reduced integral binary quadratic forms

reduced integral binary quadratic forms

Definition 1

.

A

positive

binary form is one where

F

⁢

(

x

,

y

)

≥

0

⁢

∀

x

,

y

∈

Z

.

This article deals only with positive

integral binary quadratic forms

(i.e. those with negative

discriminant

and with

a

>

0

). Some but not all of this theory applies to forms with positive discriminant.

Proposition 1

.

If

F

is positive, then

a

>

0

. If

Δ

⁢

(

F

)

<

0

and either

a

>

0

or

c

>

0

, then

F

is positive.

Proof.

If

F

is positive, then

F

⁢

(

1

,

0

)

=

a

, so

a

>

0

.

If

Δ

⁢

(

F

)

<

0

, then

4

⁢

a

⁢

F

⁢

(

x

,

y

)

=

(

2

⁢

a

⁢

x

+

b

⁢

y

)

2

-

Δ

⁢

y

2

≥

0

. Thus if

a

>

0

, then

F

⁢

(

x

,

y

)

≥

0

. The proof for the case

c

>

0

is identical.
∎

Definition 2

.

A

primitive

positive form

a

⁢

x

2

+

b

⁢

x

⁢

y

+

c

⁢

y

2

is

reduced

if

|

b

|

≤

a

≤

c

⁢

, and

⁢

b

≥

0

⁢

if either

⁢

|

b

|

=

a

⁢

or

⁢

a

=

c

This is

equivalent

to saying that

-

a

≤

b

≤

a

≤

c

and that

b

can be negative only if

a

<

c

or

-

a

<

b

. Thus

(

3

,

-

2

,

4

)

is reduced, but

(

3

,

-

2

,

3

)

is not.

It turns out that each proper

equivalence class

of primitive positive forms contains a single

reduced form

, and thus we can understand how many classes there are of a given discriminant by studying only the reduced forms.

Theorem 2

.

If

F

is a primitive positive reduced form with discriminant

Δ

≠

-

3

,

F

⁢

(

x

,

y

)

=

a

⁢

x

2

+

b

⁢

x

⁢

y

+

c

⁢

y

2

, then the minimum value assumed by

F

if

x

,

y

are not both zero is

a

. If

a

<

c

, then this value is assumed only for

(

x

,

y

)

=

(

±

1

,

0

)

; if

a

=

c

; it is assumed for

(

x

,

y

)

=

(

±

1

,

0

)

and

(

x

,

y

)

=

(

0

,

±

1

)

.

Proof.

Since

|

b

|

≤

a

≤

c

, it follows that

F

⁢

(

x

,

y

)

≥

(

a

-

|

b

|

+

c

)

⁢

min

⁡

(

x

2

,

y

2

)

Thus,

F

⁢

(

x

,

y

)

≥

a

-

|

b

|

+

c

whenever

x

⁢

y

≠

0

, while if

x

or

y

is zero, then

F

⁢

(

x

,

y

)

≥

a

. So

a

is clearly the smallest nonzero value of

F

.

If

a

<

c

, then

F

⁢

(

x

,

y

)

≥

a

+

(

c

-

|

b

|

)

>

a

if

x

⁢

y

≠

0

, and

F

⁢

(

0

,

y

)

≥

c

>

a

for

y

≠

0

, so

F

achieves its minimum only at

(

x

,

y

)

=

(

±

1

,

0

)

.

If

a

=

c

, then

|

b

|

≠

a

since otherwise

F

is not reduced (we cannot have

a

=

c

=

1

,

b

=

±

1

else we have a form of discriminant

-

3

). Thus again

c

-

|

b

|

>

0

and thus

F

⁢

(

x

,

y

)

>

a

if

x

⁢

y

≠

0

, so in this case the result follows as well.
∎

Note that the reduced form of discriminant

-

3

,

x

2

+

x

⁢

y

+

y

2

, also achieves its minimum value at

(

1

,

-

1

)

,

(

-

1

,

1

)

.

Theorem 3

.

If

F

,

G

are primitive positive reduced forms with

F

∼

G

, then

F

=

G

.

Proof.

We take the cases

Δ

≠

-

3

and

Δ

=

-

3

separately.

First assume

Δ

≠

-

3

so we can apply the above

theorem

.

Since

F

∼

G

, we can write

G

⁢

(

x

,

y

)

=

F

⁢

(

α

⁢

x

+

β

⁢

y

,

γ

⁢

x

+

δ

⁢

y

)

with

α

⁢

δ

-

β

⁢

γ

=

1

. Suppose

F

=

a

⁢

x

2

+

b

⁢

x

⁢

y

+

c

⁢

y

2

,

G

=

a

′

⁢

x

2

+

b

′

⁢

x

⁢

y

+

c

′

⁢

y

2

. Now,

F

and

G

have the same minimum value, so

a

=

a

′

.

If

a

<

c

, then

F

achieves its minimum only at

(

p

⁢

m

⁢

1

,

0

)

, so

a

=

a

′

=

G

⁢

(

1

,

0

)

=

F

⁢

(

α

,

γ

)

and thus

α

=

±

1

,

γ

=

0

. So

G

⁢

(

x

,

y

)

=

F

⁢

(

±

x

+

r

⁢

y

,

±

y

)

and thus

b

′

=

b

+

2

⁢

r

⁢

a

. Since

G

is also reduced,

b

=

b

′

and thus

c

=

c

′

and

F

=

G

.

If instead

a

=

c

, then instead of concluding that

α

=

±

1

we can only conclude that

α

=

±

1

or

γ

=

±

1

. If

α

=

±

1

, the

argument

carries through as above. If

γ

=

±

1

, then

α

=

0

,

β

=

∓

1

, so

G

⁢

(

x

,

y

)

=

F

⁢

(

∓

y

,

±

x

+

r

⁢

y

)

and thus

b

′

=

±

2

⁢

c

⁢

r

-

b

. Thus

b

′

=

-

b

. But then

c

=

c

′

since the discriminants are equal, and thus both

b

,

b

′

≥

0

. So

b

=

b

′

=

0

and we are done.

Finally, in the case

Δ

=

-

3

, we see that for any such reduced form,

3

=

4

⁢

a

⁢

c

-

b

2

≥

4

⁢

a

2

-

a

2

=

3

⁢

a

2

, so

a

=

1

,

b

=

±

1

,

c

=

1

. Thus

b

=

1

since otherwise the form is not reduced. So the only reduced form of discriminant

-

3

is in fact

x

2

+

x

⁢

y

+

y

2

.
∎

Theorem 4

.

Every primitive positive form is properly equivalent to a unique reduced form.

Proof.

We just proved uniqueness, so we must show

existence

. Note that I used a different method of proof in class that relied on “

infinite descent

” to get the result in the first paragraph below; the method here is just as good but provides less insight into how to actually reduce a form.

We first show that any such form is properly equivalent to some form satisfying

|

b

|

≤

a

≤

c

. Among all forms properly equivalent to the given one, choose

F

⁢

(

x

,

y

)

=

a

⁢

x

2

+

b

⁢

x

⁢

y

+

c

⁢

y

2

such that

|

b

|

is as small as possible (there may be

multiple

such forms; choose one of them). If

|

b

|

>

a

, then

G

⁢

(

x

,

y

)

=

F

⁢

(

x

+

m

⁢

y

,

y

)

=

a

⁢

x

2

+

(

2

⁢

a

⁢

m

+

b

)

⁢

x

⁢

y

+

c

′

⁢

y

2

is properly equivalent to

F

, and we can choose

m

so that

|

2

⁢

a

⁢

m

+

b

|

<

|

b

|

, contradicting our choice of

minimal

|

b

|

. So

|

b

|

≤

a

; similarly,

|

b

|

≤

c

. Finally, if

a

>

c

, simply interchange

a

and

c

(by applying the proper equivalence

(

x

,

y

)

↦

(

-

y

,

x

)

) to get the required form.

To finish the proof, we show that if

F

⁢

(

x

,

y

)

=

a

⁢

x

2

+

b

⁢

x

⁢

y

+

c

⁢

y

2

, where

|

b

|

≤

a

≤

c

, then

F

is properly equivalent to a reduced form. The form is already reduced unless

b

<

0

and either

a

=

-

b

or

a

=

c

. But in these cases, the form

G

⁢

(

x

,

y

)

=

a

⁢

x

2

-

b

⁢

x

⁢

y

+

c

⁢

y

2

is reduced, so it suffices to show that

F

and

G

are properly equivalent. If

a

=

-

b

, then

(

x

,

y

)

↦

(

x

+

y

,

y

)

takes

a

⁢

x

2

-

a

⁢

x

⁢

y

+

c

⁢

y

2

to

a

⁢

x

2

+

a

⁢

x

⁢

y

+

c

⁢

y

2

, while if

a

=

c

, then

(

x

,

y

)

↦

(

-

y

,

x

)

takes

a

⁢

x

2

+

b

⁢

x

⁢

y

+

a

⁢

y

2

to

a

⁢

x

2

-

b

⁢

x

⁢

y

+

a

⁢

y

2

.
∎

Let’s see how to reduce

82

⁢

x

2

+

51

⁢

x

⁢

y

+

8

⁢

y

2

to

x

2

+

x

⁢

y

+

6

⁢

y

2

:

Form

Transformation

Result

82

⁢

x

2

+

51

⁢

x

⁢

y

+

8

⁢

y

2

(

x

,

y

)

↦

(

-

y

,

x

)

8

⁢

x

2

-

51

⁢

x

⁢

y

+

82

⁢

y

2

8

⁢

x

2

-

51

⁢

x

⁢

y

+

82

⁢

y

2

(

x

,

y

)

↦

(

x

+

3

⁢

y

,

y

)

8

⁢

(

x

+

3

⁢

y

)

2

-

51

⁢

(

x

+

3

⁢

y

)

⁢

y

+

82

⁢

y

2

=

8

⁢

x

2

+

48

⁢

x

⁢

y

+

72

⁢

y

2

-

51

⁢

x

⁢

y

-

153

⁢

y

2

+

82

⁢

y

2

=

8

⁢

x

2

-

3

⁢

x

⁢

y

+

y

2

8

⁢

x

2

-

3

⁢

x

⁢

y

+

y

2

(

x

,

y

)

↦

(

-

y

,

x

)

x

2

+

3

⁢

x

⁢

y

+

8

⁢

y

2

x

2

+

3

⁢

x

⁢

y

+

8

⁢

y

2

(

x

,

y

)

↦

(

x

-

y

,

y

)

(

x

-

y

)

2

+

3

⁢

(

x

-

y

)

⁢

y

+

8

⁢

y

2

=

x

2

+

x

⁢

y

+

6

⁢

y

2

x

2

+

x

⁢

y

+

6

⁢

y

2

Theorem 5

.

If

F

⁢

(

x

,

y

)

=

a

⁢

x

2

+

b

⁢

x

⁢

y

+

c

⁢

y

2

is a positive reduced form with

Δ

<

0

, then

a

≤

|

Δ

|

3

.

Proof.

-

Δ

=

4

⁢

a

⁢

c

-

b

2

≥

4

⁢

a

2

-

a

2

since the form is reduced. So

-

Δ

≥

3

⁢

a

2

, and the result follows.
∎

Definition 3

.

If

Δ

<

0

, then

h

⁢

(

Δ

)

is the number of classes of primitive positive forms of discriminant

Δ

.

Corollary 6

.

If

Δ

<

0

, then

h

⁢

(

Δ

)

is finite, and

h

⁢

(

Δ

)

is equal to the number of primitive positive reduced forms of discriminant

Δ

.

Proof.

Essentially obvious. Since every positive form is properly equivalent to a (unique) reduced form,

h

⁢

(

Δ

)

is clearly equal to the number of positive reduced forms of discriminant

Δ

. But given a reduced form of discriminant

Δ

, there are only finitely many choices for

a

>

0

, by the

proposition

. This constrains us to finitely many choices for

b

, since

-

a

<

b

≤

a

.

a

and

b

determine

c

since

Δ

is fixed.
∎

Examples

:

Δ

=

-

4

:

b

2

-

4

⁢

a

⁢

c

=

-

4

⇒

b

even,

|

b

|

≤

|

a

|

≤

4

3

⇒

b

=

0

. So

(

1

,

0

,

1

)

, corresponding to

x

2

+

y

2

, is the only reduced form of discriminant

-

4

. Note that this provides another proof that primes

≡

1

(

4

)

are

representable

as the

sum of two squares

, since all such primes have

(

-

4

p

)

=

(

-

1

p

)

=

1

and thus are representable by this

quadratic form

.

Δ

=

-

23

:

b

2

-

4

⁢

a

⁢

c

=

-

23

⇒

b

odd,

|

b

|

≤

a

≤

23

3

⇒

b

=

±

1

. So

a

⁢

c

=

6

,

a

<

c

. This gives us

(

1

,

1

,

6

)

(

1

,

-

1

,

6

)

not reduced since

|

b

|

=

a

,

b

<

0

; properly equivalent to

(

1

,

1

,

6

)

via

(

x

,

y

)

↦

(

x

+

y

,

y

)

(

2

,

1

,

3

)

(

2

,

-

1

,

3

)

reduced since

b

|

≠

a

,

a

≠

c

There are three equivalence classes of positive reduced forms with discriminant

-

23

.

Δ

=

-

55

:

4

⁢

a

⁢

c

-

b

2

=

55

, so

b

is odd,

|

b

|

≤

55

3

⇒

|

b

|

=

1

,

±

3

. So

a

⁢

c

=

14

⁢

or

⁢

16

,

a

≤

c

. So the forms are

(

1

,

1

,

14

)

(

1

,

-

1

,

14

)

not reduced since

|

b

|

=

a

,

b

<

0

, properly equivalent to

(

1

,

1

,

14

)

via

(

x

,

y

)

↦

(

x

+

y

,

y

)

(

2

,

1

,

7

)

(

2

,

-

1

,

7

)

reduced since

|

b

|

≠

a

,

a

≠

c

(

4

,

3

,

4

)

(

4

,

-

3

,

4

)

not reduced since

a

=

c

,

b

<

0

, equivalent to

(

4

,

3

,

4

)

via

(

x

,

y

)

↦

(

-

y

,

x

)

There are four classes of forms of discriminant

-

55

.

Δ

=

-

163

:

b

2

-

4

⁢

a

⁢

c

=

-

163

⇒

b

odd,

|

b

|

≤

|

a

|

≤

163

3

≅

55

. So

b

=

±

1

,

±

3

,

±

5

,

±

7

, and

a

⁢

c

=

b

2

+

163

4

, so

a

⁢

c

=

41

,

43

,

45

,

47

. Since

a

<

c

, we must have

a

=

1

; thus

b

=

±

1

and thus we get only

(

1

,

±

1

,

41

)

. But

(

1

,

-

1

,

41

)

is properly equivalent to

(

1

,

1

,

41

)

via

(

x

,

y

)

↦

(

x

+

y

,

y

)

, so there is only one equivalence class of positive reduced forms with discriminant

-

163

.

Title

reduced integral binary quadratic forms

Canonical name

ReducedIntegralBinaryQuadraticForms

Date of creation

2013-03-22 19:18:52

Last modified on

2013-03-22 19:18:52

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

4

Author

rm50 (10146)

Entry type

Definition

Classification

msc 11E12

Classification

msc 11E16

Related topic

integralbinaryquadraticforms

Generated on Fri Feb 9 15:33:11 2018 by

LaTeXML