Binary

#20-sep

Quadratic Forms

Definition (Binary Quadratic Form).

A binary quadratic form (bqf) is a polynomial of the form
with
we may think of it as a function

In short we write

Definition (Primitive QBF).

We say that is a primitive if

Definition (Represents and Primitively Represents ).

We say that represents the integer for some
We it represent primitively or properly if and

Remark.

The same number can be represented primitively and non primitively

Definition (Disc of Binary Quadratic Form).

for we define

Proposition.

Let be a Binary Quadratic Form Binary Quadratic Form with
i) then represent 0 is a square
ii) all non zero values represented by are of the same sign

Proof.
Case I: then

also

this proves i) for this case
for ii)
then
This shows that all values of has the same sign of

Case II: then,

Clearly, is represented when

such that

which implies is a square.
for ii) if then clearly
so is of constant sign( same as ) from ^6d9892 (3.1)
conversely suppose is of constant sign then

and

are of the same sign (by hypothesis), so

Proposition.

let be a BQF BQFwith .
Then for a given there are at most a finite number of such that

Proof.
say then

which show has finitely many values
also

this show that for any value of there are finitely many values of there are finitely many values of

□

Remark.

This is false for example

Say then

#23-sep

Equivalence

set of all BQF BQF with integer coefficients
set of all Matrices with integer entries
set of all Invertable Matrices with integer entries note the determinant here is 1 or -1
set of all Matrices with integer entries and determinant
Note: and are both groups and is the sub group of

let and

Define the new BQF BQF by

This is just a linear change of variable

Proposition.

Let then

Proof.
i)

ii)

□

Definition.

let ,
we say that is equivalent to if st
we say is properly equivalent to if st
Notation
equivalent
properly equivalent

Mat of BQF

Definition (mat of Binary Quadratic Form).

Exercise.

what is

Hint:

let

Hence

Theorem.

let
i) then
ii) and represents the same set of integers the same number of times
iii) and are either both primitive or both non-primitive

Proof.
i) say with

ii) say
then

iii)

then divides
since the set of the values of and are he same it follows that divides
if is primitive then is the maximum value of which implies is primitive
if is not primitive then which implies is not primitive

Exercise (

and

are equivalence relations on

Theorem.

let and then represents properly iff

form some
moreover if then such that

Proof.
suppose
say,

then

since primitively represent n
Now the other way
assume properly represents say,

we need to show that

Now we can find such that bezout's theorem bezout's theorem
let

Note that

for some this proves the first part

let we can write

with
let where

Remark (principle forms of disc ).

can be and they can be simply represented by
if
if
these are known as principal forms of disc

Lagrange

Theorem (Lagrange).

Let . The following are equivalent

is a square
there is a BQF BQF of discriminant which properly represents

1 2
say for some
let then so properly represents
2 1
let be a BQF BQF of disc which properly represents then by ^6cf17c Theorem 3.15

for some ^216a99 Theorem 3.13
since
we get

Hence is a square mod

#26-sep

Corollary.

let let be a prime such that Then there is a BQF BQF of the disc which properly represents . This form is unique up to equivalence

Proof.
from Lagrange Lagrange
we know of disc properly represents iff then

So represents
By Lagrange, there is a bqf which properly represents . Then we know that

for some with Now converts
so we may chose thus

with so let be another form of disc which properly represents .
As above we may take

with
we need to prove that
since have the same discriminant we have

why cant 4 divide one of it and the other be odd? for both and should have the same parity or the congruence to 4 wont hold

if then we are done and
so suppose but

so in any case
it follows that

Binary Quadratic Forms with given Discriminant

An integer is called a discriminant if it is the discriminant of some BQF
recall that is a discriminant

Definition (Fundamental Discriminant).

We say that a discriminant is a Fundamental Discriminant if every bqf of discriminant is primitive

Example

is a fundamental discriminant for if is of discriminant and if then which implies .
is a fundamental discriminant.
is a fundamental discriminant.
as if then can be if is .
has a disc of The second equation is also a discriminant hence it cant be equal to .

Proposition.

let be a discriminant. Then is a fundamental discriminant if
i) and is square free
ii) and is square free and

Proof.
Let be a fundamental discriminant

Case I:
we claim that is square free suppose

with square free if then

note that:
but then the above bqf is not primitive hence should be square free
Case II:

If
is not primitive so can be congruent to 4

suppose

is a bqf which is not primitive

then we are left with
Finally we prove that is square free

#27-sep

Exercise.

What are all the bqf of disc 0?

Suppose has disc 0, say

case I:
then
hence
case II:
then

there fore

for some again

suppose

now let

is (NO!)
as
and
hence if and are complete different.
hence there are infinitely many bqf of discriminant 0 which are in equivalent

Proposition.

Let with . Then up to proper equivalence there are only finitely many distinct bqf of discriminant

Proof.
Let be of hence properly represents and hence

write with then

so there are at most distinct form up to proper equivalence

□

Lemma (Lagrange Reduction).

Let be a discriminant and not a square
Then, is properly equivalent to a form (a, ,c) with
also, such a form satisfying

Proof.

Let,

Since is not a square, let be such that .
Since represents a, we have from ^6cf17c Theorem 3.15.

□

Theorem (Lagrange).

Let be a discriminant and not a square then up to proper equivalence only finite many distinct bqf of discriminant D

Proof.
we know
with

hence there are only finitely many possibilities for and hence for as and since the possibilities for is also finite
□

Class Number

Definition.

let be a discriminant
set of proper equivalence classes of bqf of
set of proper equivalence classes of primitive bqf of disc
we have this is know as the Class Number of

Example

What is
Since we assume
then note that
than
since

thus the only possibility is hence .
Also then

since we have then

also recall that if primitively represents p
. , if , then there exist such that
Now,
Then every prime can be written as a sum of square.

Theorem.

let then
Gauss conjecture that for any other

Conjecture (UNSOLVED).

Are there infinitely many ..

Proof.
□

#30-sep

Remark.

set of proper equivalence classes of primitive bqf of discriminant with convetion that bqf are assumed to be positive if , . for with we always assume

Gaussian Reduction

From now on we shall assume that our bqf are of negative discriminant with positive values. That is with

Definition (Gaussian Reduction).

A bqf with is said to be Reduced if either
or

Lemma.

Let be reduced bqf of discriminant , then is the smallest value of
i) There are exactly 2 primitive vectors . . namely and exactly two primitive vectors . . namely unless , in which case there are four vector's . . namely
ii) if then has exactly four primitive solution namely unless in which case there are six solutions namely

Proof.

Remark.

if is primitive iff

Let then for any

Now at say first
then
now
Hence

similarly for
now
indeed in all other cases its grater than
Then unless ( ) or ( and )
or
unless or( and )

□

Theorem.

Let

be reduced bqf of the same discriminant with then

Proof.

Since minimum values of are equal hence say
Case I
in this case has exactly 2 solutions namely
hence also has exactly 2 solutions namely
as if then has >3 solutions
Also

by uniqueness we get
hence
then

So
Thus
since are reduced we have and

Adding the above 2 equations

thus

Case II

In this case so as is reduced
Subcase A:
In this case
has exactly four primitive solutions,
then will also have exactly four solution hence
then
since

we get so since both are non negative

Subcase :
thus
hence
has six primitive solution so also has six primitive solution thus thus

□

Proposition.

Let be a reduced bqf of disc if gcd( ,y)=1 and
then

Proof.
we know that for hence

if ,
when we can have or else

if and

if and
if and

if and
hence for all values of we have verified that can only be between , iff

□

Theorem.

Let be a bqf of discriminant then there is a reduced bqf such that

Proof. 1
by Lagrange reduction lemma there is a form such that and since by convention all values are positive we have
now if then we are done
case I but b<0
in this case

case II a< but
in this case

□

Proof. 2 (Constructive )

Let
if is reduced nothing to do if not either but

Case I
in this case

Case II: but
Sub-case A:
then say

sub-case :
in this case

here again
Sub-case :
here
in this case the process completes
In general continues this way we get a sequence

as the sum is reducing this will complete at some point the last step has the reduced form

Example

it is not reduced
lets reduce the b

#4-oct

Terminology

A bqf is said to be

Definition (Positive semidefinite ).

Definition (Positive Definite ).

if
and

Definition (Negative semidefinite ).

Definition (Negative Definite ).

if
and