On Some Norm Equations over Cyclotomic Fields

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On Some Norm

Equations over

Cyclotomic Fields

Peter Lombaers

Tese de Doutoramento apresentada à

Faculdade de Ciências da Universidade do Porto

Matemática

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Peter Lombaers: On Some Norm Equations Over Cyclotomic Fields , July 2019, Porto

s u p e r v i s o r: António Machiavelo

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A B S T R A C T

For an odd prime p, the equation xp+yp ₌ _zp _{can be rewritten as} (x+y)Np(x+yζp) =zp, where ζpis a primitive p-th root of unity and Npis the norm map of the fieldQ(ζp). That this norm equation does not have non-trivial integer solutions is known since Andrew Wiles’ proof of Fermat’s last theorem. However, very little is known about two natural generalisations of this equation given by the equations Np(x+yζp) =zp, and(a0+a1+ · · · +ak)Np(a0+a1ζp+ · · · +akζkp) = zp, for some integer 1≤k≤ p−3.

In this thesis, we look at the cyclotomic methods that were used to attack the equation xp+yp =zp, and we apply them to those two gen-eralisations. Naturally, this leads to a number of new observations and problems. First of all, we fix the coefficients a0, . . . , ak and investigate their behaviour as p goes to infinity. This leads to an upper bound on p in certain cases. Next, we show that Np(a0+ · · · +akζk_p)can be expressed in terms of linear recurrence sequences depending only on the coefficients ai. This allows us to show that, for fixed a, b, there are only finitely many primes p such that Np(aζ2p+bζp−a) =zphas non-trivial integer solutions.

We review Kummer’s method of attack to the equation Np(x+ yζp) =zp by the use of a logarithmic derivative, and we show that the Iwasawa p-adic logarithm can be used to obtain the same results. We review the theory of unramified extensions of degree p ofQ(ζp), and we see how this leads to a proof of Herbrand’s theorem. Combining knowledge of the units ofZ[ζp]with computational results, we show that if Np(x+yζp) = zp, then max(|x|,|y|) > 5×108, for regular primes p≥ 5. Finally, we use the p-adic logarithm to give a unified way of proving congruence identities related to Wolstenholme’s the-orem. With the same method, we can also prove a number of other congruences, which at first sight seem quite unrelated to each other.

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R E S U M O

Para um primo p ímpar, a equação xp+yp = zp pode ser reescrita como (x+y)Np(x+yζp) = zp, em que ζp é uma raiz primitiva p-ésima da unidade e Np é a norma do corpoQ(ζp). Que esta equação não tem soluções inteiras não-triviais é sabido desde a prova de Andrew Wiles do último teorema de Fermat. No entanto, pouco se sabe acerca das duas generalizações seguintes dessa equação: Np(x+ yζp) = zp e(a0+a1+ · · · +ak)Np(a0+a1ζp+ · · · +akζk_p) = zp, para algum inteiro 1≤k≤ p−3.

Nesta tese, olhamos para os métodos ciclotómicos que foram usados para atacar a equação xp+yp =zp, e aplicamo-los a essas duas gener-alizações. Naturalmente, isso conduz a uma série de novas observações e problemas. Em primeiro lugar, fixamos os coeficientes a0, . . . , ak e investigamos o seu comportamento quando p tende para infinito. Isto conduz a um limite superior para p, em certos casos. Em seguida, mostramos que a norma Np(a0+ · · · +akζk_p)pode ser expressa em termos de sequências de recorrência linear que dependem apenas dos coeficientes ai. Isso permite-nos mostrar que, para a, b fixos, há apenas um número finito de primos p tais que Np(aζ2p+bζp−a) =zp tem soluções inteiras não-triviais.

Revemos o método de Kummer para atacar a equação Np(x+ yζp) = zp usando derivadas logarítmicas e mostramos que o log-aritmo p-ádico de Iwasawa pode ser usado para alcançar os mesmos resultados. Revemos a teoria das extensões não-ramificadas de grau p de Q(ζp), e vemos como isso conduz a uma demonstração do teo-rema de Herbrand. Combinando conhecimento das unidades deZ[ζp] com resultados computacionais, mostramos que se Np(x+yζp) =zp, então max(|x|,|y|) > 5×108, para primos regulares p ≥ 5. Final-mente, usamos o logaritmo p-ádico para dar uma maneira unificada de demonstrar identidades envolvendo congruências relacionadas com o Teorema de Wolstenholme. Com o mesmo método podemos também demonstrar várias outras congruências que, à primeira vista, parecem não estar relacionadas entre si.

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A C K N O W L E D G M E N T S

This thesis would not have been possible without the help and support of a number of people. First of all I would like to thank my supervisor. António, it’s been a great pleasure to work with you. You’ve always been friendly, enthousiastic and helpful, and I feel lucky to have had you as my supervisor. Next, I would like to thank my friends here in Porto and the friends from the Netherlands that came to visit. You made my time here a great experience, whether it was to enjoy the good things, or complain about the bad things. Special mention should go to Atefeh, without you I would have been taking the stairs for the past four years, and to Nikos, you were always there, listening to my explanations of what I was working on, so that I could understand it better myself. I also want to thank my family for their love and support while I was far away from home. Finally, Saul, without you I would not survived the final year, but you turned it into a great year.

I would like to acknowledge the financial support by the FCT (Fundação para a Ciência e a Tecnologia) through the PhD grant PD/BD/128063/2016, and by CMUP (Centro Matemática - Universi-dade do Porto) (UID/MAT/00144/2013), which is funded by FCT (Por-tugal) with national (MEC) and European structural funds (FEDER), under the partnership agreement PT2020.

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C O N T E N T S

1 i n t r o d u c t i o n 1

1.1 The main problem 1

1.2 Kummer’s approach 2

1.3 Description of the chapters 4

1.4 Notation 6

2 l i m i t s o f N-th roots 9

2.1 The Mahler measure 9

2.2 An upper bound on p 11

3 r e c u r r e n c e s e q u e n c e s 15

3.1 Norms in terms of a recurrence sequence 15

3.2 A general recurrence theorem 21

4 f r o m n o r m s t o s i n g u l a r i n t e g e r s 27

4.1 Singular integers 27

4.2 Self-prime integers 28

5 s i n g u l a r i n t e g e r s 35

5.1 Idempotents and the Herbrand-Ribet theorem 35

5.2 Kummer’s logarithmic derivative 37

5.3 The p-adic logarithm 40

6 k u m m e r e x t e n s i o n s 47

6.1 Kummer extensions 47

6.2 Ramification 49

6.3 Units in cyclotomic fields 52

6.4 Unramified extensions 55

7 r e g u l a r p r i m e s 61

7.1 Theoretical results 61

7.2 Computational results 63

8 w o l s t e n h o l m e’s theorem 67

8.1 Logarithms and binomial coefficients 68

8.2 Logarithms and cyclotomic integers 73

b i b l i o g r a p h y 81

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1

I N T R O D U C T I O N

1.1 t h e m a i n p r o b l e m

In this thesis we look at two generalisations of Fermat’s last theorem. They arise naturally when we approach Fermat’s last theorem via the classical, cyclotomic method. Let us first recall the statement of Fermat’s last theorem:

Theorem 1.1(FLT). For each integer n≥3 there are no non-zero integers x, y, z such that

xn+yn=zn. (1.1)

The first succesful proof was given by Andrew Wiles, using a con-nection with elliptic curves. Before that, the traditional method to attack FLT was via cyclotomic fields. We first note that we may assume in Theorem1.1, without loss of generality, that n =4 or n = p is an

odd prime. Fermat himself already wrote a proof for the case n=4 (see chapter 1 of [29] for a good account of the early history of FLT). Therefore we can restrict our attention to the case n = p is an odd prime.

Since p is odd, we know that xp+yp is divisible by x+y. Let ζp be a primitive p-th root of unity. Then we have the equality

xp+yp = p−1

∏

i=0

(x+yζi_p).

Let us denote by Np = NQ(ζp)/Q the norm function of the p-th cy-clotomic field Q(ζp), hence we have ∏p

−1

i=1(x+yζip) = Np(x+yζp). Proving FLT is thus equivalent to showing that there are no non-zero integers x, y, z such that

(x+y)Np(x+yζp) =zp. (1.2) The first generalisation we shall consider is a slightly less general version of a conjecture by George Gras in [12]. He referred to it as ‘Strong Fermat’s last theorem’.

Conjecture 1.2(SFLT). For each prime p ≥5, the only solutions in pairwise coprime integers x, y, z to the equation

Np(x+yζp) =zp (1.3)

are given by z=1 and x+yζp ∈ {±1,±ζp,±(1+ζp)}.

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2 i n t r o d u c t i o n

Note that in [12], Conjecture 1.5 states that also N_p(x+yζ_p) = pzp only has the trivial solutions. The second generalisation we will consider is:

Problem 1.3. Let p ≥ 5 be a prime and let 1≤ k ≤ p−3 be an integer. What are the integer solutions to the following two equations?

Np(a0+a1ζp+ · · · +akζk_p) =zp (1.4)

(a0+a1+ · · · +ak)Np(a0+a1ζp+ · · · +akζk_p) =zp (1.5) For k = 1, equation (1.5) gives us back FLT and equation (1.4),

with coprime a0, a1, is precisely Conjecture 1.2. If k = p−2 then

a0+a1ζp+ · · · +ap−2ζpp−2is a generic element ofZ[ζp], and thus (1.5)

has many solutions. For example, Np(uγp)will be a p-th power of an integer if u is a unit of Z[ζp]and γ is any element of Z[ζp]. This is also the reason why we restrict to p≥5.

Though Conjecture1.2got a little attention on its own (see [2, 12, 13]), almost all results come from theorems connected to FLT which can be applied to Conjecture 1.2and Problem1.3as well. It is interesting

to compare this situation with Catalan’s conjecture, which states that the only positive integer solution to the equation

xn−1=yq

with n≥2 is given by 32₋₂3₌_{1. This has been proved completely by} cyclotomic methods by Preda Mih˘ailescu. The corresponding ‘strong’ version is given by the Nagell-Ljunggren equation

xn−1 x−1 = y

q_.

Contrary to SFLT, the Nagell-Ljunggren equation has a rich history of itself (see [3] for a survey).

In this thesis we shall look at the approach Ernest Kummer took to FLT and improvements of that approach, and we shall see how this translates to Problem1.3. Naturally this leads to a number of fresh

problems which we shall look at. To make this more concrete, we give an overview of the different steps of Kummer’s approach to FLT, and indicate how these steps relate to the chapters of this thesis.

1.2 k u m m e r’s approach

Suppose x, y, z are non-zero integers such that xn+yn =zn. We may assume without loss of generality that n = p is an odd prime. The analog of Conjecture1.2for n not a prime would be to consider the

equation

n−1

∏

i=1

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1.2 kummer’s approach 3

and we can define something similar for Problem1.3. In chapter2we

look at a way to give a lower bound for max(|x|,|y|). This method also works for n not a prime. In chapter3we look at a way to write

n−1

∏

i=0

(a0+a1ζi_n+ · · · +akζ_nik)

in terms of recurrence sequences. This method also does not require that n is a prime number.

We may assume without loss of generality that x, y, z are pairwise coprime. For FLT, we can assume this without loss of generality, but for the equation (1.3) of SFLT this is not the case. In fact if a, b are any

two integers, then we can write Np(a+bζp) =cdp for some integers c, d (with possibly d= 1). Then we see that Np(ac+bcζp) = cpdp is a solution to equation (1.3). So there are infinitely many solutions to

Conjecture 1.2if we do not assume that x, y, z are coprime. We will

not consider the case where x, y, z have a common divisor any further. We separate the proof into two cases. In the first case p-xyz, and

in the second case p|xyz. These cases are traditionally referred to as Case 1 and Case 2, and we will write FLT1 and FLT2 accordingly. We can make a similar case distinction for SFLT, and we will refer to these as SFLT1 and SFLT2.

Assume we are in Case 1, so p-xyz. Because x and y are coprime,

we find that x+yζi_p is coprime to x+yζjp if i6≡j mod p. Hence we have(x+yζp)Z[ζp] = Ip for some ideal I ofZ[ζp].The fieldsQ(ζp) do not have unique factorisation into irreducible elements for p≥23. To get around this problem, Kummer used ideals and the class group. The fact that for p-xyz and x, y coprime we have

Np(x+yζp) =zp =⇒ (x+yζp)Z[ζp] = Ip (1.6) does not carry over to the situation with more variables of Problem

1.3. In chapter4we try to find conditions under which (1.6) holds for

elements of the form a0+a1ζp+ · · · +akζkp.

Apply an annihilator η ∈Z[Gal(Q(ζp)/Q)]of the class group of Q(ζp) to get (x+yζp)η = uγp for some unit u ∈ Z[ζp] and some γ ∈ Z[ζp]. Kummer originally proved FLT for regular primes, so prime p for which p does not divide the class number ofQ(ζp). In this case, the fact that Ip is a principal ideal implies that I must be principal itself. If the prime p is irregular, then I might not be principal, but certain products of Galois conjugates of I will be principal. The two main theorems in this direction are Stickelberger’s theorem and the Herbrand-Ribet theorem. Stickelberger’s theorem was proved for cyclotomic fields already by Kummer. Both of these theorems give annihilators of the class group of Q(ζp). In chapter6we give a proof

of the easy direction of the Herbrand-Ribet theorem using class field theory and an explicit construction of an extension ofQ(ζp).

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Take the logarithmic derivative and look modulo p.The idea in this step is that the logarithmic derivative of a p-th power should be congruent to 0 modulo p. So if x+yζp = uγp, then after taking the logarithmic derivative, we only care about the unit u. We can then use the knowledge we have of the units ofZ[ζp]to derive a congruence condition modulo p that x, y and z must satisfy. For example, Kummer could conclude that if (x, y, z)is a solution to FLT1, then they must satisfy:

Bp−3

xy(x−y)

(x+y)3 ≡0 mod p (1.7)

Here Bp−3 is the (p−3)-rd Bernoulli number. In chapter 5 we will

show that instead of the logarithmic derivative, we can also use the p-adic logarithm. Although not obvious at the start, this will basically give the same information as we can obtain by using the logarithmic derivative. This method is not restricted to FLT and SFLT, but it applies to general cyclotomic elements, like in Problem1.3. In chapter8we

look closer at the condition Bp−3 ≡ 0 mod p. Primes that satisfy this condition are called Wolstenholme primes. We use the p-adic logarithm to get characterisation of these primes.

Use the symmetry of xp+yp = zp to conclude that p | xyz. If we have xp+yp = zp, then we also get xp+ (−z)p = (−y)p and yp_{+ (−}_z₎p_{= (−}_x₎p_{. From equation (}1.7) we concluded that p|x−y,

so by using the symmetry of the equation we find that p|x+z and p |y+z. So then 0≡ xp+yp−zp ≡3x mod p. So p|xyz after and we can go to Case 2. This kind of symmetry is only something we can use in the case of FLT. For SFLT we have nothing similar. Also for the more general Problem1.3we do not have this symmetry.

In Case 2, from a solution to FLT2, construct a smaller solution and use a descent argument to get a contradiction. Case 2 of FLT is considerably more difficult than Case 1. The approaches to Case 2 use a descent, and all rely on the symmetry of xp+yp ₌ _zp_{. Therefore} they do not carry over to SFLT or to Problem1.3. Though we will not

be able to proof SFLT for a specific prime, we will be able to give a lower bound on max(|x|,|y|).

1.3 d e s c r i p t i o n o f t h e c h a p t e r s

In chapter2, we fix α∈Q(_ζ_p), and look at the behaviour of

lim p→∞

p q

Np(α).

We see that if α= f(ζp)for some polynomial f ∈ Z[X], then this limit converges to the Mahler measure of f . We then use knowledge of this limit, in Theorem 2.5, to obtain the lower bound max(|x|,|y|) ≥500,

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1.3 description of the chapters 5

In chapter3, we show that we can write norms in terms of sums of

linear recurrence sequences. More specifically, in Theorem3.4, we see

that if f ∈Z[X]is of degree k, then we can write n−1

∏

i=0 f(ζi_n) = k

∑

j=0 A(nj),

where for each j, the sequence (A(nj))n∈N is a linear recurrence

se-quence, depending only on the coefficients of f . There are many results on perfect powers in linear recurrence sequences. We use these to show that for fixed a, b, there are only finitely many primes p, such that Np(aζ2p+bζp−a) =zp has a solution.

In chapter 4, we try to find out when the following implication

holds:

Np(α) =zp =⇒ αZ[ζp] = Ip for some ideal I ofZ[ζp]. This question does not arise in the case of FLT, but only arises for Problem 1.3, so there is not much previously known. We give some

conditions on α that ensure the implication is true. We also try to find out when α is coprime to its Galois conjugates. For x+yζpthis occurs precisely when x and y are coprime, and p-x+y, but for other α the

situation is more difficult.

In chapter5, we take a look at Kummer’s idea to study the equation

x+yζp=uγp, using the logarithmic derivative. We see, in detail, how this leads to a proof of FLT1 for primes which satisfy p-Bp−3. Next we see that, instead of the logarithmic derivative, we can also use the p-adic logarithm to study these kind of equations. However, in Theorem5.9, we show that the information we obtain with the p-adic

logarithm is, curiously, the same as the information we obtain with the logarithmic derivative.

In chapter6, we give a review of the theory of Kummer extensions.

We look at the ramification of primes in these extensions. Then we look at the unit group ofZ[ζp], and use these units to construct unramified extensions of Q(ζp) of degree p. This leads to a proof Herbrand’s direction of the Herbrand-Ribet theorem. All these results are well-known, but we hope that presentation of this chapter is concrete, and ties in well with the logarithmic derivative of chapter 5.

In chapter7, we look at SFLT for regular primes. If p is regular, we

have extra knowledge of the units, and this allows us to show that xy(x−y)must be divisible by a number of small primes. In Theorem

7.8we use this to show that if p≥5 is regular and N_p(x+yζ_p) =zp,

then max(|x|,|y|) >5×108_{. This method does not work for Problem}

1.3.

In chapter8, we look at Wolstenholme’s theorem and generalisations

of it. Wolstenholme’s theorem states that, for prime p ≥5, we have

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p4 _{is called a Wolstenholme prime. This is equivalent to the fact} that p divides Bp−3, and hence the link with the previous chapters. There are a number of results that generalise Wolstenholme’s theorem and characterise Wolstenholme primes modulo higher powers of p. We use the the p-adic logarithm to give easier proofs of some of these generalisations, and to extend previously known results. For example, we give a characterisation of Wolstenholme primes in terms of harmonic sums modulo p12. In the second section, we apply the p-adic logarithm to norms of certain cyclotomic integers. This gives us p-adic equations for the trace of these integers. Looking modulo specific powers of p, we obtain a number of previously known results, and a number of new results. For example, in Proposition 8.17, we

give another proof of the identity p−1

∑

k=1 1 k 2k k ≡0 mod p2.

Using the relation with recurrence sequences of chapter3, we prove

the identity p−1

∑

k=1 (−1)k−11 k 2k k ≡ (1−L 2 p)(L2p−3) 2p mod p 2_,

where (Ln)n∈N is the sequence of Lucas numbers, given by L0 = 2, L1 = 1 and Ln+2 = Ln+1+Ln. The most interesting part about this is that these results can be found using the same method, where previously each identity required different methods to proof them.

1.4 n o tat i o n

We use the following notation in this thesis. For us, p will always denote an odd prime, except when mentioned otherwise. For n≥3 an integer, we let ζnbe a primitive n-th root of unity, so it is a root of the n-th cyclotomic polynomialΦn(X). In most chapters, n= p is an odd prime. If the prime p is fixed throughout a chapter, we will drop the subscript p and just write ζ = ζp and Φ(X) = Φp(X). We will work with the cyclotomic fieldsQ(ζn). In chapters4,5,7and section

3 and 4 of chapter6, we will fix n= p throughout the chapter, then

we will denote K=Q(ζp). We assume that the reader has had a first course in algebraic number, similar to the first chapter of [16]. The ring of integers of Q(ζn)is Z[ζn] and if we use the notation K = Q(ζp), then we will write OK for the ring of integers of K. We denote the elements of the Galois group Gal(Q(ζn)/Q)by σa, for a∈ (Z/nZ)∗, where σa is determined by σa(ζn) =ζ_na. If p is fixed and K = Q(ζp), then we will also write Gal(K/Q) = G. The norm of K is given by Np(α) =∏_σ∈Gσ(α)and the trace is given by Trp(α) =∑_{σ inG}σ(α). If it is obvious which prime p we are talking about, we will drop the

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1.4 notation 7

subscript p and just write N and Tr. For us, π will always denote the element 1−ζ of K, so that we have πp−1OK = pOK as ideals. We will denote (α) for the principal ideal generated by α if it is clear in what ring we are working. If p is a prime ideal of O_K, then ordp(α)is the maximal n∈Z such that pndivides α. So, for example ord(π)(p) = p−1. The fieldQp is the field of p-adic numbers. The

p-adic logarithm will always be denoted by log_p and the ordinary logarithm will be denote by log.

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2

L I M I T S O F N - T H R O O T S

Let α ∈ Q(ζp)and write α = f(ζp)for some polynomial f ∈ Z[X]. Then we have Np(α) = p−1

∏

i=1 f(ζi_p),

and thus Np(α) =zp for some integer z∈Z if and only if

p v u u t p−1

∏

i=1 f(ζi_p) ∈Z.

In this chapter we shall fix f and investigate the behaviour of the above expression as p grows. In the first part we see how this relates to the Mahler measure of a polynomial. We give a concrete formula for the limit when the polynomial f has degree 2. In the second part of the chapter we use this to give an upper bound on p in terms of a and b (or equivalently a lower bound on a and b in terms of p) when there is a solution to Np(a+bζp) =cp. Concretely, Theorem2.4says

that if Np(a+bζp) =cp, then max(|a|,|b|) >500.

2.1 t h e m a h l e r m e a s u r e

Let f be a polynomial with integer coefficients. Note that ∏n_i₌−₁1 f(ζ_ni) makes sense for any positive integer n, but will only be equal to the norm of an element of a cyclotomic field if n is an odd prime. Also note that, since f(ζi_n)is the complex conjugate of f(ζn_n−i), the number ∏n−1

i=1 f(ζni)is always positive, so we have∏ni=−11 f(ζni) =∏in=−11|f(ζin)|. Here|x|is the absolute value of x as an element ofC.

Theorem 2.1. Suppose f(X) ∈Z[X]factors as f(X) =a(X−α1)(X− α2). . .(X−α_k)over the complex numbers, and such that a, α1, . . . , αk 6=0. Suppose|αi| 6=1 for all i=1, . . . , k. Then we have

lim n→∞ n v u u t n−1

∏

i=1 f(ζi_n) = |a|

∏

|αi|>1 |αi| = |a| k

∏

i=1 max(1,|αi|). 9

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10 l i m i t s o f N -th roots

Proof. First of all, we see n−1

∏

i=1 f(ζi_n) = n−1

∏

i=1 |f(ζ_ni)| = |a|n−1

∏

k j=1 n−1

∏

i=1 |ζi_n−αj| = |a|n−1

∏

k j=1 1−αn_j 1−αj = |a| n |f(1)| k

∏

j=1 |1−αn_j|.

Now take the n-th root and let n go to infinity. We get

lim n→∞ n s |a|n |f(1)| = |a|nlim→∞ 1 n p |f(1)| = |a|. Also, we have lim n→∞ n q |1−αn_j| =    1 if |αj| <1 |αj| if |αj| >1. Combining this we end up with

∏

i=1 f(ζi_n) = |a|

∏

|αj|>1 |αj|.

The condition|αi| 6=1 is there to ensure that this limit actually con-verges. If|α| =1 and α 6=1, then limn→∞ n

√

1−αndoes not converge. Note that, even for f(X) ∈Z[X], the condition ‘|α| 6=1 for any root of f ’ is not equivalent to the condition ‘ f(ζn) 6=0 for all n∈N’. For example, the polynomial x4₋_2x3₋_2x₊_{1 has two roots with absolute} value 1 and two roots with absolute value different from 1.

Definition 2.2. For f(X) =a(x−α1)(x−α2). . .(x−αk) ∈C[X]define the Mahler measure M(f)of f as the number

M(f):= |a| k

∏

j=1

max(1,|αj|).

In the article [21], D. H. Lehmer considered a method to find large primes. Suppose f(X)is a monic polynomial with integer coefficients and with roots α1, . . . , αk. Then, for each positive integer n, the num-bers ∆n(f) = ∏kj=1(αnj −1) will be integers. The prime factors of these numbers should satisfy certain congruence conditions, which

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2.2 an upper bound on p 11

might make it easier to find large primes. For example, if we take f(X) =X−2 then we are considering the prime factors of the number 2n−1, so this is related to Mersenne primes.

This lead Lehmer to consider the behaviour of these integers∆nas n goes to infinity. If f has no roots which are a root of unity, then

lim n→∞

∆n+1 ∆n

= M(f).

For his purposes it was important that M(f)is as small as possible. A theorem of Kronecker says that M(f) = 1 if and only if f is the product of cyclotomic polynomials. Excluding these, the polynomial with smallest Mahler measure Lehmer could find was the polynomial

f(X) =x10+x9−x7−x6−x4−x3+x+1

which satisfies M(f) ≈1.176. Lehmer’s conjectures states that there is a constant µ > 1 such that M(f) ≥ µ for all f which are not products of cyclotomic polynomials. This is still an open problem. A non-cyclotomic polynomial with smaller Mahler measure than then one above has not been found. Later, Kurt Mahler considered the problem of multivariable polynomials, and the name Mahler measure got associated to M(f).

2.2 a n u p p e r b o u n d o n p

Let f(X) ∈ Z[X]be a polynomial which does not have a root with absolute value 1. Then by Theorem2.1we know that

∏

i=1 f(ζi_n) = M(f).

If M(f)is not an integer, then we know that there is some n0 ∈N such that for all n ≥n0 the number n

q ∏n−1

i=1 f(ζin)is not an integer. If M(f) is an integer, but ∏n_i₌−₁1f(ζi_n) 6= M(f) from some n onwards, then again we can find such an integer n0. If n = p ≥ n0 is a prime then we see that qp

∏ip=−11 f(ζip) = p q

Np(f(ζp)). So we find that Np(f(ζp)) cannot be a p-th power for all primes p ≥n0.

Let us find an explicit expression for n0 in a simple example. The simplest example is the case where f(X) =aX+b is a linear polyno-mial with|a| 6= |b. In this case we have Np(f(ζp)) = a

p₊_bp

a+b . Note that if Np(aζp+b)is a p-th power, then also Np(−aζp−b)and Np(a+bζp) are p-th powers, so we can assume without loss of generality that a> |b| >0. Although the following proposition immediately follows the proof Fermat’s last theorem, we show this example anyway as an illustration of the method.

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12 l i m i t s o f N -th roots

Proposition 2.3. Let a > |b| >0 and let p be a prime such that p> log(2)

log(a+1) −log(a).

Then there is no integer c such that ap+bp=cp. Proof. By assumption we have

p> log(2)

log(a+1) −log(a) >

log(2)

log(a) −log(a−1)

and thus we see that log(2) <p log(_a₋a₁)and log(2) < p log(a+_a1). So 2(a−1)p< ap and 2ap < (a+1)p. Hence we find:

(a−1)p< ap− (a−1)p ≤ap+bp <2ap< (a+1)p. Therefore a−1 < √p ap₊_bp _< _a₊_{1. If b} ₆₌_{0 then} √p ap₊_bp ₆₌ _{a, so} we conclude that√p ap₊_bp _{is not an integer.}

A slight variation on the previous proposition is the following. Theorem 2.4. Let a> |b| >0 and let p be a prime such that

p> log(2) +log(a)

log(a) −log(a−1).

Then there is no integer c such that ap_a+₊b_bp =cp.

Proof. Note that 1−a−b≤0, and thus ap(1−a−b) +bp <0. This means that for all p we have

ap+bp a+b < a

p_.

Since we assume that p> log(2)+log(a)

log(a)−log(a−1), we find that(p−1)log(a) − p log(a−1) >log(2), and thus ₍_aa₋p−₁1₎p >2. Consequently

ap>2a(a−1)p ≥ (a−1)p(a+b+1) and therefore ap+bp> ap− (a−1)p > (a−1)p(a+b). We conclude that a−1< qp ap₊_bp a+b < a, so p q ap₊_bp a+b is not an integer. This allows us to computationally give a lower bound on max(|a|,|b|). We fix an a> 0, and we check for all primes p < _loglog₍_a(₎₋2)+_loglog₍_a(₋a)₁₎ and 1−a≤b≤a−1 coprime to a that ap_a+₊b_bp is not a p-th power. Note, for p =3, that aζ3+b is a general element ofZ[ζ3], so we will certainly find many third powers of the form a3_a+₊b_b3. The interesting case is p>3. We did these computations using a simplePari-GP[25] script. This gave us the following result.

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2.2 an upper bound on p 13

Theorem 2.5. If p ≥ 5 is a prime and a, b are coprime integers such that ap₊_bp

a+b =cp, then max(|a|,|b|) >500.

This computational process does not scale very efficiently. We will see in chapter 7 that if we assume that p is a regular prime, then

we can greatly improve this lower bound. In Lemma ?? we showed what the Mahler measure of a quadratic polynomial is. We would like to make a similar analysis like that of Theorem2.4for N_p(aζ2

p+ bζp+c). If f(X) = aX2+bX+c, can we have p

q

Np(f(ζp)) = M(f) or qp ₍_a₊_b₊_c₎_N

p(f(ζp)) =M(f)? If not, there is an integer n0such that for all primes p ≥ n0 we know that p

q

Np(f(ζp)) and (a+b+ c)qp _N

p(f(ζp))are not integers. Can we give this integer n0explicitly in terms of a, b and c? The situation is more complicated than in the two variable case, and we do not yet have an answer to these questions.

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3

R E C U R R E N C E S E Q U E N C E S

In this chapter we investigate a connection between norms of cyclo-tomic integers and linear recurrence sequences. We start with the simplest case, when the cyclotomic integer is of the form aζ2_p+bζp+c. In Theorem3.1we will see that

n−1

∏

i=0

(aζ2i_n +bζn+c) =an+cn−Hn,

where the numbers(Hn)n∈Nare given by a linear recurrence relation.

When n = p is an odd prime, this means that we have expressed

(a+b+c)Np(aζ2p+bζp+c)in terms of a recurrence sequence. The method of linear forms in logarithms can be used to investigate the perfect powers that occur in a linear recurrence sequence. As a corollary, we will see that for fixed non-zero integers a and b, there are only finitely many primes p such that bNp(aζ2p+bζp−a) = zp has a solution. In the second part of this chapter we generalise Theorem

3.1to hold for arbitrary cyclotomic elements. This does not make the

problem of finding solutions to Np(α) =zpeasier, but it is interesting in its own right. The main result is given by Theorem3.4, and we will

see some explicit examples for low variable cases.

3.1 n o r m s i n t e r m s o f a r e c u r r e n c e s e q u e n c e

We start with the simplest example of a norm that can be expressed in terms of recurrence sequences.

Theorem 3.1. For each integer n≥ 3, and all a, b, c∈ C with a 6= 0, we have:

n−1

∏

i=0

(aζ2i_n +bζi_n+c) =an+cn−Hn, where(Hn)n∈Nis defined by the recurrence relation

Hn+2= −bHn+1−acHn and the initial values H0 =2 and H1= −b.

Proof. Let f(X) = aX2+bX+c = a(X−ω1)(X−ω2), for some ω1, ω2∈ C. So we have ω1ω2= c_a and ω1+ω2= −b_a. Since Xn−1= ∏n−1 i=0(X−ζin), we find that n−1

∏

i=0 (aζ2i_n +bζ_ni +c) =an n−1

∏

i=0 (ζi_n−ω1)(ζi_n−ω2) =an(ω₁n−1)(ωn₂−1) =an+cn− (aω1)n− (aω2)n. 15

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16 r e c u r r e n c e s e q u e n c e s

Note that aω1 and aω2are roots of the polynomial(X−aω1)(X− aω2) = X2+bX+ac. Hence, if we define Hn := (aω1)n+ (aω2)n, then from the theory of linear recurrence relations, we find that Hn satisfies Hn+2 = −bHn+1−acHn, and has initial values H0 = 2 and H1= −b.

We give another proof of Theorem3.1using determinants of

ma-trices. It is a longer proof, but the recurrence relation appears very naturally when computing the determinants.

Proof. By general properties of circulant matrices (see for example [7], page 72) we have: n−1

∏

i=0 (a0+a1ζi_n+a2ζ_n2i+ · · · +an−1ζ(nn−1)i) = a0 a1 · · · an−2 an−1 an−1 a0 · · · an−3 an−2 .. . ... . .. ... ... a2 a3 · · · a0 a1 a1 a2 · · · an−1 a0 n Here|M|is the determinant of the matrix M and we use the subscript n to remember that it is an n×n matrix. In particular, we find

(−1)n−1 n−1

∏

i=0 (aζ2i_n +bζi_n+c) = n−1

∏

i=0 (aζi_n+b+cζ_n−i) = b a 0 · · · 0 c c b a · · · 0 0 0 c b · · · 0 0 .. . ... ... . .. ... ... 0 0 0 · · · b a a 0 0 · · · c b n (3.1) We will define: Gn(a,b,c)=Gn:= b a 0 · · · 0 0 c b a · · · 0 0 0 c b · · · 0 0 .. . ... ... . .. ... ... 0 0 0 · · · b a 0 0 0 · · · c b n

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3.1 norms in terms of a recurrence sequence 17

This is the same matrix as in (3.1), but with 0 in the bottom left and

upper right corners. Then we see that, for n ≥2, the determinant in (3.1) is equal to: bGn−1−a c a 0 · · · 0 0 0 b a · · · 0 0 0 c b · · · 0 0 .. . ... ... . .. ... ... 0 0 0 · · · b a a 0 0 · · · c b n−1 + (−1)n+1c c b a · · · 0 0 0 c b · · · 0 0 0 0 c · · · 0 0 .. . ... ... . .. ... ... 0 0 0 · · · c b a 0 0 · · · 0 c n−1 =bGn−1−acGn−2+ (−1)n+1a2 a 0 0 · · · 0 0 b a 0 · · · 0 0 c b a · · · 0 0 .. . ... ... . .. ... ... 0 0 0 · · · a 0 0 0 0 · · · b a n−2 +(−1)n+1c2 c b a · · · 0 0 0 c b · · · 0 0 0 0 c · · · 0 0 .. . ... ... . .. ... ... 0 0 0 · · · c b 0 0 0 · · · 0 c n−2 −acGn−2 =bGn−1−2acGn−2+ (−1)n+1(an+cn)

The numbers Gnsatisfy a recurrence relation. Indeed, for n≥2, we have

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18 r e c u r r e n c e s e q u e n c e s Gn = b a 0 · · · 0 0 c b a · · · 0 0 0 c b · · · 0 0 .. . ... ... . .. ... ... 0 0 0 · · · b a 0 0 0 · · · c b n =bGn−1−a c a 0 · · · 0 0 0 b a · · · 0 0 0 c b · · · 0 0 .. . ... ... . .. ... ... 0 0 0 · · · b a 0 0 0 · · · c b n =bGn−1−acGn−2

Together with the initial values G0 =1 and G1= b, these determine the numbers Gn. Hence, we find

n−1

∏

i=0

(aζ2i_n +bζ_ni +c) =an+cn+ (−1)n+1(bGn−1−2acGn−2).

Now define Hn := (−1)n(bGn−1−2acGn−2). It is easy to see that the numbers Hnsatisfy the recurrence relation Hn+2 = −bHn+1−acHn. Using the initial values of the sequence(Gn)n∈N, we see:

H3= −bG2+2acG1= −b3+3abc H2=bG1−2acG0= b2−2ac Hence we find: H1= − H3+bH2 ac = −b H0= − H2+bH1 ac =2 Finally we conclude that

n−1

∏

i=0

(aζ2i_n +bζi_n+c) =an+cn−Hn.

The determinants Gn from the previous proof are a specific instance of the following, more general proposition, which also follows from the results in [16].

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3.1 norms in terms of a recurrence sequence 19

Proposition 3.2. Let a, bi ∈ C for i ≥ 0. Set A0 := 1, and for n ≥ 1, define: An := b0 b1 b2 · · · bn−1 a b0 b1 ... 0 a b0 ... .. . . .. b1 0 · · · 0 a b0 For any ci ∈C, i≥0, we have the equality:

c0 c1 c2 · · · cn−1 a b0 b1 · · · bn−2 0 a b0 ... .. . . .. b1 0 · · · 0 a b0 =c0An−1−ac1An−2+ · · · + (−1)nancn−1A0.

In particular, if bi =0 for i≥ k, then the matrices Ansatisfy the recurrence relation

An=b0An−1−ab1An−2+ · · · + (−1)k−1ak−1bk−1An−k.

Proof. We use induction on n. It is obvious for n = 0. Now assume that it holds for matrices of size (n−1) × (n−1). Then we see

c0 c1 c2 · · · cn−1 a b0 b1 · · · bn−2 0 a b0 ... .. . . .. b1 0 · · · 0 a b0 =c0An−1−a c1 c2 c3 · · · cn−1 a b0 b1 · · · bn−3 0 a b0 ... .. . . .. b1 0 · · · 0 a b0 =c0An−1−a(c1An−2− · · · + (−1)n−1an−1cn−1A0) =c0An−1−ac1An−2+ · · · + (−1)nancn−1A0 The last statement of the proposition is obtained by putting ci =bi for all i ≥0.

Note that the recurrence relation of the determinants Gn follows from this proposition by putting c0 =b0, c1 =b1 and ci = bi =0 for i≥2.

Let Gnbe a non-degenerate (the coefficients of the recurrence rela-tion are non-zero) recurrence sequence of order 2. Shorey and Stewart [31], and independently Pethö [26], proved that there are effectively computable constants C1, C2, depending only on the coefficients and starting values of the sequence Gn, such that if n, y, q are integers with q>1 satisfying

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then max(|y|, n, q) <C1 if|y| >1 and n<C2 if|y| ≤1. This result is obtained by using the method of linear forms in logarithms. Generally, the constants are very large. For example, if Gn is the Fibonacci se-quence, then one finds that q<5.1×1017_{. See [}27,28] for a survey by Pethö on diophantine equations involving linear recurrence sequences. As an immediate consequence of this, we find:

Theorem 3.3. Let a, b be non-zero integers. There is an effectively com-putable constant C, depending only on a and b, such that if p is a prime and z an integer such that

bNp(aζ2p+bζp−a) =zp, then p<C.

Proof. If p is an odd prime, then p−1

∏

i=0

(aζ2i_p +bζi_p+c) = (a+b+c)Np(aζ2p+bζp+c). So if a = −c, then Theorem3.1shows that we have

bNp(aζ2p+bζp−a) = −Hn.

Here Hnis the second order linear recurrence sequence given by the initial values H0=2, H1 = −b and the recurrence relation

Hn+2= −bHn+1+a2Hn.

So, taking C=max(C1, C2), we conclude that if bNp(aζ2p+bζp−a) = zp_{, then we must have p}_<_C.

For fixed a and b, if we want to solve the equation bNp(aζ2p+bζp− a) =zp_{, this theorem tells us we only need to check a finite number} of primes p. However, in practice the constant C is too large to check all primes below C and more clever methods are necessary. In the breakthrough article [4], Bugeaud, Mignotte and Siksek show that the perfect powers in the sequence Fnof Fibonacci numbers are given by F0 = 0, F1 = F2 = 1, F6 = 8 and F12 = 144. Furthermore, they showed that the only perfect powers in the sequence (Ln)n∈N of

Lucas numbers are given by L1 =1 and L3 =4. They use a sharper version of the linear forms in logarithms method together with the modular method, which originates in Wiles’ work on Fermat’s last theorem. These two methods combined reduce the bounds on q and n far enough for the problem to be resolved computationally. Note that, because Np(ζ2_p−ζp−1) = Lp by Theorem 3.1, we find as an

immediate corollary that there are no primes p such that Np(ζ2_p−ζp−1) =zp.

It is interesting that the modular method that is used in the resolution of the Fermat equation (x+y)Np(xζp+y) = zp, and which is not cyclotomic in nature, can also be used to solve the equation Np(ζ2_p− ζp−1) =zp.

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3.2 a general recurrence theorem 21

3.2 a g e n e r a l r e c u r r e n c e t h e o r e m

Now we will show that Theorem3.1does not just hold for cyclotomic

integers of the form aζ2₊_bζ₊_{c, but can in fact be generalized to any} cyclotomic integer. For j=1, . . . , k, let ej(X1, . . . , Xk)be the elementary symmetric polynomial of degree j in k variables. So e1(X1, . . . , Xk) = X1+X2+. . .+Xk, e2(X1, . . . , Xk) = ∑1≤r<s≤kXrXs, and in general, for 1 ≤ j ≤ k, we have ej(X1, . . . , Xk) = ∑1≤i1<i2<···<ij≤k ∏jr=1Xir . We will set e0(X1, . . . , Xk) =1. Theorem 3.4. Let f(X) = akXk+ak−1Xk−1+ · · · +a1X+a0 ∈ C[X] with a_k 6=0 and let ω1, . . . , ωk be its roots. For j=0, . . . , k and n∈N we define

A(nj):= (−1)(n+1)k−jankej(ωn₁, . . . , ωn_k). For each n≥3 we have

n−1

∏

i=0 f(ζi_n) = k

∑

j=0 A(nj).

For each j, the sequence(A(nj))n∈Nis given by a recurrence relation

A(j) n+(k_j)= (k_j)−1

∑

s=0 g(sj)A (j) n+s with g(₀j), . . . , g(j)

(k_j)−1∈Z[a0, . . . , ak]and initial values A

(j)

n ∈Z[a0, . . . , ak]. Proof. Let f(X) =a_k(X−ω1)(X−ω2) · · · (X−ω_k)for certain ω1, . . . , ωk ∈ C. Then we have: a_k−1 ak = −e1(ω1, . . . , ωk) ak−2 ak =e2(ω1, . . . , ωk) .. . a0 ak = (−1)ke_k(ω1, . . . , ωk) Moreover we see: n−1

∏

i=0 f(ζi_n) =an_k n−1

∏

i=0 (ζi_n−ω1) · · · (ζi_n−ωk) = (−1)nkan_k(ω₁n−1) · · · (ωn_k −1) = (−1)nkan_k k

∑

j=0 (−1)k−jej(ω₁n, . . . , ω_kn) ! = k

∑

j=0 (−1)(n+1)k−jan_kej(ω₁n, . . . , ω_kn)

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For each j, we will show that ej(ω₁n, . . . , ωn_k)_n_∈_N and thus also

A(nj)

n∈N satisfies a recurrence sequence as in the statement of the

theorem. In general, if t

∑

i=0 ciXi =ct t

∏

j=1 (X−γj)

then from the theory of linear recurrence sequences we know that the numbers Cn := ∑tj=1γnj satisfy the recurrence relation ctCn+t = −_∑t_i₌−₀1ciCn+i. In our case, we would like the numbers Cnto be equal to ej(ωn₁, . . . , ωn_k). To achieve this, we need to set the γj equal to mr(ω1, . . . , ωk), where mr(X1, . . . , Xk), r = 1, . . . ,(kj) are the mono-mials of the polynomono-mials ej(X1, . . . , Xk). Therefore we consider the polynomial E(X) = (k_j)

∏

r=1 (X−mr(ω1, . . . , ωk)). Let hs∈ C be such that

E(X) =

(k_j)

∑

s=0 hsXs.

Then each hsis given by a symmetric polynomial in ω1, . . . , ωk. By the fundamental theorem of symmetric polynomials, this means that

hs =Ps(e1(ω1, . . . , ωk), . . . , ek(ω1, . . . , ωk)) for some polynomial Ps∈ Z[Y1, . . . , Yk].

Setting Cn :=ej(ω₁n, . . . , ω_kn), we see that

Cn=

(k_j)

∑

r=1

mr(ω₁n, . . . , ωn_k),

and therefore we conclude that the numbers(Cn)n∈N satisfy the recur-rence relation C_n₊₍k j) = − (k_j)−1

∑

s=0 hsCn+s Hence, if An := (−ak)nej(ωn₁, . . . , ωn_k)and gs = (−a)( k j)−s_h s, then we get A_n₊₍k j) = − (k_j)−1

∑

s=0 gsAn+s.

To finish the proof, we need to show that the coefficients gsand the values Anare in fact elements ofZ[a0, . . . , ak], for all s, n=0, . . . ,(k_j) − 1. Note that they are all elements of Z[a0

ak, . . . , ak−1

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power of a_k appearing in the denominator of hs is equal to the degree of the polynomial Ps. Hence if deg P₍k

j)−s ≤ s, then gs ∈Z[a0, . . . , ak]. Similarly, if

ej(ω1n, . . . , ωnk) =Q(e1(ω1, . . . , ωk), . . . , ek(ω1, . . . , ωk))

then we see that the power of a_k appearing in the denominator of ej(ω₁n, . . . , ωn_k) is equal to the degree of Q. So if deg Q ≤ n, then An∈Z[a0, . . . , ak]. Now we use the following lemma.

Lemma 3.5. Let h(X1, . . . , Xk)be a symmetric polynomial in k variables, which has degree n as a polynomial in X1. Then there is some polynomial q∈Z[Y1, . . . , Yk]of degree n (as a polynomial in k variables) such that

h(X1, . . . , Xk) =q(e1(X1, . . . , Xk), . . . , ek(X1, . . . , Xk)).

Proof of the Lemma. The proof is just the usual proof of the funda-mental theorem of symmetric polynomials, but keeping track of the degree of the polynomial q we end up with. For the usual proof, see for example [6], page 178. We use induction on the leading term of h in lexicographic ordering. Suppose the leading term is given by h0X1l1X

l2 2 · · ·X

lk

k for some h0 ∈ C. By assumption we have l1 = n. Be-cause we use lexicographic ordering, and beBe-cause h is symmetric, we have l1≥l2≥ . . .≥lk. We define

˜h(Y1, . . . , Yk) =h0Y₁l1−l2Y₂l2−l3· · ·Y_klk−−11−lkY lk k . Then we see deg ˜h=l1=n. Moreover, we see that

h(X1, . . . , Xk) − ˜h(e1(X1, . . . , Xk), . . . , ek(X1, . . . , Xk))

is a symmetric polynomial whose leading term is smaller in lexico-graphic ordering. By induction we see that h= q(e1, . . . , ek), for some polynomial q∈Z[Y1, . . . , Yk]of degree n.

Because e1(ωn₁, . . . , ωn_k) has degree n as a polynomial in ω1, the lemma tells us that Q has degree n as a polynomial in k variables. Fur-thermore, because in the definition of E(X)the monomials mr(ω₁, . . . , ω_k) have degree at most 1 as a polynomial in ω1, we find that h(k_j)−s(ω1, . . . , ωk) has degree s as a polynomial in ω1. Consequently, the lemma tells us that Pshas degree s as a polynomial in k variables. So, indeed, the coef-ficients gsand the starting values Anare elements ofZ[a0, . . . , ak].

We can give some more information on the sequences (A(ni))n∈N.

They satisfy some symmetry:

Corollary 3.6. Suppose that a0 6= 0 in Theorem 3.4, let ˆf(X) = a0Xk+ a1Xk−1+ · · · +ak−1X+ak be the reciprocal polynomial and let(Aˆ

(j)

n )n∈N

be the corresponding recurrence sequences. Then we have the following symmetrical property:

A(nj)= (−1)(n+1)kAˆ

(k−j)

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Proof. Note that ˆf(X) =a0(X− _ω1₁)(X− _ω1₂) · · · (X− _ω1_k)and thus ˆ A(nk−j) = (−1)nk+jan0ej−k 1 ω₁n, . . . , 1 ωn_k .

Because we have ω1ω2· · ·ωk = (−1)k a_a0_k we find: ej(ω₁n, . . . , ω_kn) =

∑

1≤i1<...<ij≤k ωn_i₁· · ·ω_in_j =

∑

1≤i1<...<ik−j≤k ωn₁· · ·ωn_k ω_in 1· · ·ω n ik−j = (−1)nka n 0 an_kej−k 1 ω₁n, . . . , 1 ωn_k So we conclude that A(nj)= (−1)(n+1)k−jankej(ω₁n, . . . , ω_kn) = (−1)k−jan₀ek−j 1 ωn₁, . . . , 1 ω_kn = (−1)(n+1)kAˆ(nk−j).

Furthermore, we can give(A(ni))n∈Nexplicitly for small i.

Corollary 3.7. In Theorem 3.4we have A(n0) = (−1)(n+1)kan_k, A(nk) = an₀ and for j=1 and j =k−1 we have the recurrences:

A(_n1₊)_k = − k−1

∑

s=0 (−1)k(k−s)a_kk−1−sasA(n1+)s A(_nk₊−_k1) = − k−1

∑

s=0 a₀k−1−sak−sA(nk+−s1)

Proof. From the proof of Theorem3.4we see that, by definition, A(_n0)=

(−1)(n+1)kan_k. For the case j=1 we see that

E1(X) =E(X) = k

∏

r=1 (X−ωr) =Xk+ k−1

∑

s=0 as a_kX s_.

So A(n1) = (−1)(n+1)k−1an_ke1(ωn₁, . . . , ωn_k)satisfies the recurrence

A(_n1₊)_k = − k−1

∑

s=0

(−1)k(k−s)a_kk−1−sasA(n1+)s.

The cases j=k−1 and j =k now follow by the symmetry property of the previous corollary.

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We will finish this chapter with a number of examples in the case where we have few variables. As a first example, we again look at

f(X) =aX2+bX+c. Theorem3.4tells us that

n−1

∏

i=0

f(ζi_n) = an+A(n1)+cn,

where A(₀1) = −2, A(₁1) = −ae1(ω1, ω2) = b, and where A(n1) satisfies the recurrence

A_n(1₊)₂= −bA(_n1₊)₁−acA(n1). So, as expected, we obtain Theorem3.1again.

In the case of f(X) =aX3₊_bX2₊_cX₊_{d, we find that} n−1

∏

i=0 f(ζi_n) = (−1)n+1an+A(n1)+A (2) n +dn. Here(A(n1))n∈N is given by A(01) =3, A (1) 1 = b, A (1) 2 =b2−ac and the recurrence A(_n1₊)₃= bA(_n1₊)₂−acA(_n1₊)₁+a2dA(n1). The sequence (A(n2))n∈N is given by A(02) = −3, A

(2)

1 = c, A

(2)

2 =

bd−c2 and the recurrence

A(_n2₊)₃= −cA(_n2₊)₂−bdA(_n2₊)₁−ad2A(n2).

Finally, we shall look at the example f(X) = aX4+bX3+cX2+

dX+e. Theorem3.4shows that we have

n−1

∏

i=0 f(ζi_n) =an+A(n1)+A (2) n +A (3) n +en.

For A(n1) and A(n3) we can use Corollary 3.7 to get the recurrence

relation. We see that the sequence(A(n1))n∈Nis given by the recurrence

relation

A_n(1₊)₄= −bA(_n1₊)₃−acA_n(1₊)₂−a2dA_n(1₊)₁−a3eA(n1).

Let ω1, ω2, ω3, ω4be the roots of f(X), and let us set ej(ωi):=ej(ω₁i, ω₂i, ω₃i, ω₄i) and ej := ej(ω). With this notation, we have An(j) = (−1)janej(ωn), and

thus the the initial values of A(n1) are given by: A₀(1)= −e1(ω0) = −4

A₁(1)= −ae1(ω) =b

A₂(1)= −a2e1(ω2) = −a2(e21−2e2) = −b2+2ac

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By symmetry, we find that the sequence(A(n3))n∈N is given by the

recurrence

A_n(3₊)₄= −dA(_n3₊)₃−ceA_n(3₊)₂−be2A_n(3₊)₁−ae3A(n3). The initial values are given by:

A(₀3)= −4 A(₁3)=d

A(₂3)= −d2+2ce A(₃3)=d3−3cde+3be2

To find the recurrence relation that the sequence(A(n2))n∈Nsatisfies,

we need to calculate the polynomial E(x)as in the proof of Theorem

3.4: E(x) = (x−ω1ω2)(x−ω1ω3)(x−ω1ω4)(x−ω2ω3)(x−ω2ω4)(x−ω3ω4) =x6−e2x5+ (e1e3−e4)x4 + (2e2e4−e21e4−e23)x3 + (e1e3e4−e24)x2−e2e24x+e34 =x6− c ax 4₊ bd−ae a2 x 4₊2ace−ad2−b2e a3 x 3 +bde−ae 2 a3 x 2₋ce2 a3 x+ e3 a3

As a result, we find that the recurrence relation is given by: A_n(2₊)₆ =cA_n(2₊)₅+ (ae−bd)A_n(2₊)₄+ (ad2+b2e−2ace)A(_n2₊)₃

+ (a2e2−abde)A_n(2₊)₂+a2ce2A(_n2₊)₁−a3e3A(n2)

Finally, for the initial values A(n2) = ane2(ωn) for n = 0, . . . , 5, we find:

A(₀2)=6 A(₁2)=c

A(₂2)=c2+2ae−2bd

A(₃2)=c3−3ace+3ad2+3b2e−3bcd

A(₄2)=c4+6a2e2−8abde−4ac2e− +4ad2c+4b2ce+2b2d2−4bc2d A(₅2)=c5+5a2ce2+5a2d2e+5ab2e2−5abcde−5ac3e−5abd3

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4

F R O M N O R M S T O S I N G U L A R I N T E G E R S

4.1 s i n g u l a r i n t e g e r s

In the classical approach to Fermat’s last theorem the following lemma is used.

Lemma 4.1. Let x, y be coprime integers such that p does not divide x+y. Then x+yζ is coprime to its Galois conjugates.

Proof. Suppose q is a prime ideal ofOK that divides both x+yζ and x+yζi for some i6≡0, 1 mod p. Then q also divides

x+yζ− (x+yζi) =y(ζ−ζi).

Since p does not divide x+y, we know that q 6= (1−ζ). Hence q divides y. Similarly we see that q has to divide

ζ−1x+y− (ζ−ix+y) =x(ζ−1−ζ−i)

and therefore q also divides x. But this contradicts the fact that x and y are coprime, so we conclude that x+yζ and x+yζi are coprime.

This lemma allows us to conclude that if p does not divide x+y and N(x+yζ) =zp, then (x+yζ) = Ipfor some ideal I ofO_K. Following the terminology of [29], chapter 9, we define:

Definition 4.2. An element α ∈ O_Kis called a singular integer if(α) = Ip for some ideal I ofO_K.

Now suppose we have an element α ∈ O_K satisfying N(α) = zp. The main question of this chapter is: Are there conditions that allow us to conclude that α is a singular integer? For a rational prime q, we denote the primes of O_K dividing q by qi for i = 1, . . . , g(q), where g(q)f(q) = p−1 and f(q)is the relative degree of q. With this notation, the factorisation of α into prime ideals looks like

(α) =

∏

q prime g(q)

∏

i=1 qord_i qi(α). We see that N(α) =zpif and only if

g(q)

∑

i=1

ordq_i(α) ≡0 mod p, for all rational primes q. (4.1) On the other hand we see that α is a singular integer if and only if

ordq_i(α) ≡0 mod p, for all primes q_i ofOK. (4.2) Given α = ap−2ζp−2+ · · · +a1ζ +a0 we set α(X) := ap−2Xp−2+ · · · +a1X+a0.

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28 f r o m n o r m s t o s i n g u l a r i n t e g e r s

Lemma 4.3. IfN(α) = zp, and for each rational prime q the polynomials α(X) and Φ(X) share at most one irreducible factor modulo q, then α is singular.

Proof. We want to show that ordq(α) ≡0 mod p for all prime ideals q of O_K. Since (1−ζ) is the only prime above p, condition (4.2) is

immediate if q= (1−ζ). Now let q6= p be a rational prime and let the factorisation into irreducibles ofΦ(X)modulo q be given by

Φ(X) ≡ g(q)

∏

i=1

hi(X) mod q.

Then the prime ideals above q are given by qi := (q, hi(ζ)). If qidivides α, then there are polynomials r, s∈Z[X]such that

α=r(ζ) ·q+s(ζ) ·hi(ζ). So there is some polynomial t(X) ∈Z[X]such that

α(X) =r(X) ·q+s(X) ·h_i(X) +t(X) ·Φ(X).

Because hi(X) is a factor of Φ(X) modulo q, we see that hi(X) is also a factor of α(X)modulo q. So each qi dividing α gives a distinct irreducible common factor of α(X)and Φ(X). By assumption there is at most one such common factor for each prime q, so there is at most one prime ideal qi dividing α. Since N(α) =zp, we find that α is singular.

For example, if α= aζ+b, then α(X)has degree 1 and thus α(X)

andΦ(X)can have at most one common irreducible factor per prime q. If α= aζ2₊_bζ₊_{c then α}₍_X₎_{has degree 2. So if q} _6≡_{0, 1 mod p, then} q does not split completely inQ(ζp)and thus f(q) = deg hi(X) ≥2. Therefore there can be at most one common factor of α(X)andΦ(X)

modulo q. So only at the primes which are congruent to 1 modulo p is it possible that the condition of Lemma4.3is not satisfied.

Of course, the above does not give a clear-cut way to say wether a given element α with N(α) =zpis singular or not. For a specific prime p and element α it is possible to check it, but we do not have a general condition in terms of the coefficients of α like we did in Lemma4.1.

4.2 s e l f-prime integers

In the case of α= x+yζ, we showed that α is singular by showing that αis coprime to its Galois conjugates. For a general element α∈ OK this is a sufficient condition in order to get (4.1) =⇒ (4.2). That’s

why we shall look a bit closer that this property. The goal is to find reasonably simple conditions that tell us when an element is coprime to its conjugates. Let us start with the basic definition.

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4.2 self-prime integers 29

Definition 4.4. An element α ∈ O_K is called self-prime if α is coprime to σi(α)for all i=2, . . . , p−1.

Being self-prime is a property that we can check locally, at all the prime ideals of O_K. We give some general properties of self-prime cyclotomic integers. The following lemma shows that the self-prime condition puts pretty big restrictions on α.

Lemma 4.5. If α ∈ O_K is selfprime then all prime divisors of N(α)are congruent to 1 modulo p.

Proof. Suppose q divides N(α)and q is not congruent to 1 modulo p. Then q does not split completely in K/Q. Hence for any prime q above q there is some σ ∈Gal(K/Q)such that σ(q) =q. But then if q divides α, it also divides σ(α).

If some α is self-prime, then we automatically know some other self-prime cyclotomic integers.

Lemma 4.6. Let u be a unit inO_Kand let σ be an element of Gal(K/Q). If α is self-prime, then also uα and σ(α)are self-prime.

Proof. This is immediate from the definition.

Lemma 4.7. Suppose Q ⊆ L ( K. If α is an element of L then α is not

self-prime.

Proof. Since L is a proper subfield of K, there is some non-trivial σ∈

Gal(K/Q)that fixes K. But then σ(α) =αso α is not self-prime. We will proceed by giving a characterisation of self-prime integers using the characteristic polynomial. Denote by χα(X)the charateristic

polynomial of α. So we have χα(X) =

∏

σ∈Gal(K/Q)

(X−σ(α))

=Xp−1+gp−2(α)Xp−2+ · · · +g1(α)X+g0(α).

The gi(α) are the elementary symmetric polynomials in σi(α), i = 1, . . . , p−1. They are invariant under the action of the Galois group and thus they are integers. We have

gp−2(α) = − p−1

∑

i=1 σi(α) = −Tr(α) g0(α) = (−1)p−1 p−1

∏

i=1 σi(α) =N(α) g1(α) = − p−1

∑

i=1 ∏p_j=−11σj(α) σi(α) = −Tr(α −1₎_N₍ α)

The following proposition gives a characterisation of self-prime integers.

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Proposition 4.8. An element α ∈ O_K is self-prime if and only if g0(α) = N(α)and g1(α)are coprime.

Proof. Suppose q is a prime lying above some prime integer q, which divides both α and σi(α) for some i 6= 1. Then obviously q divides N(α). Since g1(α)is the sum of the products of p−2 different σi(α), each term in this sum is divisible by q. Hence q also divides g1(α).

Conversely suppose that q divides both g0(α)and g1(α). Since it divides g0(α), there is some q above q which divides α. Since q divides g1(α)and α, it also divides∏_jp₌−₂1σj(α). Consequently q divides σj(α) for some j6=1.

We only need only p−1 powers of ζ to get a basis of K overQ. For example, we can take 1, . . . , ζp−2 as a basis and express α ∈ O_K in terms of this basis as α= a0+a1ζ+ · · · +αp−2ζp−2 for some integers ai. The linear map x7→αxthen corresponds to the matrix

Mα =          a0 a1 a2 . . . ap−2 −ap−2 a0−ap−2 a1−ap−2 . . . ap−3−ap−2 ap−2−ap−3 −ap−3 a0−ap−3 . . . ap−4−ap−3 .. . ... ... ... a2−a1 a3−a1 a4−a1 . . . a0−a1          .

Then we have the equality

det(XI−Mα) =χα(X).

However, by choosing p−1 elements out of 1, ζ, . . . , ζp−1 to form a basis, we lost some of the symmetry which these elements have. Instead we could consider the matrix

M_α0 =             a0 a1 a2 . . . ap−2 0 0 a0 a1 . . . ap−3 ap−2 ap−2 0 a0 . . . ap−4 ap−3 .. . ... ... ... ... a2 a3 a4 . . . a0 a1 a1 a2 a3 . . . 0 a0             .

In a way, this matrix corresponds to the presentation of α as a0+ a1ζ+ · · · +ap−2ζp−2+0·ζp−1. Denote α(1) =a0+a1+. . .+ap−2. By adding the top p−1 rows to the bottom row, we get only α(1)’s on the bottom row. If we then substract the last column from the first p−1 columns, we find that det M0_α = α(1)det Mα. Similarly we see

that we have

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4.2 self-prime integers 31

In particular, if we write det(XI−M0_α) = Xp₊_h

p−1(α)Xp−1+ · · · + h1(α)X+h0(α), then we have:

hp−1(α) =gp−2(α) −α(1) = −(Tr(α) +α(1)) h0(α) = −α(1)g0(α) = −α(1)N(α)

h1(α) = −α(1)g1(α) +g0(α) = (1−α(1)Tr(α−1))N(α)

It makes sense to denote α(1) =a0+a1+ · · · +ap−2when we think of it as α= f(ζ)for some polynomial f ∈ Z[X]. However, when we consider the Galois group, perhaps it makes more sense to think of it as a map σ0 :OK →Z given by σ0(α) = a0+a1+ · · · +ap−2. With this notation we have

hp−1(α) = − p−1

∑

i=0 σi(α) h0(α) = − p−1

∏

i=0 σi(α) h1(α) = p−1

∑

i=0 ∏jp=−01σj(α) σi(α)

We can use the hi(α)instead of the gi(α)in Proposition4.8.

Proposition 4.9. Let α∈ O_K. Then α is self-prime if and only if h0(α)and h1(α)are coprime.

Proof. Suppose α is not self-prime. By Proposition 4.8there is some

prime q dividing both g0(α)and g1(α). Then, using the expressions above, we see that q divides h0(α)and h1(α).

Conversely, suppose that q divides h0(α)and h1(α). From the first we see that q divides α(1) or N(α). But if q divides α(1) then the second gives us that q divides N(α). So we know that q divides N(α) and α(1)g1(α).

If q divides N(α)and g1(α)we are done, by Proposition4.8. Hence

suppose that q divides N(α)and α(1). Then there is some prime ideal q lying above q that divides both α and α(1). Write α = a0+a1ζ +

· · · +ap−2ζp−2. Then we get

q|α(1) −α= (1−ζ)(a1+a2+ · · · +ap−2).

So either q = p or q divides a1+a2+ · · · +ap−2. If q = p then we know that α is not self-prime. If q divides α(1)and a1+ · · ·ap−2then qdivides a0. Similarly, using α(1) −ζ−iαinstead of α(1) −α, we can show that q divides ai for any i. But if q divides all ai then it will divide σ(α)for any σ, so α is not self-prime.

In general the hi(α)are a bit nicer to work with, because they are more symmetrical. We can use this to give another proof of Lemma

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Lemma 4.10. Let α = a+bζ. Then α is self-prime if and only if a and b are coprime and p does not divide a+b.

Proof. If a and b have a non-trivial common divisor, then obviously α is not self-prime. Moreover, if p divides a+b, then we know by Lemma 4.5 that α is not self-prime. Hence assume that a and b are

coprime and p does not divide a+b. We have

det(XI−M_α0) = X−a −b 0 . . . 0 0 X−a −b . . . 0 0 0 X−a . . . 0 .. . ... ... . .. ... −b 0 0 . . . X−a = (X−a)p−bp.

This means that h0(α) = −(ap+bp) and h1(α) = pap−1. So if q divides h1(α)then q= p or q divides a. If q divides a and h0(α)then also q divides b, which contradicts the coprimality of a and b. If q= p then 0 ≡ −h0(α) ≡ ap+bp ≡ a+b mod p, which does not hold by assumption.

In the case of three variables we get the following.

Theorem 4.11. Let α = a+bζ+cζ2. Then α is self-prime if and only if

(a+b+c)N(α)and ap−cpare coprime and p does not divide a+b+c. Proof. By Lemma 4.9we know that α is self-prime if and only if h₀(_α)

and h1(α)are coprime. Note that we have h0(α) = −∏_ip₌−₀1(aζ−i+b+ cζi)and: h1(α) = p−1

∑

j=0 p−1

∏

i=0 i6=j (aζ−i+b+cζi) = − ∂ ∂b p−1

∏

i=0 (aζ−i+b+cζi) = − ∂ ∂bh0(α)

So if a rational prime q divides h0(α)and h1(α), then it must also divide the resultant of h0(α)and _dbdh0(α)when considered as a polyno-mial in b. But this resultant is precisely the discriminant discb(h0(α))