D(r0)c ) φ(d)δa,b(c , d)

≥ δ − 1 > 0,≤ γ 1 2δ−1−λ

3−1 2δ−1−λ

∗ 0 1 2−λ

3−1 2−λ

< δ − 1 ≤ γ 1 1− 2λ+2−δ

3−1 1

≥ δ > γ ∗ 0

≤ γ − 1 > γ 1 1− 2λ+1−γ

−1 1

≥ γ > λ ∗ 0

Table 3 : Q(√

r0) 6= Q(√

2),D ′|d ′ and δ0 > δ

λ δ (D(t)c ) φ(d)δa,b(c , d)

< δ − 1 ≤ γ 1 1− 2λ+1−δ

3 + 2λ+2+δ−2δ0

< δ − 1 ≤ γ −1 1− 2λ+1−δ

3 − 2λ+2+δ−2δ0

= δ − 1 ≤ γ 1 23 + 22δ+1−2δ0

= δ − 1 ≤ γ −1 23 −

22δ+1−2δ0

≤ γ − 1 > γ ∗ 1− 2λ−γ

≥ γ > λ ∗ 0

≥ δ > γ ∗ 0

≤ δ0 − 2 > 0,≤ min(γ, λ) 1 2δ−λ

3 + 2λ+2+δ−2δ0

≤ δ0 − 2 > 0,≤ min(γ, λ) −1 2δ−λ

3 − 2λ+2+δ−2δ0

≥ δ0 − 1 > 0,≤ γ 1 2δ−1−λ

≥ δ0 − 1 > 0,≤ γ −1 2δ−λ−1

≤ δ0 − 2 0 1 21−λ

3 + 2λ+3−2δ0

≤ δ0 − 2 0 −1 21−λ

3 − 2λ+3−2δ0

≥ δ0 − 1 0 1 2−λ

≥ δ0 − 1 0 −1 2−λB.Sury Primes, Polynomials, Progressions

Table 4 : Q(√

r0) = Q(√

2), δ ≤ 2

λ δ γ φ(d)δa,b(c , d)

0 ≤ 1 ≥ δ 17/24

0 2 ≥ δ 11/12

0 2 1 1/2

1 2 1 0

1 ≤ 1 ≥ δ 5/12

1 2 ≥ δ 5/6

≥ 2 ≤ 1 ≥ δ 2−λ/3

≥ 2 2 ≥ δ 21−λ/3

≥ 2 2 1 0

Table 5 : Q(√

r0) = Q(√

2), δ ≥ 3, λ > 0

λ δ γ φ(d)δa,b(c, d)

≥ 2 3 < δ 0

≥ δ − 1 ≥ 3 ≥ δ 2δ−1−λ

≥ 2, < δ − 1 ≥ 4 ≥ δ 1− 2λ+2−δ

≥ 2,≤ γ − 2 ≥ 4 < δ 1− 2λ+1−γ

≥ max(2, γ − 1) ≥ 4 < δ 0

1 ≥ 3 ≥ δ 1− 23−δ

1 ≥ 3 1 0

1 ≥ 3 2 1

1 ≥ 3 ≥ 3, < δ 1− 22−γ

Table 6 : Q(√

r0) = Q(√

2), δ ≥ 3, λ = 0

γ c(mod 8) φ(d)δa,b(c, d)

≥ δ 1 1− 22−δ

≤ 2 ±1 0

≤ 2 ±3 1

≥ 3, < δ 1 1− 21−γ

Illustration : a = 36, b = 1, c = 7, d = 30

Now a = 36, b = 1, c = 7, d = 30 gives a/b = 36 = 62; sor0 = 6, h = 2, λ = v2(h) = 1. As d = 30, we get δ = 1, d ′ = 15and γ = v2(c − 1) = v2(6) = 1.As r0 = 6, discriminant D(r0) = 24 = 2δ0D ′ gives δ0 = 3,D ′ = 3.

So, t = −3 and so

(D(t)c

(−37

As 3 = D ′|d ′ = 15, we need to look at table 2 or 3. Asδ0 = 3 > δ = 1, we look at table 3.

Table 3 : Q(√r0) 6= Q(

√2),D ′|d ′ and δ0 > δ

λ δ (D(t)c ) φ(d)δa,b(c , d)

< δ − 1 ≤ γ 1 1− 2λ+1−δ

3 + 2λ+2+δ−2δ0

< δ − 1 ≤ γ −1 1− 2λ+1−δ

3 − 2λ+2+δ−2δ0

= δ − 1 ≤ γ 1 23 + 22δ+1−2δ0

= δ − 1 ≤ γ −1 23 −

22δ+1−2δ0

≤ γ − 1 > γ ∗ 1− 2λ−γ

≥ γ > λ ∗ 0

≥ δ > γ ∗ 0

≤ δ0 − 2 > 0,≤ min(γ, λ) 1 2δ−λ

3 + 2λ+2+δ−2δ0

≤ δ0 − 2 > 0,≤ min(γ, λ) −1 2δ−λ

3 − 2λ+2+δ−2δ0

≥ δ0 − 1 > 0,≤ γ 1 2δ−1−λ

≥ δ0 − 1 > 0,≤ γ −1 2δ−λ−1

≤ δ0 − 2 0 1 21−λ

3 + 2λ+3−2δ0

≤ δ0 − 2 0 −1 21−λ

3 − 2λ+3−2δ0

≥ δ0 − 1 0 1 2−λ

≥ δ0 − 1 0 −1 2−λB.Sury Primes, Polynomials, Progressions

As λ = 1 = δ = γ, the first column in table 3 tells us that we arenot in the first 5 rows. The 6th and 7th rows are ruled out bylooking at the second column. Also, the last 4 are ruled out againby looking at the 2nd column as δ 6= 0. We are left with rows 8 to11. As δ0 = 3, the first column rules out rows 10 and 11. Finally,row 9 is ruled out by the 3rd column. Hence the densityφ(30)δ36,1(7, 30) = 2δ−λ+2λ+2+δ−2δ0

3 = (1 + 2−2)/3 = 5/12. Sinceφ(30) = 8, the density of primes congruent to 7 mod 30 whichdivide the sequence 36n + 1 equals 5/96.

A problem by Schinzel & Wojcik

We recall an interesting problem which can be approached similarlyto the above one. In 1976, K.R.Matthews proved on GRH asimultaneous primitive roots theorem. He proved :Given non-zero integers a1, · · · , ar , there is a constantc(a1, · · · , ar ) ≥ 0 such that if GRH holds, then

|p ≤ x :< ai mod p >= Z∗p∀i|

= c(a1, · · · , ar )Li(x) + O

(x(log log x)2

(log x)2

This has bearing on the following problem studied by Pappalardi,Schinzel, Susa, Wojcik etc. Schinzel & Wojcik posed the followingproblem :Given a, b, c ∈ Q \ 0,±1, do there exist infinitely many primes psuch that a, b, c have equal orders modulo p ?More generally, the same problem can be posed for r rationalnumbers as above instead of three. It is interesting to note thatthere are a, b, c which have no odd prime with the desiredproperty! Indeed, if b = a2 = −c , then the order of b = a2 mod p

for any odd prime p, must divideordp(a)

2 . So, it is very interestingto determine all the possible triples for which the question has anaffirmative answer.

Matthews’s theorem can be used to deduce an affirmative answerfor any a1, · · · , ar with c(a1, · · · , ar ) 6= 0, provided we assumeGRH. For instance, the triple 2, 3,−6 provides an affirmativeanswer as c(2, 3,−6) 6= 0. However, the case 4, 3,−12 is still open,for instance. In fact, Wojcik proved under Schinzel’s hypothesis Hthat whenever a1, · · · , ar 6= 0, 1 are in some number field K andgenerate a torsion-free subgroup of K ∗, then Schinzel’s hypothesisH implies that there are infinitely many prime ideals P of degree 1in K such that all the ai ’s have the same order modulo P.

The case 4, 3,−12 is thus unsolved even if we assume GRH andSchinzel’s hypothesis.

Matthews’s theorem can be used to deduce an affirmative answerfor any a1, · · · , ar with c(a1, · · · , ar ) 6= 0, provided we assumeGRH. For instance, the triple 2, 3,−6 provides an affirmativeanswer as c(2, 3,−6) 6= 0. However, the case 4, 3,−12 is still open,for instance. In fact, Wojcik proved under Schinzel’s hypothesis Hthat whenever a1, · · · , ar 6= 0, 1 are in some number field K andgenerate a torsion-free subgroup of K ∗, then Schinzel’s hypothesisH implies that there are infinitely many prime ideals P of degree 1in K such that all the ai ’s have the same order modulo P.

The case 4, 3,−12 is thus unsolved even if we assume GRH andSchinzel’s hypothesis.

Finally, I leave you with a beautiful and surprising fact which canbe explained using ideas from algebraic number theory.

Observe that the polynomial x2 + x + 41 takes prime values for theconsecutive values x = 0, 1, · · · , 39.

Observe (use a computer) that the irrational number eπ√163 has

the digit 9 to twelve decimal places.

The two facts are essentially equivalent! Can you find out why?

I want to finish by saying that:

IISER may be IISER but ....

NISER is nicer!

THANK YOU!

NISER is nicer!

THANK YOU!

NISER is nicer!

THANK YOU!

NISER is nicer!

THANK YOU!

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