Halfway to a solution of $X^2 - DY^2 = -3$

J

OURNAL DE

T

HÉORIE DES

N

OMBRES DE

B

ORDEAUX

R. A. M

OLLIN

A. J. V

AN DER

P

OORTEN

H. C. W

ILLIAMS

Halfway to a solution of X

2

− DY

2

_{= −3}

Journal de Théorie des Nombres de Bordeaux, tome

6, n

o

_{2 (1994),}

p. 421-457

<http://www.numdam.org/item?id=JTNB_1994__6_2_421_0>

© Université Bordeaux 1, 1994, tous droits réservés.

L’accès aux archives de la revue « Journal de Théorie des Nombres de Bordeaux » (http://jtnb.cedram.org/) implique l’accord avec les condi-tions générales d’utilisation (http://www.numdam.org/conditions). Toute uti-lisation commerciale ou impression systématique est constitutive d’une infraction pénale. Toute copie ou impression de ce fichier doit conte-nir la présente mention de copyright.

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

Halfway

to a

Solution of X2 - DY2

= -3

by

R. A.

_MOLLIN1,

A. J. VAN DER

POORTEN2

AND H. C.

WILLIAMS3

ABSTRACT. - It is well known that the continued fraction _expansion of

~D

readily displays the midpoint of the _{principal cycle} of ideals, that is,

the point halfway to a solution of x2 -

Dy2

= ±1. Here we notice _that,

analogously, the point halfway to a solution of x2 -

Dy2 =

-3 can be recognised. We explain what is _going on.

RÉSUMÉ - Il est bien connu _que le _{développement} en fraction continue de

~D

donne facilement le milieu du _{cycle principal} des _idéaux, c’est à dire le point à _mi-parcourt d’une solution de x2 -

_Dy2

= ±1. Nous _{montrons ici}

que de façon analogue le _point à _mi-parcourt d’une solution de x2 -

_Dy2

=

-3 _peut-être reconnu. Nous expliquons ce _qu’il en est.

1. Introduction

Let D be a

_{positive integer,}

not a _square. It is

quite

well known that

if

_{(X, Y)}

is a solution to Pell’s

_{equation X 2 - Dy2}

= 1 then there _are

integers

x , y so that

with

_x/y

a

_convergent

of

JD

_occurring

half as far

along

the continued

fraction

_expansion

of

_JD

as does the

_convergent

_{X/Y .}

Here

_{x2 -Dy2}

=

Q

with

_Q

a divisor of 4D - there is an

_ambiguous

ideal

_{halfway along}

the

period

of reduced ideals.

Similarly,

if

X2 -

Dy2

= -1 then there are consecutive

_convergents

x’/y’

and of

_JD,

once

_again

_occurring

half as far

_along

the continued

fraction

_expansion

of

_JD

as does the

_convergent

X/Y,

so that

1991 Mathematics Subject Classification. 11A55.

mots-cl6s : continued _{fraction, ideal, quadratic form, ambiguous cycle.} Manuscrit _regu le 24 mars _1994, version definitive le 16 décembre 1994.

1

Research _{supported by} NSERC Canada grant #A8484.

2 _Work

supported in _{part by grants} from the Australian Research Council and by a

research _agreement with _{Digital Equipment Corporation.}

3 _Research

Here we have

x’2 -

Dy’2

=

±Q

and

x2 -

Dy2

=

_reporting

that D

is a sum of _squares D =

Q2

₊

P2 .

There is a little more to it than

_just

that. Of course if D is not a

square there is

always

a solution for

X2 - Dy2

=

1,

but whilst it is

additionally

necessary that D - 1

(mod 4),

or D - 2

(mod 8),

for there

to be a solution to

= -1,

there is no

simple

sufficient _congruence

condition on D or on its factors.

It is also well understood that if d

_{I 4D}

_{with ]d[}

_VD

and d

_squarefree

then a solution to

X2-DY2

= d _occurs

halfway along

the

_{complete period.}

In

_particular

a solution to

X2 - Dy2

= +2

always

has the

_convergent

X/Y

occurring

halfway along

the

_{complete period.}

If

X 2 - Dy2

= ±4 then that

solution occurs a third of the _way

along

the

complete

period.

These matters are discussed in some detail in

[4]

and

[5].

As

_regards

PROPOSITION 1.

_If

X 2 -

Dy2

= k then

_{X/Y is}

a

_convergent

_of

B/D

if

Ikl I

VD.

Proof.

If k 0 then X

Y v0

so X = ₊

X)

1 /2Y,

whilst if k &#x3E; 0 then

_{1/2X .}

Hence in the first case

X/Y

is a

convergent

of

_VD

and in the second case

Y/X

is a

_convergent

of

and

_again

_X/Y

is a

_convergent

of

v0.

For

_large

K the literature is

_fairly

_silent,

_{beyond explaining}

a method

-

seemingly

reduction of a

_quadratic

form - which

relates,

for

_arbitrary

integers K,

solutions of

X 2 - Dy2

= K to

_convergents

of

_B/D.

Our

_finding

_halfway

to a solution of

X2 - Dy2

= -3 will include _our

giving

a _necessary and sufficient condition for that

_equation

to be solvable

at

_all,

and then our

_explaining

how to find the solutions

_{doing only}

half the

expected

amount of work. It will become clear that our ideas should enable one to make useful remarks about the

_equation

with

general k, including

the somewhat

_{mysterious K larger}

than

_VD.

We have

_already

alluded to the fact that if

X2-DY2

= -1 then

_halfway

to its solution we find consecutive

complete

quotients

(~+

and

(VD

+ with the fact

_Qh

=

signalling

halfway.

More-over, the well known

_symmetry

of the

cycle

of course _says that the second

half of the

_expansion

is

_exactly

the reverse of the first half.

_Similarly,

halfway

to a solution of

X 2 -

Dy2

= 1 is

_{signalled by}

_successive

_complete

We find

_{analogous signals halfway}

to a solution of

X 2 -

DY’ =

-3

and _prove that the

_expansion

up to that solution has twisted

symmetry.

Namely,

its second half is

_essentially

the reverse of its first half

heavily

disguised by having

been

_{multiplied by}

3. In any case, we

generalise

the

well known cases of

halfway

to a

solution,

and will allow them to be seen

from new

_standpoints.

2. Notation and

_Principles

Throughout

we take D a

positive

integer,

not a _square. We denote the

conjugate

of an

element -y

by

_{%y .}

The

following

remarks are of course

_just

_rappels,

and

_detailing

or

_repeating

them is is done in an

_attempt

to assist the reader. We write elements

so that

with

_integers

P and

_Q

.. I ’ , , ,

That loses no

_generality

to

_speak

of because =

(ac-vD+ be) Ic2 ,

and of course

c2

divides so we need

only replace

VD

_by

and that is tantamount to

_dealing

with an element of the order

Z[ac@]

rather than an element of the order

Ideals. The

_point

is that once

Q

+

_{P) ,}

we _may remark that

PROPOSITION 2. The Z -module

_{(Q, vlD + P),}

which

_corresponds

to the element in

_fact

an

_ideal,

that

_is,

it is a -module in the order

_of

the

_quadratic

_field

Proof.

We need

_only

check that

_{v 1--(VD- D}

+

_P)

is in the ideal. But

whence

_Q

_I

+

_P)

_completes

the verification.

Similarly,

by

constructing

we associate with

(v75 + P) I Q

a

_quadratic

form defined over Z. We note that the discriminant of this form is:

Working

with ideals of the canonical

_shape

_(Q,

_{VD-+ P) ,}

_or, all the more

with elements or with forms

QX2-2P XY +( (P2-D)IQ)y2,

is much the same

thing.

But it is rather more

_sparing

notationally

to deal with elements rather than

_forms,

and when we work with elements

(VD- +

P)/Q,

the role we

_give

to the continued fraction

algorithm

is _very

natural. Given

_that,

we need not have mentioned ideals at all. We

_do,

but in

so far as we

mostly

mention ideals

only

in their canonical form

(Q,

J5+ P)

our

_doing

so is

_mostly

an endeavour to make

_composition

within a

cycle

of forms seem the more natural. Below we abuse

language by speaking

of

composing

elements or ideals when of course we mean

_composition

of the

corresponding

forms.

Periodic Continued Fractions. We may have to be reminded that a

continued fraction is an

_expression

of the

shape

which one denotes in a

_space-saving

flat notation

by

Here the ai

(except

perhaps

ao which may be any element of

Z)

should be

_{positive integers. Nonetheless,}

the formulas are formal and do not _carry any

expectation

of the nature of the we use that in what follows. We

will

_{repeatedly apply}

the fundamental

_{correspondence, easily proved by}

induction, whereby:

PROPOSITION 3.

_{For n = 0, 1, 2,}

...

if

and

_{only if}

Of _course, the ’if’

_part

of this claim is to be read as ’for some choice of _pn

and _qn so that = ... one has ... ’.

In the

_sequel

this

_{correspondence}

will be denoted

_by

~--~ .

PROPOSITION 4. The

_general

_{step n}

=

0, 1, 2,

... in the continued

traction

expansion

of

1’0 =

(JD

₊

I’o)/Qo

=

[ao ,

_{a, , ... an,}

Proof.

The relevant formulaire is

_easily

seen to be

one verifies

by

induction that the

Pn

and

Qn

all are rational

_integers.

Do recall that

_throughout

we have the convention

whereby,

whenever we

refer to an element

(d5

+

_P)/Q,

_{always Q}

I

_{(D -}

P2).

_{So, above,}

of

course

D - 0

=

aoqo

for some

_integer

_{ao .}

It is a basic fact:

PROPOSITION 5. ’Pell’s

_equation’

has solutions in

_{integers x}

and _y, with

_{y 0 0.}

The

_argument,

see

[5],

relies on several

applications

of the box

principle,

working

with elements of

_(1/Q)Z.

PROPOSITION 6. Given a solution

(x, y) ,

with

y 0 0

to Pell’s

_equation

(2),

each

_{decomposition}

with

_integers

_{aZ ,} entails the

_{pure-periodic}

continued

_fraction

_expansion

.Remark. Here we observe that a unimodular matrix has a

decomposition

true.

_Naturally,

it cannot

_distinguish

_{between -y}

and its

_conjugate

_77.

We

see below that our

preference

is

always

for that

conjugate

which satisfies

7&#x3E;0.

The

_argument,

detailed in

_[2]

and

_[5],

is a

straightforward

application

of

the

_{correspondence}

of

_Proposition

_3,

once one recalls that

We should understand that the continued fraction

_expansion

so obtained

may not be admissible: in that the ai

might

not be

positive

integers.

Indeed

in the most

_generally

known _case,

_{if q}

=

VD

then _ar = 0 and

However,

there is a

_unique

admissible

_{decomposition}

of the unimodular

matrix,

thus with

_{positive integers ai}

_(obtained

_{by applying}

the Euclidean

algorithm

to the rows of the

matrix),

_exactly

when the first _row,

_respectively

the first

_column,

dominates the second: that

_is,

_precisely

if x _{&#x3E; y} &#x3E; 0

and y

&#x3E; x -

(~y + ~y)y &#x3E;

0. These

_inequalities

entail that x

_{and y}

be

positive

and

_{that y}

is reduced:

_{namely that 7}

&#x3E;

_1,

whilst 0 &#x3E;

_~y

&#x3E; -1. Those

_inequalities

are

precisely

the well known Galois conditions for

pure-periodicity.

There is an

_opportunity

for mild confusion

engendered by

the

frequently

seen

suggestion

that

vID

is reduced. Indeed the ideal

(1,

is

reduced,

but that is because it

_equals

_(1,

B/D

+

_{ao) ,}

and the element

_vID

+

_{ao is}

reduced. Idealists _may well see this as a

_{justification}

for their

_position;

those not troubled

_by

an occasional need for translation will take it in their

stride.

Equivalence.

Two ideals I and

J

in the order are said to be

equivalent

if there exist nonzero elements

_Q

and _{q in} such that

=

(7)~.

It is

_easily

checked that

equivalence

is _an

equivalence

relation

on the ideals

_compatible

with

_{multiplication}

of ideals. So the

_equivalence

classes

_yield

an abelian _group called the class _group of

Z [@] .

We will be

reminded below that such class groups are of finite order.

Two numbers

_Q

_{and 7}

are said to be

_equivalent

if there is a unimodular

integer

matrix so that

(3

=

(py

₊

p’)/(q7

₊

q’) .

_By

decomposing

q q )

the matrix as a

_product

of

elementary

row transformations it follows that

have ’the same tail’: that

_is,

the

expansions

differ in at most

finitely

_many

initial

_partial

_quotients.

It is a

_simple

exercise to confirm that the

_{correspondence, whereby}

an

element

_{( ~ +}

_{P) /Q}

_yields

an ideal

(Q,

~lD + P) ,

entails that

equivalent

numbers

_{yield equivalent}

ideals.

The

_import

of our discussion on

_periodicity

is that we see that each

7 E K is

_equivalent

to

_just

_finitely

_many reduced

_elements,

to wit the

complete

quotients appearing

in its

_periodic

continued fraction

_expansion.

But,

we

_recall,

(~ +

P)/Q

is reduced if and

_only

if

_~

+ P &#x3E;

_Q

and

-Q

Thus 0

_Q

_2V"D

and

_-B/D

P

B/D

shows that there are

_just

finitely

_many reduced elements and a fortiori

_just

_finitely

many

equivalence

classes of ideals.

A reduced element

_{(VT5 +}

_P)/Q

yields

a reduced ideal

(Q, v75

+

_P).

One verifies that an ideal is indeed reduced if it contains no nonzero

(3

so that

both 101

_Q

and

1(3’ 1

Q .

In this

_language

the

_period

of a

continued fraction

_corresponds

to a

period

of

equivalent

reduced ideals. Of course a

_{period yields}

a

_complete

_equivalence

class of reduced ideals. For

if an ideal is reduced it occurs in the

_cycle

of reduced ideals to which it

is

_equivalent,

and if it is

_equivalent

to some ideal then it

corresponds

to

a

_complete

_{quotient occurring}

in the continued fraction

_expansion

of the element

_{corresponding}

to that ideal.

One says that two

quadratic

forms

are

_equivalent

if there is a unimodular matrix

(q q )

so that

transforms the first form into the second. The forms are

Properly

equiv-alent if the matrix has determinant

_+1;

otherwise

_they

are

_improperly

equivalent.

It is

_{straightforward}

to

_verify

that the numbers +

_P)/Q

and are

_equivalent

_only

if the

_{corresponding forms,}

as cited

above,

are

_equivalent.

Of course

if,

with

GauB,

we restrict ourselves to

proper

equivalence

then we _may have more _proper

_equivalence

classes of

forms than we have

_equivalence

classes of ideals.

is

_{evidently generated}

over Z

by

the

_quantities

Set G =

gcd(Q,

Q’,

P ₊

P’) .

One _may

verify,

by studying

the classical

formulas or from first

principles,

that the

product

is a rational

_integer

multiple, namely G,

of

_(q,

_VD-+p)

_{where q}

=

QQ’/G2

and _p satisfies the

three _congruence conditions _{p -} P

_(mod

_{Q/G) ,}

_p = P’

_(mod

_Q’/G)

and

(P -

p) ~P~ -

p)

= 0

_(mod

The first

_pair

of _congruences determines _p modulo

_{QQ’IG(Q, Q’) .}

The last _congruence decides which of the

_remaining

_(Q,

_Q’)/G

_{possibilities}

for p

mod q

is to be taken.

Correspondingly,

the

_product

of the

_quadratic

forms

together

with a substitution

with

_integer

coefficient

_{A, ...}

and

_{A’, ... ,}

not all

_sharing

a common

factor,

yields

a form

qX2 -

_2pXY

+

~(~2 -

known as a

_compound

of the

given

forms. In fact the Grassmann co-ordinates of the substitution matrix

C A

are determined

(they

are

essentially

the six coefficients of

(

A B C D

the

_given

_forms),

so the substitution is fixed _up to

_{multiplication by}

a 2 x 2

unimodular

_integer

matrix. Note remarks on this matter

_by

Shanks at

p.182

of

_[7].

Thus the

_compound

form is defined up to

equivalence

and

we see that

_compounding

is well defined on

_equivalence

classes of forms of

the same discriminant. We will refer to the

particular

_case, where the two stated forms

_yield

the form as

_composition.

In

_particular,

one sees that the

_composite

of a form

Q~2 -

_2Pxy

+

Q’y2

and its

_opposite

_Qx2

+

_2Pxy

+

Q’y2

is

_equivalent

to the form

correspondingly

the

_product

of an ideal

_(Q,

B/D

+

_P)

and its

_conjugate

is a

_principal

ideal.

_Thus,

it turns out that

_Q

_I

2P is the condition for

a form

Qx2 -

_2Pxy

₊

Q’y2

to be

_properly

_equivalent

to its

_opposite;

the transformation is effected

_{by y--~ y, x ~--&#x3E; x +}

_{(2P/Q)y .}

In this case the

Ambiguity.

A

_quadratic

form

_{Qx2 -}

_2Pxy

+

((P2 -

is said to be

_ambiguous

if it is both

_properly

and

_{improperly equivalent}

to itself. The

surprising

equivalence

must

_interchange

the numbers

(JD + P) I Q

and its

conjugate

(- vlD- +

P)/Q.

Thus the form is

_ambiguous

if and

_only

if the element

_{(~17 +}

_P)/Q

is

_equivalent

to its

_conjugate.

In an alternative

_{interpretation}

one _says that an ideal

(Q, JD

+

_P)

is

ambiguous

if it is

_equal

to its

_conjugate.

Of course an ideal

equals

its

conjugate

only

if it contains the

_conjugate

of each of its elements. Hence

(Q,

is

_ambiguous

if and

_only

if it contains both and

+

_P)/Q

and that is so if and

only

if

We should recall that the basic condition

Hence

_2P/Q

E Z entails

_A/Q

E Z. Thus the

_ambiguity

of an ideal entails

that its norm

Q

divides the discriminant 0.

Conversely,

if an ideal has a

squarefree

norm

_dividing

the discriminant then that ideal is

_ambiguous.

An ideal

_(Q,

is

_ambiguous

if and

_only

if the continued fraction

expansion

(

has _ala2 ...

ar-, a

palindrome

and ar =

2ao -

2P/Q .

In that _sense

ambiguity

is

_symmetry;

see

[2].

One _says that an

equivalence

class of ideals is

ambiguous

if it contains

both an ideal and its

_conjugate;

_hence,

it contains the

_conjugate

of each ideal in the class.

Infrastructure. The literature _appears to

_speak

_{interchangeably}

of

com-posing

or

compounding

forms. It is not a bad idea to reserve one of these

terms for the

_original

_operation

on the

equivalence classes,

and the other

for the

_specific

manner of

_composition

-

corresponding neatly

to

multi-plication

of ideals in canonical form - which

we mention above. Thus we

may say that

compounding

is an

_operation

on the ideal or form classes but

that our formulas describe

_composition

of forms as such and therefore of

the

_{corresponding}

elements. Thanks to Shanks

_[6]

and thence to Lenstra

[1]

we now understand that in that _way

_composition

_operates

_meaningfully

within an ideal

cycle.

That is an

_operative

principle

in the

_sequel.

We

_barely

need to

_emphasise

it

_here,

because almost

_throughout

we

we are active.

_Thus,

for

_example,

were we to work in the order

_Z[8],

with 6 =

(B/D+

1)/2,

as we

_might

if D were 1

_(mod

_4),

our elements would

appear as

(8 + P)/Q -

with

Q

I

(b + P) (6 + P) -

and the

corresponding

ideals as

(Q, 6

+

_{P) .}

_{That, incidentally,}

forces a

_change

of notation from that of Perron

_[3],

as is alluded to at

Proposition

9 below.

3. Pairs of Ideals near

_Halfway

Suppose

that there is a solution to

X2 - Dy2

= -3. We shall show

the existence of

_pairs

of ideals

_{(Qh, ý’D}

+

_Ph~

and

_(3Qh,

_{B/D 2013}

_P)

which

compose to

_yield

an ideal

(3,

+

_{p) ;}

it’s because P -

_Ph

_(mod

_Qh)

that this works.

_However,

the

_point

is that the element

_{(v’D + Ph)IQh}

is

reduced and that in an _easy to understand sense the ideal

(3Qh, VD- -

P)

is ’near’ it.

_Then,

_given

that

_(3,

_v’D

+

_p)

need not be a _square, the ideals

(Qh,

v’D + Ph)

are

_pretty

well as close as we can

_get

to its square root and

in that sense lie

halfway

to it.

_Indeed,

as one _may

verify

experimentally,

the

ideals

_{(Qh, B/D}

+

_Ph)

we find below -

if

_they

lie in the

_{principal cycle}

-do lie

_{roughly halfway along}

the

_cycle

to an ideal

(3,

D-

p) .

Of course the fact of a solution to

X2 -

Dy2

= -3 entails the existence

of an ideal

(3,

d5

+

_p)

in the

_{principal cycle.}

More than

_that,

because of the minus

_sign

its immediate

_{’neighbours’}

- in

the sense of the continued

fraction

_expansion

or

_equivalently

the ideal

_cycle,

lie in the

_principal

_genus

and thus have _square roots

_lying

in

_{ambiguous cycles.}

It will become clear that the ideal

_(3,

_B/D

+

_p)

has

_appropriate

_neighbours

with norm

Q~ .

We lose no

_generality

in

_supposing

that D is

squarefree,

and _suppose

that from hereon. We may then remark that if

Dy2

= -3 has _a

solution then

_certainly

each

_prime

factor of D

_splits

in

_{~(~) .}

_Thus,

as is

easily

checked

by quadratic

_reciprocity,

the

_{only possible}

factors of

D are 3 and

_primes

_congruent

to 1 modulo 3.

_{Hence, recalling}

that the maximal order of

_~(~)

has basis

_{generated by}

1 and

₍₁

+

H)/2

it

follows that D has

_{representations}

To be

_precise,

if D has v different

_prime

factors

_congruent

to 1 modulo 3 then there are

2v-1

essentially

different such

_{representations}

_-

disregard-ing

interchange

of M and

_{N ,}

_changes

of

_sign,

and - in this

_special

case

- other actions

_by

roots of

_unity.

We can

_guarantee

the

_parity

of

_{N ;}

that

_is,

we _{may arrange} N to be odd

we _may

_interchange

M and N. If both are

odd,

then M + N is even in

D =

(-M)2 -

M(M

₊

N)

+

_(M

+

N)2.

Here too we _may

interchange

M

and N if we wish.

Nonetheless,

the cases N odd and N even are rather

different and need

separate

treatment.

Below we will

prefer

N odd.

Suppose,

however that N =

2Q

is even

-of course then M is odd. Then we have an

_equivalent

_{representation}

Conversely

D

= A2

+

3B2

entails D =

(A -

B)2

₊

(A - B) ~

2B ₊

(2B)2 .

Again disregarding

signs,

there are

2’-’

distinct such

_{representations.}

Out of contrariness we start with the even case.

Suppose

then that

D has a

representation

D =

L2

₊

3Q2

with L and

Qh

positive.

Then

it is easy to check that either the element +

_L)/Qh

or the element

(VD- + L + Qh)lQh

is reduced. We set

_{Ph = L + gQh with g = 0}

or 1 so

that is reduced. Because

Ph-L -

0

_{(mod Qh)}

the

_product

of the ideals

_{(3Qh, ~-L)}

and

_(Qh,.JÐ+Ph)

is with

some P - -L

(mod 3Q~ ) .

_Now,

consider the

_product

of

(Qh,

B/D+

_Ph)

and +

_{P) .}

It is

_{Qh (3,}

JD

+

_p)

with some _{p - P}

(mod 3)

because

Ph

+ P = 0

_{(mod Qh) .}

_Hence,

_indeed,

we have found a

pair

of

ideals and

_(3Qh,

_{@ + P)}

which _compose to

_yield

an ideal

(3,~D+p).

Since the

_reciprocal

of

_{(J5 - L)/3Qh}

is

.JÐ

+ L which is a neutral

element for

_composition,

and since we obtained the ideal +

_P)

from

_(Qh,

_by

_composing

with

_(3Q2,

the sense in which

the ideals

_{(3Q~,, ~ + P)}

and

_{(Q~,, ~ +}

_Ph)

are ’near’ to one another is

manifest.

_Moreover,

for later use we remark that

plainly, being equivalent

to a

principal

ideal,

the ideal

(3Q2,

JD -

L)

is

principal.

Finally,

we remark that on the one hand

Ph) =

and on the other hand

Ph -

D =

Ph2 -

((Ph -

₊

3Q~).

Hence

We will refer to an event

_3Qh

=

Qh-1

or

4Qh

=

Qh- 1

_+2Ph

_{as a}

_signals

-namely

a

signal

that we _may be

halfway

to a solution of

X2 - Dy2

= -3.

Before

_summarising

we had best deal with the cases L or

Qh negative.

In so far as we are

_dealing

with ideals a

change

in

sign

of

Qh

is irrelevant -so we assume

throughout

that

yh

&#x3E; 0 . On the other hand

_(Qh,

continued fraction formulas reminds us that the reduced element

_conjugate

to is Our

_argument

above should

_change

to start with the ideal

_(3Qh,

Vi5

+

_{L) .}

It is

_{multiplied by}

_(3Qh,

_{v75 - L)}

as

above,

we

might

_just

as well _say

’composed

with’,

yielding

an ideal

(Qn,,

~ + P~,+1 ) .

That

_product

is then

_composed

with

_(3Qh,

_{v75 + L)}

to

yield

an ideal

(3,

~D + ~~ .

We will have set

_Ph+,

=

L + gQ

with g

= 0 _or

1. Now we conclude with the remarks that

once

_again

_{with 9}

= 0 or 1. These then are the

_conjugate

_signals.

PROPOSITION 7. Each

_{representation}

D =

L2

₊

3Q2h

with

Qh

_&#x3E; 0

yields

a pair

of reduced

elements

_and,

in

some

_{ambiguous cycle,}

one or other

_of

a

_conjugate

_pair

of signals

Remark. We _say

_only

_{that, given}

the

_{representation,}

we find one or other

of the

_pairs

of

_{conjugate signals}

in some

ambiguous cycle

of the order

Of course we can find them in the continued fraction

_expansion

of

@ itself,

thus in the

_{principal cycle, only}

if = -3 has _a solution

in

_integers

_{(X, Y) .}

Our ultimate purpose, hence the Theorems

below,

is inter alia to show that the appearance of such a

_signal

_{is necessary} as well as sufficient. Thus we hasten to

emphasise

that the

_Proposition

does not

rely

on there

_being

a solution to

X2 - Dy2

= -3. We _use

only

the fact

of the

_{representation}

of D and deduce from it that there then is an ideal

(3,

B/D+p) 2013

namely

that there is a rational

_integer

_p so that 3

I D - p2 .

We now turn to the

_{representations}

D =

M2+MQh+Qh

still

_supposing

that

_Q~,

&#x3E; 0. Moreover we shall also _suppose that

_Qh

is

_odd,

as we _may

without loss of

_generality.

If M &#x3E; 0 _{it is easy} to

_verify

that the element

(B/D

+

_M)/Qh

is reduced. In

_analogy

with the

_foregoing

case we _propose

to compose the ideal

(Qh, v’Ï5

+

_M)

with an

_appropriate

ideal of norm

3Q~,

indeed with the ideal

_(3Q2,

v’Ï5 -

M

_{+ 2 (3Qh -}

_{1)Qh) .}

By

the _way, we have not

_pulled

this ideal

_quite

out of thin air. We noted

that 4D -

_(2M

+

Qh)2

=

3Q2

and then with some mild effort rewrote the

throughout.

In _{any case,} as before we obtain an ideal

(3Qh,

P) .

We

could have deduced that with less

_turgid

detail since the

_only

_point

is that

M - 2 (3Qh -

1)Qh -

M

_{(mod Qh) .}

Hence P == -M

_{(mod 3Qh) .}

Now

composing

the ideals

_(Qh,

JD

+

_M)

and

_(3Qh,

_JD

+

_P)

_yields

an ideal

(3,

J5

+ p)

just

as above. We set M =

Ph .

Of _course D -

Ph

=

and D -

_Ph

=

Phoh

₊

Q~

so we obtain

Qh- i

=

Qh

₊

Ph,

an attractive

signal

indeed.

In the

_conjugate

case we start with the ideal

_(3Qh,

M -

_{Qh) .}

The first

_composition

as above

yields

an ideal

(Qh,

Ph+i)

_{corresponding}

to a reduced element

( ~ +

and when it is

composed

with

(3Qh,

VD

+ M -

_{Qh )}

we once

_again

obtain an ideal

(3,

JD

+

_{p) .}

All this

is tantamount to our

_setting

M = and

yields

the

signal

Qh+l

=

~’

PROPOSITION 8. Each

_{representation}

D =

M2

₊

MQh

₊

Q~

with

Qh

_&#x3E; 0

and odd

_yields

reduced elements and and

a

_{conjugate pair}

of signals

Once

_again,

this means that

_given

the

_{representation}

we

find,

in some

ambiguous cycle

of the order the

_pair

of

_conjugate

_signals

-

to

wit,

neighbouring

P’s and

_{Q’s satisfying}

the cited relations.

We will notice below that when =

Qh

+ Ph

then =

Qh

and

Qh-2

=

Ph . Hence,

automatically, Qh-1

=

Ph-1

₊

Qh-2 ,

which is the

conjugate signal slightly displaced.

Thus it

_actually

suffices to

_emphasise

only

one of the

_present

_pair

of

signals.

We reiterate that the

Proposition

does not

_rely

on there

_being

a solution to = -3. We _use

only

the

fact of the

_{representation}

of D . Our purpose in

alluding

to a solution at all was to be able to _say that a solution is

equivalent

to the _presence of an ideal

(3,

JI5

+

_p)

in the

_{principal cycle.}

However we have the

_{representations}

as soon as 3

I

D -

p2

and that last is silent as to the class of the ideal

(3,~ + ~).

4.

_{Halfway Along}

the Continued Fraction

_Expansion

In this section we _suppose that the

_halfway

ideal

(Qh, JI5

+

_{Ph) ,}

and therefore also its

_conjugate,

lies in the

_{principal cycle.}

We set

and denote the

_convergents

_{variously by}

or We write

It will be useful to set

So

_{Mn J}

is

_symmetric,

and since

we have

-with the eccentric notation

_Nn

for the

_transpose

of

_Nn

_congenially

remin-ding

us that it is

Here we have said

nothing startling

other than

_perhaps

for our

_allegation

that

PROPOSITION 9.

This remark is central and warrants an extended

explanation.

Explanation.

We commence

_{by observing}

that

_{’multiplication by x -}

.JÏ5y’

is a

Q-linear

_map of the

_Q-vector

_space into itself. With

_respect

to the basis

_{2013B/D,

₁₁

its matrix is

Thus M has determinant =

x2-Dy2

and in

particular

Mn

has determinant =

compatible

with our

allegation.

More to the

_point,

we

_apply

the

correspondence

of

Proposition

3 between

matrices and continued fractions after first

_noticing

that

Then

explaining

that the constants and do indeed define the next

complete quotient

exactly

as our notation had

suggested.

It therefore seems

_appropriate

to consider the

Mn

to define the _sequences

and

_{(Q~,) ,}

via the statement of the

_proposition.

Were we

_working

in

an order different from

Z[B/D]

this would force a

_change

of notation from

that of Perron

_[3].

In _{any case,} with that cleared _up, we next remark that the

product

of

any two matrices of the

_shape

M is

_again

of that

_shape.

That is

That’s

_plain

without

_computation

on

remarking

that

We shall

_interpret

our idea for

_finding

the

_halfway

_point

to a solution of

X2 - Dy2

= -3 in terms of the observation that

_{multiplication}

of matrices

M

_corresponds

to

_composing

the

_{corresponding forms,}

as is

_clear,

of _course,

in the

_light

of the remark

_just

made.

_Accordingly

we are now in a

_position

Representations

D =

M2+MN+N 2with

N odd. It seems convenient

to suppress

subscripts

as

long

as

practicable.

Our

_argument

runs as follows.

Suppose Q

odd. The

_{representation}

D =

p2

₊

PQ

₊

Q2

yields

the ideal

(Q, B/D

-f-

_P)

and it

_corresponds

_to,

say,

Expanding

V75

we

mayl

write

corresponds explicitly

to the

matrix

The

_conjugate

of the ideal

_corresponds

to the

_adjoint

of the

matrix2.

Hence the first

_composition

_yields

the matrix

_product

the last line after a little work. We

recognise

the intermediate matrices as

_reciprocals

and _may therefore

_interpret

the

product

as

telling

us of the

following configuration

in the

_expansion:

1

"No, you may not", we hear you say.

"P+ ~(Q -1)

is not the _{integer part} of

VT3".

To that one of us _{replies "Up} a _{gumtree" (an Antipodean response} to _nonsense). Our

assertion is an _{identity showing} that a certain ideal is _principal. The fact that it is not

the usual continued fraction expansion of

VD

is _quite irrelevant.

2 Strictly

speaking, it _corresponds to the _inverse, but since the correspondence with continued fractions _{disregards multiplication by} nonzero constants we _effectively obtain

More

_compactly

Of course we have recalled that .

Whatever,

we _may now _compose the ideal

(3Q,

vT5

+

Q -

P)

with

(Q,

VD

+

_{P) .}

Here one

might

notice,

the absurd _way is

by

the use of

false

_{transposition3,}

that also

We obtain

corresponding

to some ideal

(3,

VD

+

_{p) .}

The first matrix

_corresponds

to the continued fraction

and since it is well

_known,

and

_easily

seen

by

the matrix

correspondence,

that

the second matrix

_corresponds

to that

_expansion

_{’multiplied symmetrically’}

by

3,

to wit to the continued fraction of

3That is, transposition in the wrong diagonal. This should not be confused with "students’ transposition" where one over-enthusiastically transposes in both _diagonals. Mind _you, the students have a _point. The "double transpose" of a _product of 2 x 2

matrices is the _product of the double _transposes. There is none of this nonsense of

Of course the matrix

I - _

,

expands

as a

_product

of matrices of the

shape

I -B I

’

-’- , -’- ,

-The

_{decomposition}

is tantamount to

_performing

the euclidean

_algorithm

on the rows of the 2 x 2 matrix. That is

_{plain given}

that the continued

fraction

_expansion

of a rational is

_effectively

the euclidean

_algorithm

on its

numerator and denominator. We

_give

numerical details in our

_examples

below. In

_particular,

_above,

- y

To ensure that we have

_expressed

matters

_correctly

we had better confirm

that all the entries

_{2y’ -}

_{y ,} 2x -

_{x’ ,}

... , are

positive.

We note that is so if

and

_only

if the

_partial

_quotient

= 1.

Fortunately, always Qh

=

and so _ah-1 =

1,

because the element

(VT)+

Qh)/(Qh

₊

Ph)

is reduced

when D =

Ph

_{+ +}

Qh .

THEOREM I.

Let x2 -

_Dy2

= and _suppose that

Qh-1

=

Qh

₊

Ph

in the

_{principale cycle.}

Then

X2 -

Dy2

=

-3,

where

Moreover the continued

_fraction

_expansion

_of

_XIY

is twisted

_symmetric.

Its

_{first half}

is the

_expansion

_of

_xly,

say

[ao,

w~ ,

and its second

part

is

the

_expansion

_of

_{(2(u ~ -}

_{1)/(2 -}

_~w~)

- thus the _reverse

of

that

_first

half ,

omitting

the zero-th

_partial

_quotient,

twisted

_{by multiplication by}

a

transformation of

determinant 3.

Proof. Having

been

_{signalled by Qh-1}

=

Qh

_+-

Ph

we have

and thus the

_{representation}

D =

Ph

₊

PhQh

₊

Q2.

But then _we have _seen

explicitly

that

yielding

the claims.

We will not need the

_conjugate

_case; see _{however the discussion below}

An

_example.

One can see that

1228312 - 1729 . 29542

= -3 after a brief

computation.

In

_detail,

Here

_Q7

= 48 = 23 ₊ 25 =

Q8

₊

P8.

Incidentally,

a7 = 1 whence

P7

=

Q8

as in the continued fraction

configuration

we mentioned above. That’s

clear of course because an element

(JD

+

_Q)/(Q

+

_P)

is reduced if D =

p2

+

_PQ

+

Q .

We read off that

We

_purport

that the continued fraction

_expansion

_{[41 , 1 , 1 , 2 , 1 , 1 , 2 , 1~}

continues with the

_expansion

_{corresponding}

to

Note that all entries in the

_product

are

positive

exactly

because a7 = 1.

Indeed,

the

_product

_expands

to

_yield

that is

_55/19

=

[2,1,8,1,1]

and

2287/790

=

(2 , 1 , 8 , 1 , 1 , 40/3~ ,

precisely

as

predicted4.

We should also confirm that

and

Representations

D =

M2+MN+

N2

with N even. We now return to

_{representations}

D

= M2 + MN + N2

with N even.

Appropriately

expanding

VD

we obtain

immediately yielding

the ideal

_{( 4 N2, ~lD}

+ M

_{+ 2 N) .}

_Accordingly

we

set N =

2Q

and L = M ₊

Q

and note that one or other of the elements or

( fD+L+Q)/Q

is reduced. As

before

we write P =

L+gQ

with g

= 0 or 1. The first

_composition

_corresponds

to

If g

= 0 this

_just

reports

that

_Qh-1

=

3Qh,

in the _proper continued

fraction

_expansion.

Of course we have recalled that D =

(Ph - gQh) 2+ 3Q2

and

₍₃

+

_g2)Qh

=

Qh-1

₊

2gPh .

Now,

the second

_composition

with the ideal

_(Q,

_v75

+

_P)

corresponds

to

Firstly

suppose

that g

= 1. Then the first two matrices

correspond

to the

continued fraction

whence the last two matrices

_correspond

to the

_expansion

Thus the continued fraction

_expansion

of

_X/Y,

where

X2 - Dy2

=

-3,

is twisted

_symmetric

with the

_symmetry

twisted in that the second half of the

_expansion

is the reverse of the first half -

omitting

the zero-th

partial

quotient,

and is divided

_by

3.

_{If g}

= 0 _we

_again

have twisted

symmetry.

The first half of the

_expansion

is

_{~ao , w]}

and the second half

is the continued fraction

_expansion

of the rational

_[+W-

_]

_{multiplied by}

3.

THEOREM II.

_Suppose

that either _g = 0 _or

g = 1 and

(3

₊

Qh-1

+

_2gPh

in the

_{principale cycle,}

and x2 -

_Dy2

= Then

X2 - DY2 = -3,

where

Moreover the continued

_fraction

_expansion

_of

_XIY

is twisted

_symmetric.

Its

_{first half}

is the

_expansion

_of

_x/y

with _g

_appended

and its second

_half

is the reverse

of

that

first half

-

omitting

the zero-th

_partial

_quotient,

and

the

_expansion

_of

_xly

and its second

_half

is the reverse

_of

that

_{first half}

-omitting

the zero-th

_partial

_quotient,

and is twisted

_{by multiplication by}

_3.]

Proof. Having

been

_{signalled by}

₍₃

+

g2)Qh

=

Qh-1

₊

2gPh

_we have

and thus the

_{representation}

But then we have seen

_explicitly

that

yielding

the claims.

Another

_example.

It _{is easy} to see that

34798636~ -1891’ 800233~

= -3

where

_4Q6

= 4 . 25 = 58+2.21 =

Q5 + 2P6,

and also

Q12

= 49 =

15 + 34 =

Q13

₊

_P13.

We should also notice that it

happens

_to

happen

that

4Q6=4~25=42+2~29=Q7+2P7.

We have that

Our

_arguments

_pretend

to show that the

_expansion

_{[43 ,}

_{2 , 16 ,}

_{1 ,}

_{8 ,}

_1]

continues with the

_partial

_quotient

1 and then with the

_{partial quotients}

corresponding

to

According

to our

description

we should have

This is indeed the case. The

_displayed

detail of this

_{decomposition}

is

and

The continued fraction

_expansion

_following

the

_{signal Q12}

= 49 = 15+34 =

Q13

+

P13

should

_correspond

to

Indeed,

in accordance with

_prediction

and

_showing

that the

_present

_{identity signals}

the other solution to

X2 _ 189ly2 -

-3. Indeed our first

signal

arose

from the

_{representation}

1891 =

4 2+3.25 2

=

212 - 21.50 + 502

whilst this

second

_signal

arises from the

_{representation}

1891 =

342

+ 34 . 15 + 152.

It

is not

_unamusing

to

_compound

these two

_{representations obtaining}

both

D2

=

2792 -

279 ~ 2015 ₊

20152

and

D 2

=

14642 -

1464 . 2135 ₊

21352 .

It follows that 1891 = 31 .61. That’s clear because the

greatest

common

divisor of 279 and 2015 is

_31,

y and the

greatest

common divisor of 1464

and 2135 is 61.

The

_{signal 4Q6}

=

Q7

₊

2P7

is in fact

conjugate

to a

_{signal happening}

to coincide in

_position

with the

_signal

_conjugate

to

_4Q6

=

Q5

₊

2P6 .

We

shall see in the next section that it of course leads

_naturally

to the first

solution,

by exactly

computations

already provided.

The

_conjugate

cases. We now deal with the formulas

_appropriate

to the

conjugate

cases. To obtain these cases we consider the ideal

(3Qh,

v75+L)

composed sequentially

with

_{(3Qh, ~ - L)}

and then with

_(3Qh,

_{L) .}

The

_compositions

_{correspond sequentially}

to a matrix

_product

The last

_pair

of matrices _may be

_{’simplified’}

to

_yield

Setting

L =

gQh,

these matrices detail

_respectively

the

_following

equivalent

configurations

in a notional continued fraction

_expansion:

, ,,--- L n v · . , --,-" ---..

,

-expansion

just

reports

that

_{Qh+l = 3Qh .}

Of course we have recalled that

- - - .. 1""1. - --1’"B - n...- -- _ -.

The second

_composition

now

yields

corresponding

to some ideal

(3,

B/D

+

_{p) .}

Moreover,

the continued

_{fraction expansion of}

_XIY

is twisted

_symmetric.

Its

_first

_half

is the

_expansion

_of

_xly

and its second

_half

is the reverse

of

that

first half

-

omitting

the zero-th

_partial

_quotient,

and is twisted

_by

division

by

3.

Proof.

We have seen

explicitly

that

yielding

the claims.

This is all _very well if

_x/y

is a

_convergent,

but it is not if ₉ = 1.

However,

above,

we saw the continued fraction

configuration

so

Hence we have the alternative

Moreover,

the continued

_fraction

_expansion

_of

_XIY

is twisted

_symmetric.

Its

_{first half}

is the

_expansion

_of

_{followed by}

the

_partial

_quotient

g and its second

half

is the reverse

of

that

first half

-

omitting

the zero-th

_partial

_quotient,

and is twisted

_by

division

_by

3.

Moreover,

the continued

_fraction

_expansion

_of

_XIY

is twisted

_symmetric.

Its

_{first half}

is the

_expansion

_of

_xhlYh

and its second

_half

is the reverse

_of

that

_{first half}

-

omitting

the zero-th

_partial

_quotient,

and is twisted

_by

the linear

_{fractional transformation}

(3:2 7 ) .

(

-g 1

Yet a further

example.

It’s clear that

13522342 -

5719

178812

= -3

and

5453976942 -

5719 .

72119592

= -3. Here

where

_Q7

= 27 = 3 - 9 =

3Q6

and

Q9

= 75 = 3 . 25

= 3Qio .

These

signals belong respectively

to the

_{representations}

5719 =

74 2+3 _ 92

=

652

+ 65 .18 + 18~

and 5719 =

622

+ 3.252

=

372

+ 37.50 ₊

502 .

One _notes

that

_{Q13 = Q17 = 3.}

This and the next

_example

have been ’concised’ on the advice of the

referee. The reader can

readily

_compute

the relevant table

[a

_spreadsheet

program is

optimal].

The first

_signal

is a

_conjugate

case. We have that

Our

_arguments

_suggest

that the

_expansion

_{[75 ,}

_{1 , 1 , 1 , 1 , 1 ,}

_16]

contin-ues with the

partial quotients

corresponding

to

This is indeed the case. We also note

that,

_again

as

_suggested,

In

_respect

of the second

_signal

we have that

Our claim is that the

_expansion

_{~75 ,}

_{1 , 1 , 1 , 1 , 1 , 16 , 5 , 1 ,}

_1]

continues with the

_partial

_quotients

_{corresponding}

to

We also note

_{that just}

as

suggested,

And

yet

another

example.

It takes one

only

a few moments to confirm that

1699118998912 -

9139

17773561342

= -3 and that

similarly

_we have

541447255636762 -

9139 -

5663785771692

= -3. Here

One may see the

_{signal 4Q16}

= 4.49 = 10 ₊ 2 - 93 =

Q17

₊

2Pl7

_arising

from the

_{representation}

D =

442

₊ 3 .

492

=

52 -

5 ~ 98 ₊

982 .

There is

also

_Q13

= 107 = 30 ₊ 77 =

Q14

₊

P14

_arising

from the

_{representation}

D =

_{302 + 30 . 77 + 772 .}

will have the

_{partial quotient}

2 = 3 - 1

_appended.

In matrix terms that

yields

Now

_dividing

the reversed continued fraction

_by

3

_supplies

the matrix

It will be more instructive to

_display

the _sequence of calculations

_providing

the

_{decomposition}

of this

matrix,

rather than to

_allege

the result. One has

in detail

Recall that the first two columns are the _sequence of 2 x 2 matrices obtained

as we

split

off each

partial

_quotient

matrix

by employing

the euclidean

algorithm

on the rows of the matrices. now

_properly

seen to be

the

_quotients

of that euclidean

_algorithm,

are

displayed

in the third column.

The last 2 x 2 matrix

_displays

_P25

= _p = 95

and,

of _course,

Q25

= 3.

Evidently

all is

_well,

with truth

_prevailing.

We were

halfway

to the

verifying

our formulas.

According

to the

_signal

( the continued

fraction

_expansion

continues with the

_expansion

_{corresponding}

to

I I I

The

_{decomposition}

details are

U is

precisely

as

predicted.

The last 2 x 2 matrix

displays

P23 =

_p = 94

and,

of _course,

_Q23

= 3.

We also confirm that

5.

_Signals

in

_{Arbitrary Cycles}

Our

_preceding

discussion shows that if there is a solution to = -3

then D has

_{representations}

D =

M2

+MN+N 2

and there are

signals

That

_is,

there are reduced elements

(~lD +

Ph)lQh,

or +

yielding

the

_signal,

with

_Qh

= N _or

2Qh

= N

according

_as N is odd _or

even,

occurring

in some continued fraction

cycle.

A

_priori,

_signals

_may occur in any

cycle

whatsoever. Our

_arguments

only

entail that if a

signal

occurs in some ideal

class,

or

cycle

- thus if there is

a

reduced element

_{corresponding}

to an ideal

(Q h ,

in

that

_{cycle, respectively}

a reduced element

corresponding

to an ideal

(Qh,

B/D

+

_Ph+1)

in that

_cycle

if we are

_referring

to a

conju-gate

signal

from the second line in the list above -

then there is an ideal

(3,

B/D

+

_p)

in the ideal class that is the square of the

signalling

class.

However,

if there is a solution to

X 2 - Dy2

= -3

then,

in

particular,

the ideal

_(3,

_B/D

+

_p)

must lie in the

_principal

class.

_Hence,

if there is a

solution,

the

_signals

must occur in

ambiguous classes,

for those are

precisely

the classes whose _{square is}

_principal.

_Conversely,

if there is a

_signal

in some

ambiguous

class then there is a

_principal

ideal

(3,

thence a solution

to

X2 -

Dy2

=

-3,

and all

signals

must occur in

_ambiguous

classes.

We shall address the

_question

whether there are

signals

in the

principal

class, given

that there is a solution to

X2 - Dy2

= -3. We also

briefly

explain

how a

signal

in an

arbitrary

cycle containing

an

_ambiguous

ideal

permits

one to construct a solution to

Dy2 -

-3.

Counting ambiguous

classes and

_{representations.}

We will be

_dealing

with

_squarefree

D divisible

_by

_primes

_congruent

to 1

_{(mod 3),}

and

_possibly

by

3. If D is divisible

_{by exactly v}

different

_primes

_congruent

to 1

_{(mod 3)}

then there are

_exactly

2"

_ambiguous

_ideals,

_respectively

2"+1

_ambiguous

ideals,

in the order

_{Z[~/D] 2013}

_according

as

3~

D or 3

I

D. Half of these

ambiguous

ideals have odd norm S

dividing

D and half have even norm

2S

_dividing

2D. These matters are well

known;

there is a fresh discussion

in

_[2].

We have to

_distinguish

the cases D - 1

(mod 4)

and D - -1

(mod 4).

In the latter case we have

_nothing

to add at this

_point.

In the former case we note that the maximal order of is in fact

_7G~(~

+

_1)/2].

We