representation of integers by equivalent integral binary quadratic forms

representation of integers by equivalent integral binary quadratic forms

Theorem 1

.

If

F

,

G

are

equivalent

integral binary quadratic forms

, then

F

and

G

represent the same set of integers.

Proof.

Write

G

⁢

(

x

,

y

)

=

F

⁢

(

α

⁢

x

+

β

⁢

y

,

γ

⁢

x

+

δ

⁢

y

)

where

det

⁡

(

α

γ

β

δ

)

=

±

1

Then

m

=

G

⁢

(

r

,

s

)

⇒

m

=

F

⁢

(

α

⁢

r

+

β

⁢

s

,

γ

⁢

r

+

δ

⁢

s

)

, so if

G

represents

m

, so does

F

. Since the matrix has

determinant

1, it is

invertible

and its

inverse

is another

integer matrix

, so the reverse statement follows as well.
∎

Lemma 2

.

F

properly represents an

integer

m

if and only if

F

is properly equivalent to a form

m

⁢

x

2

+

B

⁢

x

⁢

y

+

C

⁢

y

2

.

Proof.

⇐

: It is obvious by the above that

F

represents

m

; the problem is to show that it represents

m

properly. Write

G

⁢

(

x

,

y

)

=

m

⁢

x

2

+

B

⁢

x

⁢

y

+

C

⁢

y

2

; then

G

⁢

(

x

,

y

)

=

F

⁢

(

α

⁢

x

+

β

⁢

y

,

γ

⁢

x

+

δ

⁢

y

)

, where

α

⁢

δ

-

β

⁢

γ

=

1

. Then

m

=

G

⁢

(

1

,

0

)

=

F

⁢

(

α

,

γ

)

. But clearly

(

α

,

γ

)

=

1

since otherwise we cannot have

α

⁢

δ

-

β

⁢

γ

=

1

. So

F

represents

m

properly.

⇒

: Write

F

⁢

(

p

,

q

)

=

m

, where

(

p

,

q

)

=

1

. Since

(

p

,

q

)

=

1

, we can find integers

r

,

s

such that

p

⁢

s

-

q

⁢

r

=

1

, and then

F

⁢

(

p

⁢

x

+

r

⁢

y

,

q

⁢

x

+

s

⁢

y

)

=

a

⁢

(

p

⁢

x

+

r

⁢

y

)

2

+

b

⁢

(

p

⁢

x

+

r

⁢

y

)

⁢

(

q

⁢

x

+

s

⁢

y

)

+

c

⁢

(

q

⁢

x

+

s

⁢

y

)

2

=

(

a

⁢

p

2

+

b

⁢

p

⁢

q

+

c

⁢

q

2

)

⁢

x

2

+

(

2

⁢

a

⁢

p

⁢

r

+

b

⁢

p

⁢

s

+

b

⁢

q

⁢

r

+

2

⁢

c

⁢

q

⁢

s

)

⁢

x

⁢

y

+

(

a

⁢

r

2

+

b

⁢

r

⁢

s

+

c

⁢

s

2

)

⁢

y

2

=

F

⁢

(

p

,

q

)

⁢

x

2

+

(

2

⁢

a

⁢

p

⁢

r

+

b

⁢

p

⁢

s

+

b

⁢

q

⁢

r

+

2

⁢

c

⁢

q

⁢

s

)

⁢

x

⁢

y

+

F

⁢

(

r

,

s

)

⁢

y

2

=

m

⁢

x

2

+

B

⁢

x

⁢

y

+

C

⁢

y

2

∎

Definition 1

.

If

F

is a

binary quadratic form

, its

discriminant

,

Δ

⁢

(

F

)

is

b

2

-

4

⁢

a

⁢

c

.

Note that

Δ

⁢

(

F

)

is always either

congruent

to

0

or

1

mod 4, and that

b

is even (odd) exactly when

Δ

⁢

(

F

)

≡

0

⁢

(

1

)

(

mod

4

)

.

Theorem 3

.

If

F

,

G

are equivalent integral

quadratic forms

, then

Δ

⁢

(

F

)

=

Δ

⁢

(

G

)

.

Proof.

For any form

F

, define

M

F

=

(

2

⁢

a

b

b

2

⁢

c

)

Then

2

⁢

F

⁢

(

x

,

y

)

=

(

x

⁢

⁢

y

)

⁢

M

F

⁢

(

x

y

)

Note further that

Δ

⁢

(

F

)

=

-

det

⁡

(

M

F

)

.

Now in our particular case, if

G

⁢

(

x

,

y

)

=

F

⁢

(

α

⁢

x

+

β

⁢

y

,

γ

⁢

x

+

δ

⁢

y

)

, then

2

⁢

G

⁢

(

x

,

y

)

=

(

α

⁢

x

+

β

⁢

y

⁢

⁢

γ

⁢

x

+

δ

⁢

y

)

⁢

M

F

⁢

(

α

⁢

x

+

β

⁢

y

γ

⁢

x

+

δ

⁢

y

)

=

(

x

⁢

⁢

y

)

⁢

(

α

γ

β

δ

)

⁢

M

F

⁢

(

α

β

γ

δ

)

⁢

(

x

y

)

Hence

M

G

=

(

α

γ

β

δ

)

⁢

M

F

⁢

(

α

β

γ

δ

)

But

Δ

⁢

(

F

)

=

-

det

⁡

(

M

F

)

, so since

det

⁡

(

α

γ

β

δ

)

=

det

⁡

(

α

β

γ

δ

)

=

±

1

,

Δ

⁢

(

G

)

=

-

det

⁡

(

M

G

)

=

-

det

⁡

(

α

γ

β

δ

)

⁢

det

⁡

(

M

F

)

⁢

det

⁡

(

α

β

γ

δ

)

=

-

det

⁡

(

M

F

)

=

Δ

⁢

(

F

)

∎

Note that this proof shows that applying a set of

transformations

amount to multiplying by the transform matrix on the left and its

transpose

on the right.

Example

: In the previous example, note that

Δ

⁢

(

F

)

=

1

-

4

⋅

1

⋅

6

=

-

23

, and

Δ

⁢

(

G

)

=

51

2

-

4

⋅

82

⋅

8

=

2601

-

2624

=

-

23

.

The

converse

of this

theorem

is not true - that is, there are forms of the same discriminant that represent different numbers. For example,

x

2

+

5

⁢

y

2

and

2

⁢

x

2

+

2

⁢

x

⁢

y

+

3

⁢

y

2

both have discriminant

-

20

, yet the second form represents

2

while the first clearly does not. However,

equivalence classes

of forms under arbitrary (proper or improper) equivalence represent

disjoint sets

of primes:

Theorem 4

.

Let

p

be an

odd prime

. Suppose

F

,

G

both represent

p

and

Δ

⁢

(

F

)

=

Δ

⁢

(

G

)

. Then

F

and

G

are equivalent (but perhaps not properly equivalent).

Proof.

Since

p

is prime,

F

obviously represents

p

properly. So

F

∼

p

⁢

x

2

+

b

⁢

x

⁢

y

+

c

⁢

y

2

. Note that the transformation

(

x

,

y

)

↦

(

x

+

d

⁢

y

,

y

)

results in a form whose middle term is

2

⁢

p

⁢

d

+

b

, so by an appropriate choice of

d

we can arrange that

-

p

<

b

≤

p

. Similarly,

G

∼

p

⁢

x

2

+

b

′

⁢

x

⁢

y

+

c

′

⁢

y

2

with

-

p

<

b

′

≤

p

. Note also that since

b

2

-

4

⁢

p

⁢

c

=

b

′

⁣

2

-

4

⁢

p

⁢

c

′

, it follows that

b

≡

b

′

(

2

)

(i.e.

b

,

b

′

have the same parity).

Since

Δ

⁢

(

F

)

=

Δ

⁢

(

G

)

, we see that

b

2

-

4

⁢

p

⁢

c

=

b

′

⁣

2

-

4

⁢

p

⁢

c

′

⇔

b

2

≡

b

′

⁣

2

(

p

)

⇔

b

≡

±

b

′

(

p

)

, so

b

=

±

b

′

+

k

⁢

p

for some

k

. Since

b

,

b

′

have the same parity and

p

is odd,

k

is even; since

-

p

<

b

,

b

′

≤

p

,

k

=

0

(since otherwise

b

,

b

′

would be

separated

by at least

2

⁢

p

, which is impossible).

We are left with two cases. If

b

=

b

′

, then

Δ

⁢

(

F

)

=

Δ

⁢

(

G

)

implies that

c

=

c

′

and hence

F

∼

G

. If

b

=

-

b

′

, then again

Δ

⁢

(

F

)

=

Δ

⁢

(

G

)

implies that

c

=

c

′

. Then

F

and

G

are equivalent via the transformation

(

x

,

y

)

↦

(

x

,

-

y

)

.
∎

Note that

F

⁢

(

x

,

y

)

=

a

⁢

x

2

+

b

⁢

x

⁢

y

+

c

⁢

y

2

and

G

⁢

(

x

,

y

)

=

a

⁢

x

2

-

b

⁢

x

⁢

y

+

c

⁢

y

2

are always improperly equivalent via the transformation

(

x

,

y

)

↦

(

x

,

-

y

)

. They are sometimes properly equivalent, and sometimes not. For example,

2

⁢

x

2

+

2

⁢

x

⁢

y

+

3

⁢

y

2

and

2

⁢

x

2

-

2

⁢

x

⁢

y

+

3

⁢

y

2

are properly equivalent while

3

⁢

x

2

+

2

⁢

x

⁢

y

+

5

⁢

y

2

and

3

⁢

x

2

-

2

⁢

x

⁢

y

+

5

⁢

y

2

are not. (See the article on

reduced integral binary quadratic forms

for details).

In summary, we have proved the following:

F

,

G

⁢

equivalent

⇒

F

,

G

⁢

represent the same set of integers

F

,

G

⁢

equivalent

⇒

Δ

⁢

(

F

)

=

Δ

⁢

(

G

)

Δ

⁢

(

F

)

=

Δ

⁢

(

G

)

⁢

and

⁢

F

,

G

⁢

both represent some odd prime

⁢

p

⇒

F

⁢

and

⁢

G

⁢

are equivalent

We conclude with the following lemma and corollary, which provide concrete criteria for when an integer is representable by a class of forms.

Lemma 5

.

If

D

≡

0

,

1

(

4

)

is an integer, and

m

is an

odd integer

relatively prime to

D

, then

m

is properly represented by a primitive form of discriminant

D

if and only if

D

is a

quadratic residue

mod

m

.

Proof.

If

F

⁢

(

x

,

y

)

properly represents

m

, then by the preceding lemma, we may assume that

F

⁢

(

x

,

y

)

=

m

⁢

x

2

+

b

⁢

x

⁢

y

+

c

⁢

y

2

. Then

D

=

b

2

-

4

⁢

m

⁢

c

, being the discriminant of

F

, so that

D

≡

b

2

(

D

)

. Conversely, if

D

≡

b

2

(

D

)

, we may assume

D

≡

b

(

2

)

(if they have different parities, replace

b

by

b

+

m

; since

m

is odd, the condition now holds and

D

≡

(

b

+

m

)

2

(

D

)

as well). Since

D

≡

0

,

1

(

4

)

, it follows that

D

≡

b

2

(

4

)

and thus

D

≡

b

2

(

4

⁢

m

)

. Hence

D

=

b

2

-

4

⁢

m

⁢

c

for some integer

c

. But then

m

⁢

x

2

+

b

⁢

x

⁢

y

+

c

⁢

y

2

represents

m

and has discriminant

D

; it is primitive since

gcd

⁡

(

m

,

b

)

=

gcd

⁡

(

m

,

D

)

=

1

.
∎

Corollary 6

.

Let

n

be an integer, and

p

an odd prime not dividing

n

. Then

(

-

n

p

)

=

1

if and only if

p

is represented by a primitive form of discriminant

-

4

⁢

n

.

Proof.

By the preceding lemma,

p

is represented by a primitive form of discriminant

-

4

⁢

n

if and only if

1

=

(

-

4

⁢

n

p

)

=

(

-

n

p

)

∎

Title

representation of integers by equivalent integral binary quadratic forms

Canonical name

RepresentationOfIntegersByEquivalentIntegralBinaryQuadraticForms

Date of creation

2013-03-22 19:18:48

Last modified on

2013-03-22 19:18:48

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

7

Author

rm50 (10146)

Entry type

Topic

Classification

msc 11E12

Classification

msc 11E16

Related topic

integralbinaryquadraticforms

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