Polynomial cycles in certain rings of rationals

J

OURNAL DE

T

HÉORIE DES

N

OMBRES DE

B

ORDEAUX

W

ŁADYSŁAW

N

ARKIEWICZ

Polynomial cycles in certain rings of rationals

Journal de Théorie des Nombres de Bordeaux, tome

14, n

o

_{2 (2002),}

p. 529-552

<http://www.numdam.org/item?id=JTNB_2002__14_2_529_0>

© Université Bordeaux 1, 2002, tous droits réservés.

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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques

Polynomial cycles

in

certain

_rings

of rationals

par WLADYSLAW NARKIEWICZ

To Professor Michel France on his 65th birthday

RÉSUMÉ. On montre que la méthode

développée

dans

_[HKN3]

peut être

_appliquée

pour l’étude des

cycles

polynomiaux

dans cer-tains anneaux, notamment les anneaux Z

[1/N]

pour

lesquels

nous décrivons les

_{cycles polynomiaux lorsque}

N est

_impair

ou le double

d’un nombre

premier.

ABSTRACT. It is shown that the methods established in

_[HKN3]

can be

_effectively

used to

_study

_polynomial

_cycles

in certain

_rings.

We shall consider the

_rings

Z

_[1/N]

and shall describe

_polynomial

cycles

in the case when N is either odd or twice a

_prime.

0. Let R be an

integral

domain of zero characteristic. A _sequence

of elements of R is called a

polynornial

_sequence if there exists a

_polynomial

f

E

_R[X]

such that

_{f (xi) _}

holds for i =

_{0,1, ... , n -1.}

A

polynomial

sequence ~

is called a

_{polynomial cycle}

of

_length

n

(or

an

_n-cycle

for

short)

if the elements _{xo, xl, ... ,} Xn- 1 are distinct and _Xn = It is well-known

(see

e.g.

[HKN1])

that if R is

finitely generated

then the

length

of a

polynomial

cycle

in R is bounded

_by

a number

_{depending only}

on the

_ring

in

question

but not on the

polynomial

realizing

it. In the case when R is the

ring

of

integers

in a

_p-adic

field or an

_algebraic

number

field,

such

explicit

bounds were

_{given by}

T. Pezda

( (Pe~ ) .

Two

_{cycles ~ =}

xl, ... ,

and 1/ =

(yo7

y, 7 ...

yo )

of the

same

_length

are called

equivalent,

if for some a E R and u in the _group

U(R)

of invertible elements of R

_(which

we shall call units of

R)

one has

has its coefficients in R and realizes _q.

_Obviously

_every

_cycle

is

_equivalent

to a

cycle

with xo = 0. A

cycle ~

is called

normalized, if xo = 0

and _xl = 1.

It is _easy to show

_(see

_e.g. the

_proof

of Lemma 12.7

_(ii)

in

_[Nl])

that if

~ =

x 11 - - - i Xn- 1)

xo)

is an

n-cycle

and we

_{put Yj =}

(~~ -

xo)

then _{yo, yl, ...} all lie in R and _fJ =

(yo,

_{yl, ... , i Yn- 1)}

yo)

is a normalized

cycle.

We shall

_{call 77}

the nominalization

_{of .}

Note that

_{if Xl}

is not invertible then the

_{cycles ~ and 77}

are not

_equivalent.

A

_cycle

is called

_linear,

if it can be realized

_by

a linear

polynomial.

It is a trivial task to describe all linear

_cycles.

It has been shown in

_[HKN3]

that

in a

finitely

generated

integral

domain R of zero characteristic in which

every non-zero element lies in

_{only finitely}

_many

_principal

ideals there are

only finitely

many

non-equivalent

non-linear

polynomial

cycles.

The _purpose of this note is to show that the

_arguments

_presented

in

[HKN3]

can be used to obtain an

_explicit

construction of all

polynomial

cycles

in certain

_rings.

We shall consider

_finitely

_generated

_subrings

of the field of rational

_{numbers, containing}

1.

_Clearly

every such

ring

has the

form

where N is a

positive

square-free integer.

Denote

by

C(N)

the set of

_lengths

of

_{polynomial cycles}

in

_0~).

Our aim is to obtain some information about these sets and to

_compute

them in the case when N is either odd or twice a

prime.

The calculations were

performed using

Borland Pascal for Windows

_7.0,

PARI 1.38.62

_[Ba]

and KASH 1.9

_[Da].

I am

_grateful

to an _anonymous

_referee,

whose

_{suggestions helped}

to

im-prove the

presentation.

1. We recall first certain

_simple

facts about

_polynomial

_cycles:

Lemma 1. Let R be a domain and denote

by

U(R)

its group

of

units.

(i)

([N],

Lemma

_12.8)

_If

_{(XO,Xl,X2,...,Xn-l,XO) (xo}

=

0, xl

=

1)

is _a

normalized

_cycle

in R

_{of length n}

and zue extend _Xj

_{by putting}

_xj =

for j

&#x3E; _n, then E

_U(R)

holds

_for

i =

1, 2, ....

Moreover

for

_every

i 0 the elements xi

_{and Xj}

are

associated,

which means that their ratio is a

unit,

and

if

( j,

n)

= 1 then

xj is a unit.

(ii)

([N],

Corollary

2 to Lemma

_12.8)

_{If R}

contains an ideal I

_{of finite}

norm N =

#(R/I)

_&#x3E;

1,

then

prime

divisors

of cycle-lengths

in R cannot

exceed N.

Corollary

1.

_If

in a domain R there is a

polynomial cycle of

odd

length

n &#x3E;

_1,

then the

_{equation x}

+ y = 1 has _a solution with

Conversel y, if

units x, y

satisfy

this

equation

then

(0,1,

x,

0)

is a

c ycl e of

t ength

three,

realized

_{b y}

the

_polynomial

Proof.

Assertion

_(i)

of the lemma shows that if

_{(0, 1 , z2 , ... )}

is a normalized

cycle

of odd

_length

n, then X2 and 1 - X2 are units. The

_{polynomial f}

is the

_{Lagrange interpolation polynomial}

_realizing

the

_cycle

_{(0,1, x, 0).}

0

Corollary

2.

_If

there is a normalized

cycle

(0,1,

_{~2, ... ,}

0)

of length

n &#x3E; 3 in a

_ring

R and I is an ideal in R

_of

norm

N(I) =

m _n, then some non-zero element

of

that

cycle

lies in I.

Proof.

The residue

classes xi

mod I

_(i

=

0,1, 2, ... ,

n -

1)

cannot be all

distinct,

and if for some i

&#x3E; j

we

_{have Xi - Xj}

E

_I,

then

_by

Lemma 1

_(i)

the element _xi-j lies in I. 0

Lemma 2. Let R be are

integral

domain.

(i)

([HKN2],

Lemma

₅₎

E R then

_{0,1,~,~,0)

is a normalized

cycle of length 4 for

a

polynomials

in

R[X]

if and only if the

elements

#,

_{1 -a,}

a -

/3,

a/(l -

(3)

are units

of R.

If

these conditions are

satisfied

then the

Lagrange interpolation polynomial realizing

the

_{cycle equals}

where

(ii)

([NPe],

Lemma

₄₎

_If

there is a

polynomial cycle of length

4 in R then either R contains a root

_of

the

_{polynomial X2+1}

or there exist units _{u, v,} w

of

R,

distinct

_from

_1,

_satisfying

u + v + w = 1.

(iii)

3 be a

prime

and denote

by

L(R)

the Lenstra constants

of R.

There exists a

_{cycle of length}

_{p in} R

if

and

only if

p :5 L(R).

If

there is a

polynomial cycle of length

n in R then adl

prime

divisors

of

n are bounded

by

L(R).

(Recall

that

_L(R)

is

_defined

as the maximal number n such that there exist

elements

_0,1,

_{x2, ... ,}

_{of R}

whose all non-zero

differences

are units

(see

[Len],

[LN]).

Proof

of

(iii).

The

_necessity

of the stated condition is

_{implied by}

Lemma 1

(i)

and its

_sufficiency

follows from the observation that the

_Lagrange

inter-polation polynomial

for the

_cycle

_{(0,1, x2, ... , xp_1, 0)}

has its coefficients

in R. The last

_part

is a _consequence of the trivial observation that if

poly-nomial

_f

has a

cycle

of

length n

and

din

then the d-th iteration of

f

has a

Corollary

1. Let n be a

_{positive integer}

and denote

by

p(n)

the minimal

prirrze

not

_dividing

~.

If

there exists a

polynomial cycle of

length

rn in

Q(n),

then the

_prime

_{factors of}

m cannot exceed

_p(n).

Proof.

Follows from

_part

_(iii)

of the lemma and the observation that the

Lenstra constant of

_Q(n)

_equals

_p(n).

D

Corollary

2. Let n be a

_positive

_integer.

_If

_a,

_b,

_c, d are non-zero

_integers

such that all

_{prime factors of}

the

_product

abcd divide _n,

_(a,

_b,

_c,

_d)

=

_1,

the

equation

is

_satisfied

and moreover the ratio

(b+

c)/(d

+

_c)

is a unit

of

Q(n)

then

is a

4-cycle

in the

ring

Conversely,

every normalized

4-cycle

in

Q(n)

leads to a solution

of

(1)

satisfying

the above conditions.

Proof.

The first assertion is a _consequence of

(i).

To _prove the converse let

(0,l,pi/~p2/~0) ((PI,P2,q)

=

1)

be _a

4-cycle

in

Q(n).

Then

(i)

implies

that the numbers

_p2/q,

1 -

_pl/q,

_{(P2 -}

_pi)/q

and

_{pI/(q - P2)}

are all units of

_Q~n~.

Thus the numbers a = _{q - pal,} b _{= pi - P2, c} = _p2, d = _-q lie in

Q~"~,

satisfy

(1),

(b + c) / (d + c) =

-pl/(q - p2)

is a unit and

(a, b, c, d)

= 1

holds. 0

Corollary

3.

_If

Q(n)

contains two

_{units 54}

-1,

-2 with

_difference

_2,

then

4 E

_C(n).

Proof.

If

_A,

A + 2 are units of

_Q~n~

then the assertion results from

_part

_(i)

of the lemma with

_{Q!==1+A,/3=2+A.}

D

We shall utilize also a result of T. Pezda about

_cycles

in

_{p-adic rings:}

Lemma 3

_([Pe,

Theorem 2

_(ii)]).

Let

_Zp

be the

_{ring of integers}

_of

the p-adic

_field

_Qp

and

_put

The

_{set of}

all

_{lengths of polynomial}

_cycles

in

_Zp

_equals

_Ap

_if

_{P &#x3E;}

5 and

Corollary.

If n

has r &#x3E; 2 distinct

_prime

_factors,

then the

_{length of}

a

polynomial cycle

in

_{O(r2 log2,r).}

Proof. If p

is the smallest

_prime

not

_dividing

n, then p =

O(r

log

r)

and it

The next lemma

_provides

a method for

_constructing

all

non-equivalent

non-linear

_(i.e."

not realizable

_by

a linear

polynomial)

_n-cycles

in a

_domain,

provided

one has a

_complete

list of all normalized

n-cycles.

It is contained

implicitly

in the

_proof

of Theorem 2 of

_[HKN3].

Lemma 4.

_{Let 17 =}

_(0,1,

_{Y2,... ,}

_{Yn-l, 0)}

be a normalized non-linear

n-cycle

in an

integral

domain R and let A be the

deading

term

of

the

Lagrange

interpolation polynomials

(of

degree

M,

with 2 M n -

₁₎

_realizing

_77. Let also be an

n-cycle,

whose normalization

equals

77. Then

for j

=

1, 2, ... , n -

1 we have _~~ =

yjxl and divides A.

Corollary. If

the

_{Lagrange interpolation polynomials. of}

all normalized non-linear

_n-cycdes

in a domain R have their

leading coefficients

invert-ible,

then _every non-linear

_n-cycle

in R is

_equivalent

to a normalized

cycle.

We shall also often use the _easy fact

(see

_e.g.

[HKN2])

that if R is a domain and a

polynomial cycle

in R is realized

by

a

_polynomial

with coefficients in

_R,

then the

_{Lagrange interpolation polynomial}

_realizing

that

cycle

has its coefficients in R.

3. Now we describe the sets

_C(p)

for

_prime

_p: Theorem 1.

_{If p is}

a

_prime

then

Proof.

If

_{p &#x3E;}

3 then

Q(P)

C

_Z2

and it follows from Lemma 3 that

_C(p)

c

{1,2,4}.

has in

Q(3)

the

_cycle

(-1,0,1,2, -1)

it remains to show that in case

_{p &#x3E;}

5 there are no

_cycles

of

length

4. If there would be such a

cycle,

then

by

Lemma 2

(ii)

the

equation

u + v + w = 1 would have a solution in units _{u, v,} w of

Q(P)

distinct from

1. Write u =

Elpa, v

=

E2pb

and _w =

f2pc,

with suitable rational

_integers

thus a &#x3E; 0. This shows that v + 2v must be a rational

_integer.

The

equality

v + w = 0 would lead to u =

_1,

which is _not

possible

thus

iv

+ wi ~

1,

and

hence,

because

_{of p &#x3E; 2,}

we

_get

c &#x3E; 0. Since c &#x3E; 0

_implies

we must

have c = 0 and so

finally

with b &#x3E; 0. In view of 2 we

_get

b = 0 thus _p =

3,

which is a contradiction.

We are thus left with the case _p = 2. Since

Q2

_C

Z3nZ5,

Lemma 3

implies

C(2)

C

_{{1, 2, 3, 4, 6}.}

Lemma 5.

_(i)

All solutions

_of

the

_equations

u + v = 1 and u + v = 3 in

units

_of

_Q(2)

are

_given

by

and

respectively.

(ii)

There are three normalized

cycles of length

3 in

Q(2),

namely

The

_{corresponding Lagrange interpolation pol ynomials}

are

(iii)

Every cycle of length

3 in

_Q(2)

is

_equivalent

either to one

_of

the

cycles given

in

_(ii)

or to one

of

the

cycles

The

_{corresponding Lagrange interpolation pol ynomials}

are

Proof.

Assertion

_(i)

is

_immediate,

_(ii)

follows from

_(i)

and

_Corollary

1 to

Lemma

_1,

whereas

_(iii)

results from

_(ii)

and Lemma 4. D

Corollary. If

there is a

_cycle

(0,1, x2, ... ,

~5, 0)

_{of length}

6 in

Q(2),

then there exist

_{A, IL}

E

_{{1/2, -1, 2}}

such that X4 =

Az2

and

x5 = 1 +

~C(x3 - 1).

Moreover x2 is eitker a unit

_of

Q~2~

or is

_of

the

_form

3e with a unit e and

the same

applies

to X3 - 1.

Proof.

The

_assumption

_implies

that

_(0,~2,~4,0)

and

_{(0, X3 -}

_{1, 0)}

are

3-cycles,

hence it suffices to

apply

(iii).

D

One can also

easily

list all

_4-cycles

in

Q(2):

Lemma 6.

_(i)

All solutions

_of

the

_equation

u + v + w = 1 in units

of

Q(2)

distinct

_from

1 are

given by

and every

4-cycle

in

Q(2)

is

equivalent

to one

of

them. The

corresponding

Lagrange interpolation polynomials

are

Proof.

Write u =

_{+2’, v}

=

f26

and w = :i:2c with a &#x3E; b &#x3E; _c. If _{u, v, w}

are

_integers

then

clearly

_{c = 0, w = -1 so u + v = 2}

and

dividing by

2 and

_applying

Lemma 5

_(i)

we arrive at the solution u =

4, v = -2, w

= 1.

If however w is not an

_integer,

then c 0 hence all terms in u2-c +

v2-c t 1 = 2-c _are

integers

and _{moreover must}

equal

fl. Thus

u2-c +

_{(-2-°) =}

~2

and

_dividing

_by

_ip2

and

_{using again}

Lemma 5

_(i)

we

obtain the

_remaining

two solutions. This settles

_(i).

To obtain

_(ii)

observe first that

_Equation

₍₁₎

does not have solutions

a,

b, c, d

in

Q~2~

with a + b = c + d = 0 such that

(b

+

c)/(d

+

c)

is a unit of and then use

(i),

Lemma 2

(i),

(iii)

and the

corollary

to Lemma 4. 0 Now we shall _prove that there are no

6-cycles

in

Q~2~.

Assume thus that

_(0,1,

_{~2?" - ?} is such a

_cycle.

The

_corollary

to Lemma 5 shows that with suitable units _{E, 77} we have x2 E

le7 3el

and _{X3 E}

{1

+

_{q, 1}

+

_3q).

Moreover X4 =

Az2

and _x5 = 1 ₊

1)

with

A,

_{it E}

{-I, 1/2, 2}

and

from

_Corollary

2 to Lemma 1 we infer that _X3 is divisible

_by

5. We have four case to consider:

In cases

(A)

and

(B)

the difference X2 - 1 is a

_unit,

hence Lemma 5

_(i)

shows that x2 E

_{{-I, 1/2, 2}}

and we

_get

Moreover in case

(A)

_{x3 - X2} = 1

_{+ ~ 2013 6}

is a unit hence

_using

Lemma 6

(i)

we

_get

and we see that _X3 = 1 _{+ 7y} is not divisible

by

5,

contradiction. Thus _case

(A)

cannot occur. The case

(B)

is also not

_possible,

since

_{x5 -1}

=

~(x3 -1)

must be a

_unit,

_being

associated to X4 =

e/A,

and so

3q

=

~3 - 1 must be

a unit. This contradiction shows that the case

(B)

is not

_possible.

The

_{impossibility}

of case

(C)

follows from the observation that

_{x3 -1 = ~}

It remains thus to deal with case

(D).

Here _E1 =

_{x2 -1}

= 3e-1 is a unit and the

_equality

leads,

in view of Lemma 5

_(i)

to e E

_{{-1,1,1/4,1/2},}

thus

Moreover E2 = _{X3 - X2} is also _a

unit,

and in view of

El + E2 = _{X3 -} 1 =

₃₇₇

we

_get

thus

_E.l/1J

_{~ {-1,1,2,4}.}

Since x3 = 1 +

3q

_we

_get

thus 16

_{possibilities}

for

x3, however

only

in the

following

five cases _X3 is divisible

_by

5:

So we are left with the

_following

_possible

_cycles:

Using

the fact that _{z5 must} be a unit

and p

E

_{-1,1/2,2}

we see that in

al and a3 one

has &#x3E;

=

1/2,

in

a2 and a4 one

_{has &#x3E;}

E

_{{-I, 2}}

and in a5 one

has p

= 2. Furthermore

x4 - X3 is a unit and thus in _{al one} has A

_{= -1,}

in _a2 one has A =

1/2,

in

A = -2. This leaves us with the

_following

ten cases:

Now a short

computation

shows that the

leading

coefficients of the

cor-responding Lagrange interpolation polynomials

are

_equal

to

_{1/40, -52/35,}

-4/5, -1/160, 1/35, -2/5, -2/35, 3/11, -2/5

and

_-417728/5355

respec-tively,

hence do not lie in

Q(2).

This establishes our claim. 0

4. Now we consider the case of odd

_composite

n.

Theorem 2.

_{If n is}

an odd

_composite

number then either

_{C(n) =}

_{1,2}

or

C(n) =

~1, 2, 4}.

The second

possibility

occurs

if

and

only if Equation

(1)

has a non-trivial

(i.e.,

without a

_vanishing

subsum

_of

the

_left-hand

_side)

solution in

_{coprime integers}

_a,

_b,

_c, d whose all

_prime

divisors divide n and moreover the ratio

(b

+

_c)/(d

+

_{c) is}

a unit

of

Q~n~.

In

_particular

one has

C(n) = {1, 2, 4}

in the

_following

cases:

(a) n is

divisible

_by

_3,

(b) n

is divisible

_by

a

product

u(u+2),

where both u and u+2 are

primes

or

_prime

_powers,

(c) n is

divisible

_by

a

_product

u(2u

db

_1),

where both u and 2u db 1 are

primes

or

_prime

_powers.

Proof.

The first assertion is an immediate consequence of

Q~n~

E

Z2

and Lemma 1

_(iii).

_{To prove} the second we use Lemma 2

(iii)

and the

_following

simple

lemma:

Lemma 7.

_{If n is odd,}

then no solution

of the

equations

(1)

with a

_vanishing

subsurrc can lead to a

4-cycle

in

Proof.

Let _a,

_b,

_c, d be non-zero

_integers

in

Q~n~

satisfying

a+b+c+d = 0 and

with

_(b

+

_c)/(d

+

_c)

_being

a unit of

Q(n).

Assume also that c = _-a, d = -b

with

_{r, s &#x3E;}

1 and odd

_{A, B.}

Our

_assumptions

_imply

that

_{(b - a)/(b}

+

_a)

is a unit of

Q~n~

and thus we must have r = s. But this

_implies

contradiction. 0

Note that for even n this lemma

fails,

as the

_example

n = 30 shows.

In that case

(1)

has two trivial

_{solutions, given by}

2 + 3 +

_(-2)

+

_{(-3) =}

3 + 5 + (-3) + (-5)

= 0 which

_satisfy

the

_divisibility

condition in Theorem 2 and thus lead to

_4-cycles

_{(0,1, 2/5, -3/5, 0)}

and

_{(0,1, -1/3, -2/3, 0).}

The last assertion of the theorem follows from the

_simple

observation that

mln

implies

Q(’)

C

Q(n),

thus

_C(m)

C

_C(n).

In case

(a)

it is sufficient to recall that 4 E

_C(3),

as shown in Theorem 1. In case

(b)

our condition for the existence of a

_cycle

of

_length

4 is satisfied with a = _{-u -}

2,

b = d =

1,

c = u and in case

(c)

that condition is satisfied with a =

-(2u:l: 1),

7

b=d=u,

c = :1:1. 0

A direct check shows that at least one of the conditions

(a), (b), (c)

is satisfied

_by

every

square-free

composite

odd number below 100

except

65 = 5 .13, 77 = 7.11, 85 = 5.17 and 95 = 5 .19.

We shall now show that

C(77)

=

C(85)

=

C(95) =

{1, 2}

and

C(65) = {1, 2, 4}.

For this

purpose

we shall use the

following

result,

which is an immediate

corollary

of the main results of

_[MDT]:

Lemma 8.

_{If n}

E

_{{65, 77, 85, 91}}

then the

_{only solutions,}

_up to

permuta-tions,

of

in

_integers

_a,

_b,

_c, d

_{composed of}

_prime

_{factors of}

n are

given by a = b = 13,

c =

-52,

d = -1

for n = 65

b =

_-5,

c=-1,

d = -19

_forn=95.

One checks without

difficulty

that in case n = 95 the

_divisibility

condi-tion of Theorem 2 is

_violated,

whereas in case ~e = 65 we have the

_cycle

(0,1,12/13, 25/13, 0)

realized

_by

the

_polynomial

It follows from known results about

exponential

diophantine equations

that for _any

_given

n there are

only finitely

_many non-trivial solutions of

(1),

i.e.,

solutions with no _proper subsum of the left-hand side

vanishing.

Hence one can use the

_existing

bounds

(see

[MDT], [Sk])

for non-trivial solutions of

₍₁₎

to determine whether there is a

4-cycle

in

Q(n).

5. Our final

_example

deals with

_C(2p)

with

_{prime p.}

Here we _prove the

There exists an

effective procedure

to

_distinguish

between these cases

_for

any

given

prirne

p.

Proof.

At first we shall assume

_{p &#x3E;}

5

leaving

for a while aside the case

p = 3. Observe that

Q(2oS)

_C

Z3

and for

_{p &#x3E;}

7 we have C

_Z3

n

_Z5.

Lemma 1

_(iii)

_implies

thus

_C(10)

C

_{{1, 2, 3, 4, 6, 9}}

and for _p &#x3E; 7 we

_get

C(2p) C {1, 2, 3, 4, 6}.

The

_following

lemma contains an

algorithm leading

to a list of all

non-equivalent

cycles

of

_length

six and nine for a

_large

class of domains. It is in certain cases

simpler

than the

algorithm

which can be deduced from

[HKN3].

Lemma 9. Let R be a

_{finitely generated integrals}

domain and assume that we have a

compdete list,

_say

of

pairwise

non-equivalent cycles of length

3 in R. Without restriction we

may assume,

multiplying, if

necessary, all elements

of

a

_{cycle by}

a

_unit,

that

_if

_{aj and} ak are associated then

_they

are

equal.

(i)

If

is a normalized

cycle of length

6 in

_R,

then there

_{exists j}

and k such that

ai = _ak and

where E =

1/u,

1] =

-1/z,

_u is _a solution

of

the unit

_equation

_{u -I- v} =

aj,

and z is a solution

_of

the unit

_equation

z + w =

(3k.

(ii)

If

is a normalized

cycle of l ength

9 in

R,

then there exist a unit e such that

1 - e is also a

_unit,

and

_integers

_j,

_k,

1 such that _aj =

ak = _al and

where El =

1/u, u

is _a solution

of

the

equation u

_{+ v} =

a-j, e2 =

-1/w,

w is a solution

of

the

_equation

w + z = s is a solution

_of

the

_equation

s + t

= ,Ci~

and e3 =

-e/s.

Proof.

(i)

Assume

_{that ~}

is a normalized

_cycle

of

_length

6 in R. Then

(0,

X2, X4,

0)

is a

cycle

of

length

3. It is

equivalent

to one of the

_cycles

in our

list,

hence with a

_{suitable j}

and a unit e we have _{X2 = fO:j} and

X4 =

~7-

Also

(l,x3,xs,l)

is _a

_cycle

of

length

3,which

is

_equivalent

to

X3 =

_{1 +qak}

and _x5 = Lemma 1

(i)

implies

that A =

x2-1

=

and x5 = 1 +

r¡(3k

are both

_units,

hence

and

It remains to establish the

_equality

_a~ = _ak. Since

by

Lemma 1

(i)

the

elements

_{x3 -1}

and x2 are associated we obtain the

_equality

with a certain unit u. This shows that _a~ and _ak are associated and our

assumption

implies

now

(ii)

Assume

_{that ~}

is a normalized

_cycle

of

_length

9 in R. Then

(0,

X3, X6,

0)

is a

cycle

of

length 3,

hence with a

suitable j

and a unit _el we have _X3 =

claj and X6 =

Elpj.

Applying

the same

_procedure

to the

3-cycles

(1, x4, x7,1)

and we obtain the existence of

integers

k, l

and units _{E2, e3} such that

Lemma 1

_(i)

_implies

that X2 and 1- X2 are both units. Since X8 is a

unit,

we

_get

with both summands

_being

units and since _X4 is also a unit we

_get

with unit summands.

_Finally

=

_{X3 - 1 =qX2}

with a

_{unit 17}

and hence

again

with unit summands. To obtain the

_equality

_a~ = _{ak = al} it suffices

to observe that Lemma

_{1 (i)}

_implies

that the elements

_{x3, x4 -1}

and _{x5 - X2} are associated and thus _a~ and _ad are also associated. D The last assertion of the theorem follows

_immediately

from the first

_part

of the lemma and the observation that a

_complete

family

of

_{non-equivalent}

3-cycles

in

_Q(’)

can be determined

_{using Corollary}

1 to Lemma

_1,

Lemma 4

and an effective

procedure

of

solving

the

equation u

+ v = 1 in units of any

finitely generated

number

_ring

_(see

_[Sch]).

Our

_argument

for the non-existence of

_9-cycles

in

_Q~2~~

did not cover the case _p =

_5,

_{so we} have _{now to} address this _case.

To obtain this we shall use the second

_part

of the

preceding

lemma and to

do that we have to find first a

_complete

_system

of

_{non-equivalent}

_3-cycles.

This is done in a

slightly

_greater

generality

in the

following

lemma: Lemma 10. Let _p be an odd

_prime

and

(0,1,

_a,

0)

a normalized

3-cycle

in

~~2P) .

.

(a)

If p

&#x3E; 3 is a Fermat

_prime

then

(b)

If p is

a Mersenne

prime

then

(c)

If p

= 3 then

(d)

In all other cases a E

_{-1,

~,2}.

In case

(d)

the

Lagrange

polynomials realizing

normalized

cycles of length

3 are those

_given

in Lemmas 5

(ii).

Proof.

Lemma

_2(i)

shows that we have to determine all solutions of the

equation u

+ v = 1 in units of

Q~2P~.

_Multiplying

this

_equation

be the common denominator of u and v we

_get

an

_equation

of the form

and one sees

_easily

that

_apart

from the trivial case 1 + 1 = 2 this

equation

is reducible to

It is well-known

_(see

_e.g.

_{[Si], [Wa])}

that this is

_{possible only}

if either Fermat

_prime,

or _p = 2y - 1 is _a Mersenne

prime,

_or

finally p

= 3 and the solutions are

32 - 23

=

1, 22 - 1 = 3, 3 - 2 = 1.

Now

note that each solution of

₍₇₎

leads to the

_following

_companion

solutions of the unit

_equation

u + v = 1:

It remains to write down all units

_appearing

in these

_equations.

In

partic-ular in case

(d)

the

only

solutions of the unit

equation

are

1/2 + 1/2 = 1

and 2 +

_(-1)

=

1,

leading

to _a

= -1,1/2

and 2. 0

Corollary

1.

_{If p}

&#x3E; 3 is neither a Fermat

_prime

nor a Mersenne

_prime,

then in

_Q(2P)

we have the

following non-equivalent cycles of length

3:

Proof.

It follows from the lemma that the first three listed

_cycles

form a

complete

set of normalized

_3-cycles

in Lemma 5

_(ii)

_implies

that the

leading

terms of the

_Lagrange

_{interpolation polynomials}

are

3/2,3/2

and

3,

respectively,

hence

_{they have,}

_up to a unit

factor, only

1 and 3 for divisors in

_Q~2P~.

The assertion follows now from Lemma 4. D

The same

_approach

leads to the

_following

two assertions:

Corollary

2.

_{If p}

=

22" + 1

_&#x3E; 3 is _a Fermat

_prime,

then the list

of

all

non-equivadent cycles of lengths

3 in

Q~2P~

consists

_of

_{(0, 3, 3a,}

_{0) (a}

_{= -1,1/2, 2)}

and all

_{cycles of}

the

_form

_(0,

_{d, ad,}

₀₎

where a is a number listed in Lemma

10 (i)

except

a

= -1,1/2, 2

and d is a divisor

of

p2 -~+

1.

Corollary

3.

_{If p}

= 2q - 1 is _a Mersenne

_prime,

then the list

of

all

non-equivalent

cycles of lengths

3 in

Q~2~~

consists

_of

_{(0, 3, 3a, 0)}

_(a

_{= -1,1/2, 2)}

and all

_{cycles of}

the

_form

_{(0, d,}

_ad,

₀₎

where a is a nurrcber listed in Lemma 10

_(ii)

_except

a

= -1,

1/2, 2

and d is a divisor

of

p2

_+p

+ 1.

Now we can

dispose

of

9-cycles

in

Q(10).

Corollary

2 to the last lemma shows that the set

forms a

_complete

_system

of

_{non-equivalent 3-cycles}

in

Q )(10)

Let

_{(o,1, x2, ... , XS, 0)}

be a normalized

cycle

of

length

9 in

Q(10)

and let be defined as in Lemma 9

(ii),

thus a~

~ {1,3,7,21}.

Lemma 11. One has _aj = 21.

Proof.

Since x2, x4, x5, X7 and X8 are

_units,

hence

_{by Corollary}

2 to Lemma

1 we have 3 -

7~3:r6.

Now

(0,~3,

is a

3-cycle

which is

equivalent

to

some

cycle

listed in

(9).

Thus with some unit E and

d, a

listed in

(9)

we have _X3

_{= Ed,}

xs = aEd and _{as a} is _a unit _{we must} have 3 - = 0

Lemma 12. The

_equation

has

_exactly

three solutions in

_{coprime positive integers}

_{x, y,} z such that the

product

xyz has

onl y

2 and 5 ,

for

its

prime

divisors.

They

are

_given

by

Proof.

In view of

_(x,

_{y, z) =}

1 at most one of the numbers _{x, y, z} can be divisible

_by

_{2 resp.}

_5,

thus at least one of them must

_equal

1.

_By

_symmetry

we _may assume x &#x3E; _y. Assume first z = 1. Then _we have _to solve the

following

four

_exponential

_equations:

The first has X =

4,

Y = 1 for its

only solution,

leading

_to

24

₊

51

= 21

and the second has the solution

2 2.51

+ 1 = 21. To solve the

remaining

two

equations

one uses Theorem 3 of

[MDT]

which shows that the summands in the left-hand sides of these

equations

do not exceed

_294,

so it suffices

to

_perform

a

simple

_computer

check. It shows that the obvious solution

52 - 22

= 21 is the

only

one.

In the

_remaining

case

(y

=

_{1, z}

_&#x3E;

1)

_we have _to consider the

following

four

_equations:

However it follows from the table

_given

in

_[Le]

that none of these

_equations

has a solution. 0

Corollary.

Numbers u, v E

Q(10)

which are solutions

of

the unit

_equation

u + v = 21

form

the

following

set

of

six elements:

Proof.

This follows from the observation that _{in every} solution of

₍₁₀₎

one

hasz=1. 0

Now we

_apply

Lemma 9

(ii)

which

_permits,

with the use of the last

two corollaries to establish a list of all

_9-tuples

of elements of

Q(10)

which

may form normalized

9-cycles

and a

_computer

check shows that for all of them the necessary condition

given

in Lemma 1

_(i)

is violated. Since the

polynomial

realizes the

_6-cycle

_{(0,1, 2, 3, 4, 5, 0)}

in one

gets C(10) =

{I,

2, 3, 4,

61.

0

6. It remains to consider the

_ring

_Q(6).

Since it is contained in

_Z5

and

Z7

hence it follows from Lemma 3 that the

_lengths

of

_{polynomial cycles}

in

(~~6~

lie in the

_{set {I, 2, 3,4,5,6,8, 10, 12}.}

Because of

_Q(2)

C

~~6~

Theorem

1

_implies

the existence of

_cycles

of

_lengths

_1,2,3

and 4. The existence of

cycles

of

_length

5 follows from the observation that the

_polynomial

has the

_cycle

_{(o,1, 2, 3, 4, 0) .}

Lemma 13.

_(i)

All solutions = 7z in

positive

integers

_{x, y,} z

having

no

prime

divisor

exceeding

3 and

satisfying

(x,

_y,

z)

=1 are

_given

by

(ii)

All solutions

_of

x ± _y = 13z in

positive integers

_{x, y,} z

having

no

prime

divisor

_exceedircg

3 and

_satisfying

_(x,

_y,

_z)

=1 are

_given

by

(iii)

All solutions

_{of x}

_{f y} = 73z in

positive

_integers

_{x, y, z}

having

no

prime

divisor

_exceeding

3 and

_{satisf ying}

_(x,

_{y, z)}

= 1 are

given by

Proof.

We use standard methods. Our

assumptions

imply

that in our

equa-tions z + y

= _az

(a

=

7, 13, 73)

_one of the numbers

x, y, z must be

equal

to 1. Consider first the case a E

_{7,13}.

If z &#x3E;

_1,

_{thus y}

= 1 and a

_glimpse

at

the tables

_provided

in

_[Le]

shows that the

_only

solutions are

26 -1=

7 ~ 32,

33 + 1

=

7 ~ 22

and

22 ~ 3 + 1

= 13. If z = 1 then we have to solve the

exponential equations

12x

+

_I

=

7,

2X

₊

3y

I

=

_13,

12x -

3y

I

= 7 and

~2X -

3y

=13. The first two are trivial and the solution of the

remaining

two is

_accomplished

with the use of the main result of

[MDT]

which

im-plies

that

_3Y}

_294,

hence a short

_computer

calculation suffices

to

_complete

the list of solutions in cases

(i)

and

(ii).

If a = 73 then the case z =1 is resolved in the same _way as

above, using

the bound

_given

in

_[MDT].

_{If y =1}

then we

apply

the standard

procedure,

going

back to St6rmer

_([St]),

of

_reducing

_{exponential equations}

to Pell’s

equations.

We have to solve the

_equations

Consider the first of them. If

X,

Y are both _even, then

putting u

=

2X12

and v =

3Y~2

we

_get

Since the fundamental unit of the field

_{Q( 73)}

_equals

1068 -125B/73

hence

in every solution _{u, v} of

₍₁₃₎

the number v is divisible

_by

_5,

so

513,

contra-diction.

If X is odd and Y is even then write ~c =

2~X -1)~2

and v =

3Y~2,

, thus

2u2 - 73v2

= :1::1 and

(2u)2 -146v2

= ~2. Since the fundamental unit of

the field

_{Q( 146)}

_equals

E = 145 ₊

12Ý146

and the

_{only prime}

ideal of norm 2 of that field is

principal

and

generated by ~

= 12 -

146

we

_get

2u +

_{v 146}

=

:I::~(145:1:

12ý146)N

with N =

0, 1, 2, ....

For N = 0 we

_get

u =

_{6, v}

= 1 which

provides

_us with the solution

2~ - 32

_{+ 1 =} 73 and _one sees

_easily

that for

_positive

N we

_get

12~v,

thus v cannot be a _power of 3.

If X is even and Y is

odd,

a similar

procedure

leads us to the

_equation

u2 -

3 -

73v2

= fl with v

being

a _power of 3. However this is not

_possible

as the fundamental unit of the field

equals

74 +

5B/219

and this forces v to be divisible

_by

5.

Finally

let both X and Y be odd. Then we are led to the

_equation

2u2 -

3 -

73v2

=

_~l,

thus

(2~)2 -

438v2

=

_~2,

hence 2u +

v 438

is _an

integer

of norm 2 or -2 in the field

Q(,/43-8).

This is however not

_possible

as the

only

ideal of norm 2 in this field is

non-principal.

It remains to consider the second of the

_equations

_(14).

If

_X,Y

are both even then it reduces to

_Equation

₍₁₅₎

with v

_being

a _power of 2 but we have seen

_already

that _any solution of

(15)

satisfies

51v.

If X is odd and Y is even then we are led to

3u2 - 73v2

=

_fl,

hence

(3u)2 -

219u2

=

t3 but the

_only

ideal of norm 3 in the field

C~( 219)

is

_{non-principal,}

making

the last

_equality

_impossible.

If X is even and Y is even then we

_get

u2 - 146v2

= fl with v =

2~Y-1»2.

But the fundamental unit of

Q( 146)

equals f

= 145 ₊

12V-14-6,

hence

31v,

contradiction.

Finally

let both X and

Y be odd. Then we

_{get 3u2 -}

2.

73v2

= fl and

(3u)2 -

438v2

= ~3. Since

the fundamental unit of

_{l~( 438)}

_equals

293 +

14 438

and the

_only

ideal of norm 3 is

generated by ~

= 21 +

438

_{we must} have

Denote

_by

P the

_unique

_prime

ideal in the field

_Q(ý438)

_dividing

73. Then

_{438 -}

_{0 (mod P)}

and 293 -

_{1 (mod P)}

thus 3u - t21

_(mod

_P),

3u - ~21

_{(mod 73)}

and u - f7

_{(mod 73).}

This is however not

_possible,

as no _power of 3 is

_congruent

to f7 mod 73. This concludes the

_proof

of

the lemma. 0

Corollary.

(i)

Numbers _{u, v E} which are solutions

of

the unit

_equation

u + v = 7

form

the

following

_set

of

14 elements:

(ii)

Numbers _u, v E

_Q(6)

which are solutions

of

the unit

_equation

u + v =

13

_form

the

_following

set

_of

six elements:

(iii)

Numbers _u, v E

Q(6)

which are solutions

_of

the unit

_equation

u+v =

73

_form

the

_following

set

_of

six elements:

Proof.

Follows from the observation that u =

xlz, v

=

y/z,

where

~, y, z

are

given

in the

corresponding

_part

of the lemma. D Lemma 14. There are no

_{cycles of length}

6 or 12 in

(~~6~.

Proo, f.

Lemma 10

_provides

us with a list of all normalized

cycles

of

length

three and to deduce the list of all

_{non-equivalent}

such

_cycles

with the use of Lemma 4 we have to find the

_leading

coefficients of

_polynomials

_realizing

the normalized

_cycles.

A

_{simple application}

of

_Corollary

1 to Lemma shows that these coefficients form the set

_{I,

_{7, 13,}

_73}.

This

_implies

that a

_complete

set of

_{non-equivalent 3-cycles}

in

_C~~6~

is

_{given by}

where

_by

A we denote the set of numbers listed in Lemma

_10(c).

It suffices now to

_perform

the

_{algorithm given}

in Lemma 9

_(i),

all

prelimi-nary information

being

contained in the corollaries to the

preceding

lemma.

This has been made

_by

a Pascal _program, which checked the 257 985 can-didates for a

6-cycle

and found that

_only

six of them

_satisfy

the _necessary conditions

_given

in Lemma 1

_(i).

However it turned out that their

_Lagrange

interpolation polynomials

do not have all coefficients in the

_ring

Q~6~,

in

fact,

in each case at least one coefficients has its denominator divisible

_by

seven. The non-existence of

_6-cycles

_{obviously implies}

the non-existence of

cycles

of

_length

12. 0

We need to have a list of all solutions of the

_equation

_{a + b + c + d = 0}

in

_integers

_a,

_b,

_c, d whose

_prime

divisors lie in

_{2,3}

under the conditions

Since at most two of the numbers a,

b,

c, d can be divisible

_by

2 _resp. 3 one sees

easily

that this task consists in

finding

all solutions in

_nonnegative

integers

of the eleven

_{following equations:}

under the condition that no cancellation occurs. We shall call non-trivial

every solution

satisfying

this condition.

Lemma 15.

_(i)

_([Pil)

All non-trivial solv,tions

_{of Equation}

₍₁₅₎

in

non-negative integers

are

_given

_by

(ii)

([Pi])

All non-trivial solutions

_{of Equation}

₍₁₆₎

in

_nonnegative

inte-gers are

given

by

(iii)

([Pi],

[Wg],

Theorem

₂₎

All non-trivial solutions

_{of Equation}

₍₁₇₎

in

_{nonnegative integers}

are

_given

_by

(iv)

([Pi],

[Wg],

Theorem

₁₎

All non-trivial solutions

_{of Equation}

₍₁₈₎

in

nonnegative

integers

are

_given

by

(v) (Known

essentially

since medieval

_times,

see _e.g.

[Al],

Lemma

2.1)).

Equation (19)

has

for

its

_only

solutions.

(vi)

([AFl],

Lemma

_3.2;

_[TW],

Theorem

₂₎

All non-trivial solutions

_of

Equation

(20)

in

_nonnegative

_integers

are

_{given by}

(vii)

([A2],

Lemma

_2.3;

_[TW],

Theorem

₃₎

All non-trivial solutions

_of

Equation (21)

in

_{nonnegative integers}

are

_given

_by

(viii)

([AF2],Theorem

1-A.1;

[Wg],

Theorem

₃₎

All non-trivial solutions

(ix)

([AF2],

Theorem

_2.A.2;

_[Wg],

Theorem

₄₎

All non-trivial solutions

of Equation

(23)

in

_{non-negative integers}

are

given

by

(x)

([AF2],

Theorem

_2.A.3;

_[Wg],

Theorem

₅₎

All non-trivial solutions

of Equation

(24)

in

_{nonnegative integers}

are

given

by

(xi)

([TW],

Theorem

_1;

_[Sk],

Theorem

₃₎

All non-trivial solutions

_of

Equation

(25)

in

_{nonnegative integers}

are

_given

_by

To obtain all

_possible

normalized

_4-cycles

in

_Q~6~

a

_computer

test was made to check which solutions of the

_{equation (1)}

_{implied by}

the

_preceding

lemma

_satisfy

the condition stated in Lemma 2

_(iii).

_Finally

the

_computer

used Lemma 4 to obtain a

complete

set of

_pairwise

_{non-equivalent 4-cycles.}

It turned out

_unexpectedly

that all non-zero elements of normalized

_4-cycles

are units of

Q(6).

The calculations are summarized in the

_following

lemma: Lemma 16.

_(i)

There are 114 distinct normalized

4-cycles cycles

in

Q~6~ .

All non-zero terms

_of

these

_cycles

are units

_of

Q(6),

the

_{corresponding}

La-grange

polynomials

are all

of degree

3 and the ideals

of

Q~6~

generated by

the

_{leading coefficients of}

these

_polynomials

have the

_form

aQ(6)

with

(ii)

There are 8 distinct non-normalized

_4-cycdes

in

~~6~

_containing

0. Lemma 17. There are no

cycles of length

8 in

Q(6).

Proof.

Assume that

_{(0,1,~2~3?" -}

X7,

0)

is a

cycle

of

length

8.

According

to

_Corollary

2 to Lemma 1 for some i _xi is divisible

_by

7. We shall now show that this

_implies

_7~4.

Observe first that the sequence

(O,1, x4/x2, xs/x2, 0)

forms a normalized

4-cycle,

thus

by

Lemma 2

(i)

E = _must be _a unit

and

_by

Lemma 1

_(i)

the same

_applies

_{to q}

=

x4/(xs - x2).

If thus _X2 is

divisible

_by

7 then 7 divides X6 = _ez2 and hence also

however ~2 is not divisible

by

7 then 7 does not divide X6 and hence it must divide X4. Now note that

(0,

X2, X4, X6,

0)

is also a

4-cycle

and hence

by

Lemmas 4 and 16 we must have _X4 =

~u, where ~

is a unit and

u2

divides the

_leading

term of the

_{Lagrange polynomial corresponding}

to one of the

_cycles

determined in Lemma 16. That lemma shows that this can

happen

only

if u is either a unit or is associated with

_5,

since

52

is the

only

non-squarefree

number listed in

_(26).

Thus X4 cannot be divisible

by

7,

contradiction. ll

It remains to exclude the

_possibility

of a

_cycle

of

_length

10 in

Q(6).

Lemma 18.

_(i)

There are 240 normalized

_{5-cycles in}

Q~6~.

In

_five

cases the

_{corresponding Lagrange interpolation polynomials}

care

_cubic,

in the

re-mdining

cases

they

are

quartic.

(ii)

Every

non-normalized

_5-cycle

_containing

zero in

_by

a unit

_{factor from}

one

of

the normalized

_5-cycles.

Proof.

The first

_part

is _pure

_computing,

based on Lemma 2

(ii).

To _prove the second one has to observe that the

principal

ideals in

generated

by

the

_leading

terms of the

_quartic

_{polynomials realizing 5-cycles}

do not

have _any non-trivial fourth power

divisors,

and the

leading

terms of cubic

polynomials

realizing 5-cycles

are all units of

Q~6~.

Thus

(ii)

follows from

Lemma 4. 0

It has been shown in

_[Mo]

that no

_{quadratic polynomial}

with rational coefficients can have a

cycle

of

length

four in the rational field and one could

naively

conjecture

that a rational

_polynomial

of

_degree

N cannot have a

cycle

of

_length

N+2 in the field of rationals. The five

_polynomials

in the last lemma refute this

_hope.

In

_particular

the

_{polynomial 3X3 -}

_2X2

+

6X

+ 1

has the

_5-cycle

_{(0,1, -1/3,1/3, 2/3, 0).}

Corollary.

There are no

_{cycles of length}

10 in

Proof.

Assume

_{that 6 =}

_{(0, ~i, ~ ~3?."}

_{i X97}

₀₎

is a

_10-cycle

in

Q(6).

Multi-plying

all elements of it

_by

a suitable _power of 2 and 3 we _may assume that all ~Z are

_integers.

As

(0,

_{X2 X4, ~s, xs,}

0)

is a

_5-cycle

in

Q~s~,

_part

(ii)

of the lemma shows that it is

_equivalent

to a normalized

_cycle,

hence Lemma 2

(ii)

implies

that ~2?~4?~6?~8 are all units of

Q~s~.

By

Corollary

2 to Lemma

1 some non-zero elements

of 6

must be divisible

_by

5 and as are

units,

we

_get

51xs

and

_similarly

so

finally 351x.5,

so _x5 = 35N with some

_integer

N. Now observe that

(~5,~7,~9,1,~3,~5)

is a

_5-cycle,

thus

is a

5-cycle containing

0. Since all non-zero elements of a

_5-cycle

are

_units,

the same

_applies

to non-zero elements of _17. Hence

holds with suitable units fie Note that ei are

_integers.

Lemma 19.

_(i)

_If

a number in

Q(6)

has at least two distinct

representa-tions as a sum

of

two units then the

principal

ideal that it

generates

has the

_form

_{NQ (6)}

with

(ii)

The number 35 has three distinct

_{representations}

as a sum

of

two units

_of

Q(6),

_namely

Proof.

(i)

Let 0 a E

Q(6)

and assume that with suitable _{fi, fli}

E 11,

-1 ~

and

_integral

_{x, y, z, w, s,}

_t,

_{r, q} one has

Multiplying

both sides

_by

suitable _powers of 2 and 3 we _may assume that all

_exponents

on the

right-hand

side are

non-negative,

the left-hand side al

is an

_{integer, generating}

in

~~6~

the same ideal as a and

min(z,

_{z, s,}

r}

= 0.

Subtracting

we

_get

an

equation

of the form a + b + c + d = 0 with

integral

a,

b,

c, d E

Q~6~

_satisfying

_(a,

_b,

_c,

_{d) =}

1. This is an

equation

having

one of the form considered in Lemma 15 hence that lemma

_provides

us with a list of all solutions and now it is a trivial

_computer

task to list all ideals which can be

_generated

by

_{al ,} hence also

_by

a.

(ii)

A

_simple

_computer

calculation on basis of the list of solutions

_given

in Lemma 15 shows that there are no other

_{representations}

of 35 as a sum of two

_integers

invertible in

_{~~6~ .}

It remains to show that there is no such a

representation

with

_non-integral

summands. If there is such

_{representation}

then it must have the form 35 = +a ₊ ~b with either a =

2"/3 k,

b =

3Y/21

or a =

2~/3~,

b =

1/213~

_or

finally

_a =

3x /2k,

b =

1/2l3’’~

(with

x, y,

k, I,

m

being

non-negative

integers).

In the first case we

_get

which is

_{possible only}

if k = I = 0. In the second _case, if k m then we

get

35. 3m =

±2’3’-k ±

1 and if k _{&#x3E; m} then 35 ~

3~ =

~2~ ~

3k-m.

This

forces in both cases k = _m and thus 35 -

3 k =

~2~ ~

1,

but it follows from

[Le]

that this

_equation

has no solutions. The third case can be handled

similaxly.

0

Corollary. If

a E

Q(6)

is divisible

_by

35 and has at least two

representa-tions as a sum

of

two units then a = 35 with

_{integral k,}

l.

Proof.

It suffices to observe that 35 is the

_only

element of the list

₍₂₈₎

If the

_{representations}

_given

in

₍₂₇₎

of 35N as a sum of two

_integers

invertible in

~~6~

all

_coincide,

_then,

in view

_{of xi 34}

x,

(i

=

3, 7, 9),

we must have E3 - f7 = xZ thus El - _{x3 - X7,} which is

clearly

im-possible.

Hence there must be at least two distinct such

_{representations}

and the

_corollary

to Lemma 19 shows that N is a unit and

_using

the

equalities

(27)

and Lemma 19

_(ii)

we see that for i =

_{1, 3, 5, 7}

_one has

Xi E

_{-N, 3N, 8N, 27N, 32N, 36N.}

However

_by

Lemma

_l(i)

the

differ-ences Xi - xl are units for i =

_{3, 5, 7}

but _a direct calculation shows that

this is not

_possible.

This shows that there cannot be a

polynomial cycle

of

length

10 in

Q~6? .

D

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