Binary and Ternary Quadrati Forms

S. E. Payne

February 4, 2000

1 The Dis riminant of a Binary Quadrati

Form

Let A =

a b=2

b=2

!

and B = A + A

T

=

2a b

b 2

!

, where a; b; 2 Z.

Put

f(x; y) = (x; y)A

x

y

!

= ax

2

+ bxy + y

2

: (1)

Then f(x; y) is a binary quadrati form over Z. The dis riminant of

f(x; y) is de�ned to be

dis (f) = d = b

2

� 4a = �det(B) = �4det(A): (2)

An easy omputation shows that

4af(x; y) = (2ax + by)

2

� dy

2

: (3)

Theorem 1.1 f(x; y) fa tors over Q if and only if f(x; y) fa tors over Z if

and only if its dis riminant d is a square in Z.

Proof: First suppose that

f(x; y) = (Cx +Dy)(Ex+ Fy)

= CEx

2

+ (CF +DE)xy +DFy

2

; (4)

1

with C;D;E; F 2 Z. Then d = (CF + DE)

2

� 4CDEF = (CF � DE)

2

.

Hen e if f(x; y) fa tors over Z its dis riminant is a square.

Conversely, if d = D

2

, from Eq. 3 we see that

f(x; y) =

(2ax+ (b�D)y)(2ax+ (b+D)y)

4a

= a(x� r

1

y)(x� r

2

y); (5)

where r

i

=

�(b�D)

2a

2 Q. Here ar

1

r

2

= 2 Z. It follows that there are integers

h

1

; h

2

for whi h h

1

h

2

= a, h

1

r

1

2 Z, h

2

r

2

2 Z. Hen e

f(x; y) = (h

1

x� h

1

r

1

y)(h

2

x� h

2

r

2

y) fa tors over Z: (6)

Theorem 1.2 Suppose d is not a square, so in parti ular a 6= 0. Then

f(x; y) = 0 if and only if x = y = 0.

Proof: Here 0 = f(x; y) if and only if 0 = 4af(x; y) = (2ax + by)

2

� dy

2

if and only if dy

2

= (2ax + by)

2

. Sin e d is not a square, this last equation

an hold if and only if y = 0 and 2ax = 0, i.e., x = 0 (sin e a 6= 0).

positive de�nite The binary quadrati form f(x; y) is de�ned to be positive

de�nite provided f(x; y) > 0 whenever (x; y) 6= (0; 0).

positive semide�nite If f(x; y) � 0 for all (x; y) 2 Z � Z, we say f is

positive semide�nite.

If >; � are repla ed with <; �, respe tively, in the above de�nitions,

we obtain the de�nitions of negative de�nite and negative semide�nite,

respe tively.

inde�nite If f(x; y) yields both positive and negative values as (x; y) ranges

over Z � Z we say f is inde�nite.

2

Theorem 1.3 Let f(x; y) = ax

2

+ bxy+ y

2

be a binary quadrati form over

Z with dis riminant d. Then

d > 0 =) f is inde�nite;

(7)

d = 0 =) f is semide�nite (and NOT de�nite); (8)

(9)

d < 0 =) f is de�nite.

Proof: Case 1. d > 0. Here f(1; 0) = a, f(b;�2a) = �ad. So f(1; 0)

and f(b;�2a) have opposite signs unless a = 0. Similarly, f(0; 1) = ,

f(�2 ; b) = � d. These have opposite signs unless = 0. So suppose that

a = = 0. Then d = b

2

> 0 implies b 6= 0, so f(x; y) = bxy. Hen e

f(x; x) = b, f(1;�1) = �b, implying f is inde�nite.

Case 2. d = 0. If a 6= 0, then from Eq. 3, 4af(x; y) = (2ax + by)

2

,

implying that all values of f(x; y) have the same sign as a does, so that at

least f(x; y) is semide�nite. Sin e f(b;�2a) = �ad = 0 and �2a 6= 0, learly

f is NOT de�nite. Now suppose that a = 0. So b

2

= d = 0 and f(x; y) = y

2

.

Here f(1; 0) = 0, so f(x; y) is semide�nite and not de�nite.

Case 3. d < 0. Here 4af(x; y) = (2ax + by)

2

+ jdjy

2

> 0 ex ept when

(x; y) = (0; 0). Hen e f is de�nite and the sign of f(x; y) is the same as the

sign of a. Note that sin e d = b

2

� 4a < 0, b

2

< 4a , so that a and must

have the same sign.

Theorem 1.4 Let d 2 Z. Then there exists a binary quadrati form f with

dis riminant d if and only if d � 0 or 1 (mod 4), i.e., if and only if d is a

square modulo 4.

Proof: Suppose f has dis riminant d = b

2

� 4a � b

2

(mod 4). Then

learly d � 0 or 1 (mod 4).

For the onverse, �rst suppose that d � 0 (mod 4). Then x

2

�

�

d

4

�

y

2

has

dis riminant d. Se ond, suppose that d � 1 (mod 4). Then x

2

+xy�

�

d�1

4

�

y

2

has dis riminant d.

De�nition: We say that f represents n properly provided there are integers

(x

0

; y

0

) for whi h g d(x

0

; y

0

) = 1 and f(x

0

; y

0

) = n. More generally, if

k = g d(x

0

; y

0

) and f(x

0

; y

0

) = n, then f(

x

0

k

;

y

0

k

) =

n

k

2

.

3

Put Q(d) = ff(x; y) = ax

2

+ bxy + y

2

: d = b

2

� 4a g.

Theorem 1.5 Let n; d 2 Z, n 6= 0. Then there is an f 2 Q(d) for whi h f

represents n properly if and only if x

2

� d (mod j4nj) has a solution.

Proof: First suppose b

2

� d (mod j4nj), say b

2

�d = 4n . Then f(x; y) =

nx

2

+bxy+ y

2

has dis riminant b

2

�4n = d, and f(1; 0) = n, so f represents

n properly.

Conversely, suppose f 2 Q(d) and f(x

0

; y

0

) = n with g d(x

0

; y

0

) = 1.

Then there are integers m

1

; m

2

for whi h

m

1

m

2

= 4jnj; g d(m

1

; m

2

) = 1; g d(m

1

; y

0

) = 1; g d(m

2

; x

0

) = 1: (10)

For example,

m

1

=

Y

p

�

jj4n

pjx

0

p

�

; m

2

=

4jnj

m

1

:

Then

4an = 4af(x

0

; y

0

) = (2ax

0

+ by

0

)

2

� dy

2

0

� 0 (mod m

1

):

As g d(y

0

; m

1

) = 1, there is a �y

0

for whi h y

0

�y

0

� 1 (mod m

1

). Then

(2ax

0

+ by

0

)

2

�y

2

0

� dy

2

0

�y

2

0

� d (mod m

1

):

So u

1

= (2ax

0

+ by

0

)�y

0

solves u

2

1

� d (mod m

1

). Then

4 n = 4 f(x

0

; y

0

) = (2 y

0

+ bx

0

)

2

� dx

2

0

:

As g d(x

0

; m

2

) = 1, there is an �x

0

for whi h x

0

�x

0

� 1 (mod m

2

). Then

[(2 y

0

+ bx

0

)�x

0

℄

2

� dx

2

0

�x

2

0

� d (mod m

2

):

So u

2

= (2 y

0

+ bx

0

)�x

0

solves u

2

2

� d (mod m

2

). By the Chinese Re-

mainder Theorem we an �nd a w 2 Z for whi h w � u

1

(mod m

1

) and

w � u

2

(mod m

2

). Then w

2

� d (mod m

1

m

2

= 4jnj).

Corollary 1.6 Suppose d � 0 or 1 (mod 4) and p is an odd prime. Then

there is an f 2 Q(d) for whi h f represents p properly (why properly?) if

and only if

�

d

p

�

= +1.

4

Proof: We know f(x

0

; y

0

) = p (ne essarily properly) if and only if x

2

�

d (mod 4p) has a solution; So in parti ular, f(x

0

; y

0

) = p implies

�

d

p

�

= +1.

Conversely, suppose

�

d

p

�

= +1. Sin e d � 0 or 1 (mod 4), d must be a

square modulo 4. Hen e d is a square modulo 4p. By Theorem 1.5, f(x; y)

represents p properly.

2 TheModular Group & Equivalen e of Quadrati

Forms

There is a general theorem in linear algebra that says that if M is a square

matrix with entries from some ommutative ring R with 1, then M has a

multipli ative inverse with entries from R if and only if the determinant

det(M) (whi h is an element of R) is a unit in R, i.e., has a multipli ative

inverse in R. In the present ontext our ring R is just the ring Z of integers

with the usual addition and multipli ation of integers. Hen e we have the

following theorem.

Theorem 2.1 The matrix M =

m

11

m

12

m

21

m

22

!

has a multipli ative inverse

with entries whi h are integers if and only if det(M) = �1.

A omplete proof of the theorem in this spe ial ase of 2� 2 matri es of

integers is given in Se tion 3.5 as part of the proof of the following theorem

(Theorem 3.5 in our text).

Theorem 2.2 Let M be a 2� 2 matrix with integer entries as above. Then

the mapM : Z�Z ! Z�Z :

"

x

y

#

7!

"

u

v

#

=M

"

x

y

#

permutes the points

of Z �Z ( alled latti e points) if and only if M has integral oeÆ ients and

det(M) = �1.

We re ommend that the reader study the proof in Niven, Zu kerman &

Montgomery, but here we assume the result without going through the proof

of it. Also re all from linear algebra that if M and N are two n�n matri es

with real entries (or in fa t with entries from any ommutative ring), then

5

det(MN) = det(M) � det(N). It is easy enough to write out a proof for 2� 2

matri es (see the text), and we also skip the proof of this result. The above

results provide most the details of proving the following:

Theorem 2.3 The set GL(2; Z) of 2�2 matri es with integer entries whi h

have determinant �1 form a (non ommutative) group with binary operation

being ordinay multipli ation of matri es.

Put � = SL

2

(Z) = fM 2 GL(2; Z) : det(M) = +1g. Then it is easy

to show that � is a subgroup of GL(2; Z), whi h will be alled the Modular

Group.

Sin e we are using matri es to multiply times olumn ve tors (latti e

points), we will now hange the notation f(x; y) to f

x

y

!

.

De�nition: If f and g are two (not ne essarily distin t) binary quadrati

forms over Z, we write f � g and say that f and g are equivalent provided

there is an M 2 � for whi h g

x

y

!

= f(M

x

y

!

).

If f

x

y

!

= ax

2

+ bxy+ y

2

, we denote the matrix A =

a b=2

b=2

!

of

f by [f ℄. So f

x

y

!

= (x; y)[f ℄

x

y

!

. Then note the following: if f � g,

so there is a matrix M 2 � for whi h g

x

y

!

= f(M

x

y

!

), then

g

x

y

!

= (x; y)[g℄

x

y

!

= f(M

x

y

!

) = (x; y)M

t

[f ℄M

x

y

!

for all (x; y) 2 Z � Z. This easily for es

M

T

[f ℄M = [g℄: (11)

Sin e (M

T

)

�1

= (M

�1

)

T

, we may simply write M

�T

for this matrix.

Moreover, the following is now an easy exer ise to prove.

Theorem 2.4 The relation \�" on binary quadrati forms is an equivalen e

relation.

6

The next theorem shows that if f and g are \equivalent binary quadrati

forms over Z" they have mu h in ommon.

Theorem 2.5 Suppose that f and g are equivalent quadrati forms, say

M

T

[f ℄M = [g℄. Then the following hold.

f

x

y

!

= (x; y)M

�T

M

T

[f ℄MM

�1

x

y

!

= (x; y)M

�T

[g℄M

�1

x

y

!

=

"

M

�1

x

y

!#

T

[g℄

"

M

�1

x

y

!#

= g(M

�1

x

y

!

): (12)

This is readily interpreted as:

f

x

y

!

= n i� g(M

�1

x

y

!

) = n: (13)

g d

x

y

!

= g d(M

�1

x

y

!

); (14)

implying that f represents n properly if and only if g represents n properly.

det[g℄ = det[f ℄; so dis (f) = dis (g): (15)

De�nition: Let f 2 Q(d), d not a square. Then f is redu ed provided

(i) �jaj < b � jaj < j j;

or

(ii) 0 � b � jaj = j j.

Theorem 2.6 Let d 2 Z, d not a square. Then ea h equivalen e lass of

binary quadrati forms in Q(d) ontains at least one redu ed form.

7

Proof: Put M =

0 1

�1 0

!

. Then M transforms the quadrati form

(a; b; ) to the form ( ;�b; a).

Put M =

1 m

0 1

!

. Then M transforms (a; b; ) into (a; 2am+ b; am

2

+

bm + ). (Here am

2

+ bm + = f(m; 1).)

There is a unique integer m su h that �jaj < 2am + b � jaj. (a > 0 =)

�a�b

2a

< m �

a�b

2a

= 1 +

�a�b

2a

. a < 0 =)

�a�b

2a

� m <

a�b

2a

= 1 +

�a�b

2a

. So

from (a; b; ) we an get a new form (A = a; B; C) with �jAj < B � jAj, but

C might be hanged. If jCj < jAj, use the �rst of the two transformations

given above.

Theorem 2.7 Let f 2 Q(d) be redu ed (and d not a square). If f is indef-

inite (d > 0), then 0 < jaj �

1

2

p

d. If f is positive de�nite, (d < 0), then

0 < a �

q

�d

3

. In all ases the number of redu ed forms with �xed non-square

d is �nite.

Proof: Clearly a 6= 0. If a > 0, then d = b

2

� 4a = b

2

� 4ja j �

a

2

� 4a

2

< 0. So if d > 0, then a < 0 and d = b

2

� 4a = b

2

+ 4ja j � 4a

2

,

implying jaj �

p

d

2

. Finally, if d < 0, then f is de�nite (and we assume positive

de�nite), so a > 0, > 0. Then d = b

2

� 4a � a

2

� 4a � a

2

� 4a

2

= �3a

2

.

Hen e a

2

�

�d

3

, implying 0 < a �

q

�d=3. Clearly if a is bounded then b is

bounded, and as soon as d; a; b are all �xed, there is at most one value of

that works.

Example 1. For an odd prime p, there are integers x and y for whi h

p = x

2

� 2y

2

if and only if p � �1 (mod 8).

Proof: Let f 2 Q(8), f redu ed. Sin e d = 8 > 0, we have 0 < jaj �

1

2

p

8 =

p

2 =) a = �1. Then �jaj < b � jaj =) 0 � b � 1. But

b � d (mod 2) =) b = 0. Then d = 8 = 0

2

� 4a =) = �2a. So

f = x

2

� 2y

2

and �f = �x

2

+ 2y

2

are the only redu ed forms in Q(8).

Put M =

1 �2

1 �1

!

2 �. Then an easy he k shows M

T

[f ℄M = [�f ℄. So

f � �f , and x

2

� 2y

2

represents p properly if and only if x

2

� 8 (mod 4p)

has a solution, i.e., if and only if x

2

� 8 (mod p) has a solution. But

�

8

p

�

=

8

�

2

p

�

= 1 if and only if p � �1 (mod 8).

Note that what was shown in the pre eding example is that Q(8) ontains

only one equivalen e lass of quadrati forms. Consequently what was shown

is that a prime p is represented (properly, of ourse) by some (and hen e

ea h) quadrati form with dis riminant 8 if and only if p � �1 (mod 8).

So for example, given an odd prime p, there are integers x and y for whi h

p = x

2

+ 6xy + 7y

2

if and only if p � �1 (mod 8).

Example 2. We investigate the forms f with dis riminant d = �4 < 0, so

f is de�nite. We may on entrate on the positive de�nite ase, and without

loss of generality may assume that f is redu ed. By Theorem 2.7 we know

that 0 < a �

q

�d

3

=

q

4

3

, whi h for es a = 1. Then d = �4 = b

2

�

4a � b

2

(mod 4) implies b � 0 (mod 2). And jbj � jaj = 1 implies that

b 2 f�1; 0; 1g. Hen e b = 0. Then �4 = �4 determines = 1, so

x

2

+ y

2

is the unique redu ed positive de�nite form in Q(�4). By Theorem 1.4

there is a form f 2 Q(�4) properly representing n (n 6= 0) if and only if

x

2

� �4 (mod j4nj) has a solution. Hen e by Theorem 2.5, with n > 0,

x

2

+ y

2

= n properly if and only if x

2

� �4(mod 4n).

Note that x

2

� �4 (mod 8) has a solution but x

2

� �4 (mod 16) does

not. So if x

2

� �4 (mod 4n) has a solution, 2jn is possible, but not 4jn.

If p is a prime with p � 1 (mod 4), then x

2

� �4 (mod p) has a solution.

Then h(x) = x

2

+ 4 � 0 (mod p) has a solution; h

0

(x) = 2x 6� 0 (mod p),

so any solution of x

2

+ 4 � 0 (mod p) lifts to all powers of p. Then by

the Chinese Remainder Theorem we see that if n = ek, where e = 1 or 2

and k is a produ t of powers of primes all ongruent to 1 modulo 4, then

x

2

+ 4 � 0 (mod n) is solvable, so that n is properly representable in the

form n = x

2

+ y

2

.

But if p � 3 (mod 4), then x

2

� �4 (mod p) has no solution. Hen e

if n is divisible by a prime p with p � 3 (mod 4), then n is not properly

representable in the form n = x

2

+y

2

. We have proved the following theorem.

Theorem 2.8 A positive integer n is preperly representable as n = x

2

+ y

2

if and only if ea h prime p dividing n is ongruent to 1 modulo 4, or p = 2

and 4 does not divide n.

9

We have now all the information needed to give another proof of the

following theorem whi h we proved in lass.

Theorem 2.9 If n is a positive integer, then n = x

2

+ y

2

has a solution if

and only every odd prime dividing the square-free part of n is ongruent to 1

modulo 4.

Proof: We note that if n = x

2

+y

2

and g is any integer, then g

2

n = (gx)

2

+

(gy)

2

. Conversely, if g = g d(x; y) and n = x

2

+ y

2

, then

n

g

2

=

�

x

g

�

2

+

�

y

g

�

2

properly. Now re all that if m and n are integers that an be represented as

sums of two squares, then mn is also su h an integer. The theorem follows.

3 The Number of Representations by Quadrati

Forms

We adopt the following notation:

R(n) = jf(x; y) 2 Z � Z : x

2

+ y

2

= ngj

r(n) = jf(x; y) 2 Z � Z : x

2

+ y

2

= n and g d(x; y) = 1gj

P (n) = jf(x; y) 2 Z � Z : x

2

+ y

2

= n and g d(x; y) = 1 and x > 0; y � 0gj

N(n) = the number of solutions to s

2

� �1 (mod n)

Theorem 3.1 Suppose n > 0. Then the following hold:

r(n) = 4P (n) = 4N(n); (16)

R(n) =

X

d:d

2

jn

d>0

r(

n

d

2

): (17)

Proof: Consider any solution of x

2

+ y

2

= n, where n > 0. Of the four

points (x; y), (�y; x), (�x;�y), (y;�x), exa tly one of them has positive

�rst oordinate and non-negative se ond oordinate. As P (n) is the number

of proper representations x

2

+ y

2

= n for whi h x > 0 and y � 0, this shows

10

that r(n) = 4P (n). We now show that P (n) = N(n) by exhibiting a one-

to-one orresponden e between proper representations of n with x > 0 and

y � 0, and solutions s of the ongruen e s

2

� �1 (mod n).

Suppose that

(x; y) satis�es x

2

+ y

2

= n; x > 0; y � 0; g d(x; y) = 1:

Then g d(x; n) = 1, so there is a unique integer �x modulo n for whi h x�x �

1 (mod n. Put s � �xy. Then sin e x

2

� �y

2

(mod n), s

2

� �x

2

y

2

� �x

2

�x

2

�

�1 (mod n). So F : (x; y) 7! s de�nes a fun tion F from the set of proper

representations (x; y) of n with x > 0; y � 0, to the set of s with s

2

�

�1 (mod n).

Next we show that F is one-to-one. So suppose n = x

2

i

+y

2

i

, x

i

> 0, y

i

� 0,

g d(x; y) = 1, x

i

s

i

� y

i

(mod n), i = 1; 2. Suppose that s

1

� s

2

(mod n).

Then x

1

y

2

s

1

� y

1

y

2

� x

2

y

1

s

2

(mod n). Hen e x

1

y

2

� x

2

y

1

(mod n), sin e

s

1

� s

2

(mod n) and g d(s

i

; n) = 1. But 0 < x

2

i

� n =) 0 < x

i

�

p

n, and

0 � y

2

i

< n =) 0 � y

i

<

p

n. This for es 0 � x

1

y

2

< n and 0 � x

2

y

1

< n

and thus x

1

y

2

= x

2

y

1

. But then x

1

jx

2

y

1

and g d(x

1

; y

1

) = 1 for es x

1

jx

2

.

Similarly, x

2

jx

1

, implying x

1

= x

1

=) y

1

= y

2

.

At this point we know that F is well-de�ned and one-to-one. We want to

show that it is onto. So suppose that s

2

� �1 (mod n). Say s

2

+1 = n . Then

(2s)

2

�4n = �4 so that the binary quadrati form g(x; y) = nx

2

+2sxy+ y

2

has dis riminant �4. Hen e g 2 Q(�4) and g is positive de�nite. In the

pre eding se tion we showed that the unique redu ed positive de�nite binary

quadrati form in Q(�4) is x

2

+ y

2

. This implies that there is a matrix

M =

m

11

m

12

m

21

m

22

!

for whi h

m s

s

!

= [g℄ =M

T

1 0

0 1

!

M:

Writing out this matrix produ t gives

m

2

11

+m

2

21

m

11

m

12

+m

21

m

22

m

11

m

12

+m

21

m

22

m

2

12

+m

2

22

!

=

m s

s

!

:

Also 1 = det(M) = m

11

m

22

�m

12

m

21

, implying g d(m

11

; m

21

) = 1. Hen e

m

11

s = m

2

11

m

12

+m

11

m

21

m

22

11

� �m

2

21

m

12

+m

11

m

21

m

22

(mod n)

= �m

2

21

m

12

+m

21

(1 +m

21

m

12

)

= m

21

Exa tly one of (m

11

; m

21

), (�m

21

; m

11

), (�m

11

;�m

21

), (m

21

;�m

11

) has

positive �rst oordinate and nonnegative se ond oordinate. Also, start-

ing with m

11

s � m

21

(mod n) and s

2

� �1 (mod n), we have (�m

21

s �

�s

2

m

11

� m

11

(mod n). It follows that xs � y mod n) for any one of the

four pairs given above. Hen e F is onto, implying that F is a bije tion and

proving that r(n) = 4P (n) = 4N(n).

Now we prove that R(n) =

P

d:d

2

jn

r

�

n

d

2

�

. But this follows immediately

from the observation that

u

2

+ v

2

=

n

d

2

properly i� (du)

2

+ (dv)

2

= n with d = g d(du; dv):

Theorem 3.2 Fa tor the positive integer n as

n = 2

�

�

Y

p

�

jjn

p�1 (mod 4)

p

�

�

Y

q

jjn

q�3 (mod 4)

q

:

(1) If � 2 f0; 1g and = 0 for all q, then r(n) = 2

t+2

, where t is the

number of primes p � 1 (mod 4) dividing n. Otherwise r(n) = 0.

(2) If all are even, then R(n) = 4

Q

p

�

jjn

p�1 (mod 4)

(� + 1). Otherwise

R(n) = 0.

Proof: Using the Chinese Remainder Theorem we �nd that N(n) =

N(2

�

)

Q

p

N(p

�

)

Q

q

N(q

). Note that N(2) = 1 and N(4) = 0 (sin e s

2

�

�1 (mod 4) has no solution). So N(2

�

) = 0 for all � � 2. SimilarlyN(q) = 0

and N(q

) = 0 for all � 1. Hen e we have

N(n) =

(

2

t

; if � = 0 or 1 and all = 0;

0; otherwise.

This implies r(n) = 4N(n) = 2

t+2

if � = 0 or 1 and all = 0, and

equals 0 otherwise. We know R(n) = 4

P

d

2

jn

N

�

n

d

2

�

. If n = m

1

� m

2

with

12

g d(m

1

; m

2

) = 1, then d

2

jn if and only if d = d

1

� d

2

with d

2

1

jm

1

and d

2

2

jm

2

.

Moreover, d$ (d

1

; d

2

) is a bije tion. Hen e we have the following:

X

d

2

jn

N

�

n

d

2

�

=

X

d

2

1

jm

1

d

2

2

jm

2

N

m

1

d

2

1

�

m

2

d

2

2

!

=

X

d

2

1

jm

1

d

2

2

jm

2

N

m

1

d

2

1

!

�N

m

2

d

2

2

!

=

X

d

2

1

jm

1

N

m

1

d

2

1

!

X

d

2

2

jm

2

N

m

2

d

2

2

!

:

So for n = 2

�

Q

p

�

Q

q

as above, we have

X

d

2

jn

N

�

n

d

2

�

=

X

d

2

j2

�

N

�

2

�

d

2

�

Y

p

0

�

X

d

2

jp

�

N

p

�

d

2

!

1

A

Y

q

0

�

X

d

2

jq

N

�

q

d

2

�

1

A

:

Now onsider the ontributions of the three main fa tors above. For the

�rst fa tor, if � is even, the only nonzero ontribution omes from d

2

= 2

�

.

If � is odd, the only nonzero ontribution omes from d

2

= 2

��1

. So the �rst

fa tor is 1 in any ase. For the se ond fa tor, if � is even, N

�

p

�

d

2

�

= 2 for

d = 1, p; p

2

; : : : ; p

�

2

�1

, and N

�

p

�

d

2

�

= 1 if d

2

= p

�

. So the prime p ontributes

2

�

�

2

�

+ 1 = � + 1. If � is odd, then N

�

p

�

d

2

�

= 2 for d = 1; p; : : : ; p

��1

2

. So

again the prime p ontributes 2

�

1 +

��1

2

�

= � + 1. For the third fa tor, if

is odd, then q divides

q

d

2

for all appropriate q, so N

�

q

d

2

�

= 0 always. If is

even, then N

�

q

d

2

�

=

(

1; d

2

= q

;

0; otherwise.

This implies that q ontributes 1 if is even, and ontributes a fa tor of zero

if is odd. Hen e

R(n) = 4

X

d

2

jn

N

�

n

d

2

�

= 4 � 1 �

Y

p

(� + 1) �

(

1; all even;

0; otherwise.

13

Re apitulation: n = x

2

+ y

2

has a proper solution if and only if

n = 2

�

Y

p�1 (mod 4)

p

�

jjn

p

�

with � = 0 or 1, and in that ase r(n) = 4N(n) = 2

t+2

, where t is the number

of p � 1 (mod 4), pjn.

4 Ternary Quadrati Forms

Let

F =

3

X

k;l=1

a

kl

x

k

x

l

= (x

1

; x

2

; x

3

)A

0

B

�

x

1

x

2

x

3

1

C

A

; A = (a

kl

) with a

kl

= a

lk

: (18)

a

11

F = (a

11

x

1

+ a

12

x

2

+ a

13

x

3

)

2

+K(x

2

; x

3

); (19)

where

[K(x

2

; x

3

)℄ =

a

11

a

22

� a

2

12

a

11

a

23

� a

12

a

13

a

11

a

23

� a

12

a

13

a

11

a

33

� a

2

13

!

: (20)

Now a bit of routine omputation shows that

det([K(x

2

; x

3

)℄) = a

11

� det(A): (21)

Theorem 4.1 F is positive de�nite if and only

(i) a

11

> 0;

(ii) b = det

a

11

a

12

a

21

a

22

!

> 0;

and

(iii) d = det(A) > 0:

.

14

Proof: F (1; 0; 0) = a

11

, so if F is de�nite, then a

11

> 0. F (x

1

; x

2

; 0) =

(x

1

; x

2

)

a

11

a

12

a

21

a

22

!

x

1

x

2

!

, so if F is de�nite then b > 0 by the theory of

binary forms.

So assume that (i) and (ii) hold. Then by Eq. 19, if K is positive de�nite,

learly F is positive de�nite. And if K is not positive de�nite, we an �nd

x

2

; x

3

not bot 0, but both divisible by a

11

, for whi hK(x

2

; x

3

) � 0. And sin e

x

2

� x

3

� 0 (mod a

11

), we an solve for x

1

su h that a

11

x

1

+ a

12

x

2

+a

13

x

3

= 0.

Then F (x

1

; x

2

; x

3

) � 0 with at least one of s

1

; x

2

; x

3

not zero. So if (i) and

(ii) hold, then F is positive de�nite if and only K is positive de�nite. And

this holds if and only if b > 0 and a

11

det(A) > 0, whi h is if and only if

d = det(A) > 0.

Theorem 4.2 If g d(

11

;

21

;

31

) = 1, the six remaining integers

kl

an be

hosen in su h a way that det(

kl

) = 1.

Proof: Put g = (

11

;

21

), so (g;

31

) = 1. We an hoose x

12

;

22

; u; v so

that x

11

22

�

12

21

= g, and gu�

31

v = 1. Then

det

0

B

�

11

12

11

g

v

21

22

2

1

g

v

31

0 u

1

C

A

=

31

12

21

v �

11

22

v

g

!

+ u(

11

22

�

21

21

)(22)

= �

31

v + ug = 1: (23)

Theorem 4.3 Ea h equivalen e lass of positive de�nite ternary forms on-

tains at least one form for whi h a

11

�

4

3

3

p

d, 2ja

12

j � a

11

; 2ja

13

j � a

11

.

Proof: Let F be a �xed form belonging to the given lass. Let a

11

be

the smallest positive integer represented by F , and hen e by any form of the

same lass. Then for suitable

11

;

21

;

31

, we have a

11

= F (

11

;

21

;

31

). Here

g d(

11

;

21

;

31

) = 1, sin e otherwise

a

11

g d(

11

;

21

;

2

31

)

would be representable.

Let C =

0

B

�

11

12

13

21

22

23

31

32

33

1

C

A

be the matrix with det(C) = 1 onstru ted in

Theorem 4.2, and de�ne the form G by

G(�x) = F (C�x) = �x

T

C

T

[F ℄C�x:

15

Then if we say [G℄ = B = (b

ij

) = C

T

[F ℄C, we have b

11

= G(1; 0; 0) =

F (C

0

B

�

1

0

0

1

C

A

) = F (

11

;

21

;

31

) = a

11

. To G apply the transformation (d

kl

) =

0

B

�

1 r s

0 t u

0 v w

1

C

A

where tw � uv = 1. Then for arbitrary r; s we have det(d

kl

) =

1. Put D = (d

kl

), and put H(�y) = G(D�y), so A = [H℄ = D

T

[G℄D =

D

T

BD. The oeÆ ient of y

2

1

in H(�y) is the (1; 1) entry of D

T

BD, whi h is

(1

st

row of D

T

) � B � (1

st

ol of D) = G(1; 0; 0) = a

11

, i.e., (A)

11

= a

11

. IF

A = (A

IJ

) for general i; j, 1 � i; j � 3, then

a

12

= (row 1 of D

T

)(b

ij

)( ol 2 of D

T

) =

= (1; 0; 0)(b

ij

)

0

B

�

r

t

v

1

C

A

= (b

11

; b

12

; b

13

)

0

B

�

r

t

v

1

C

A

= ra

11

+ tb

12

+ vb

13

(sin e a

11

= b

11

): (24)

a

13

= (row 1 of D

T

)B( ol3 of D

T

) = (1; 0; 0)(b

ij

)

0

B

�

r

t

v

1

C

A

= sa

11

+ ub

12

+ wb

13

: (25)

From Eq. 19 we have

a

11

G(x

1

; x

2

; x

3

) = (b

11

x

1

+ b

12

x

2

+ b

13

x

3

)

2

+K(x

2

; x

3

);

a

11

H(y

1

; y

2

; y

3

) = (a

11

y

1

+ a

12

y

2

+ a

13

y

3

)

2

+ L(y

2

; y

3

);

where K and L are de�nite in their respe tive variables.

Sin e H(�y) = G(D�y), the transformation �x = D�y takes G into H. Put

�

D =

t u

v w

!

, then �x = D�y implies

x

2

x

3

!

. SoK(x

2

; x

3

) = K

�

D

y

2

y

3

!!

must equal L

y

2

y

3

!

.

16

By Eq. 21,

det[K(x

2

; x

3

)℄ = a

11

det[G℄ = a

11

det(B);

and

det[L(y

2

; y

3

)℄ = a

11

det(a) = a

11

d:

the �rst oeÆ ient of L is a

11

a

22

� a

2

12

. So we may hoose

t u

v w

!

so

that tw � vu = 1, and

a

11

a

22

� a

2

12

�

2

p

3

q

a

11

d:

Re all: In ea h lass of positive de�nite binary quadrati forms there is

at least one form ax

2

+ 2bxy + y

2

with 2jbj � a �

2

p

3

p

a � b

2

.

From a

12

= ra

11

+ tb

12

+ vb

13

, we may hoose r so that ja

12

�

1

11

2

.

Similarly, from a

13

= sa

11

+ ub

12

+ wb

13

we may hoose s so that ja

13

�

a

11

2

.

Sin e a

22

= H(0; 1; 0) is representable by H, therefore a

22

� a

11

. Then

a

2

11

� a

11

a

22

= (a

11

a

22

� a

2

12

) + a

2

12

�

2

p

3

p

a

11

d +

a

2

11

4

, whi h implies that

3

4

a

2

11

�

2

p

3

p

a

11

d, implying a

3

2

11

�

8

3

p

3

p

d, and �nally a

11

�

4

3

d

1

3

.

Theorem 4.4 Every positive de�nite ternary form f(�x) = �x

T

A�x with A =

A

T

and det(A) = 1 is equivalent to the form f(�x) = x

2

1

+ x

2

2

+ x

2

3

. Conse-

quently, every number representable by su h a form an be written as a sum

of three squares.

Proof: By Theorem 4.3 the given form is equivalent to a form �x

T

(a

ij

)�x

in whi h a

11

�

4

3

, 2ja

12

j � a

11

�

4

3

, 2ja

13

j � a

11

�

4

3

. Hen e a

11

= 1,

a

12

= a

13

= 0. The lass therefore ontains a form

G = x

2

1

+ a

22

x

2

2

+ 2a

23

x

2

x

3

+ a

33

x

2

3

;

where

K(x

2

; x

3

) = a

22

x

2

2

+ 2a

23

x

2

x

3

+ a

33

x

2

3

17

is positive de�nite and has determinant 1. Hen e we know K is equiva-

lent to the form x

2

2

+ x

2

3

by a suitable M =

t u

v w

!

2 �. Consequently

0

B

�

1 0 0

0 t u

0 v w

1

C

A

takes G into x

2

1

+ x

2

2

+ x

2

3

.

Lemma 4.5 If n = x

2

1

+x

2

2

+x

2

3

, n > 0, then n is not of the form 4

a

(8b+7),

a � 0, b � 0.

Proof: Clearly n = 8b + 7 � 7 (mod 8), but ea h perfe t square is on-

gruent to one of 0, 1 or 4 modulo 8. So the sum of three squares is ongruent

to one of 0, 1, 2, 3, 4, 5, 6 modulo 8. So 8b+7 is not a sum of three squares.

Suppose 4

a+1

(8b+ 7) = x

2

1

+ x

2

2

+ x

2

3

� 0 (mod 4). Clearly ea h of x

1

, x

2

and x

3

must be even. Hen e 4

a

(8b+ 7) =

�

x

1

2

�

2

+

�

x

2

2

�

2

+

�

x

3

2

�

2

. The result

follows.

Theorem 4.6 If n > 0 is not of the form 4

a

(8b + 7), a � 0, b � 0, then n

an be written as the sum of three squares.

Proof: If n is not the sum of three squares, neither is 4n. So we may

assume that n is odd or is twi e an odd number. So in fa t it suÆ es to

assume that n � 1; 2; 3; 5 or 6 (mod 8), and for su h an n it suÆ es to

show that there is a positive de�nite ternary form of dis riminant 1 whi h

represents n. So we have to spe ify nine integers a

ij

, 1 � i � j � 3; x

1

; x

2

; x

3

whi h satisfy:

(i) n = a

11

x

2

1

+ 2a

12

x

1

x

2

+ 2a

13

x

1

x

3

+ a

22

x

2

2

+ 2a

23

x

2

x

3

+ a

33

x

2

3

;

(ii) a

11

> 0;

(iii) a

11

a

22

� a

2

12

> 0;

(iv) det(a

ij

) = 1 (where a

ij

= a

ji

for i > j):

We may even take a

13

= 1, a

23

= 0, a

33

= n, x

1

= 0, x

2

= 0, x

3

= 1. Then

(i) automati ally holds, and the three remaining unknown a

ij

must satisfy:

(ii) a

11

> 0;

(iii) b = a

11

a

22

� a

2

12

> 0;

18

(iv) 1 = det

0

B

�

a

11

a

12

1

a

12

a

22

0

1 0 n

1

C

A

=

�

�

�

�

�

a

12

a

22

1 0

�

�

�

�

�

+n

�

�

�

�

�

a

11

a

12

a

12

a

22

+

�

�

�

�

�

= �a

22

+nb,

i.e., a

22

= bn� 1.

Sin e the ase n = 1 is trivial, we assume n > 1. Then b > 0, so

a

22

= bn� 1 > 0. Here b > 0 implies a

11

a

22

> a

2

12

� 0, hen e a

11

> 0. So (ii)

follows from (iii) and (iv). So we need: b = a

11

a

22

�a

2

12

> 0 and a

22

= bn�1.

This is equivalent to saying: b > 0 and �b is a quadrati residue modulo

bn� 1.

Case 1. n � 2 or 6 (mod 8). Sin e g d(4n; n�1), by Diri hlet's Theorem

there is a prime p � n � 1 (mod 4n), say p = 4nv + n � 1 = (4v + 1)n� 1.

If we set 4v + 1 = b, then b > 0, and sin e p � 1 (mod 4),

�b

p

!

=

�

p

b

�

=

bn� 1

b

!

=

�

�1

b

�

= +1 sin e b � 1 (mod 4):

Case 2. n � 1; 3; or 5 (mod 8). Put = 1 if n � 3 (mod 8) and

= 3 if n � 1 or 5 (mod 8). So n � 3 or 7 (mod 8) and

n�1

2

is odd.

Hen e g d(4n;

n�1

2

) = 1. By Diri hlet's Theorem there is a prime p �

n�1

2

(mod 4n), say p =

n�1

2

+ 4nv =

1

2

[(8v + )n� 1℄.

Set b = 8v + . then b > 0 and 2p = bn� 1.

n � 1 (mod 8 =) 2p � b � 1 � � 1 � 3 � 1 � 2 (mod 8) and

p � 1 (mod 4), b � 3 (mod 8).

n � 3 (mod 8) =) 2p � 3b � 1 � 3 � 1 � 3 � 1 � 2 (mod 8), for ing

p � 1 (mod 4) and b � 1 (mod 8).

n � 5 (mod 8) =) 2p � 5b� 1 � 5 � 1 � 6 (mod 8) =) p � 3 (mod 4)

and b � 3 (mod 8).

�

�2

b

�

=

�

�1

b

��

2

b

�

=

(

(�1)(�1) = 1 if b � 3 (mod 8);

(+1)(+1) = 1 if b � 1 (mod 8):

And

�

�b

p

�

= (�1)

[

�b�1

2

℄[

p�1

2

℄

�

p

b

�

=

�

p

b

�

(in all three ases) =

�

p

b

� �

�2

b

�

=

�

�2p

b

�

=

�

1�bn

b

�

=

�

1

b

�

= +1.

So �b is a quadrati residue modulo p. Sin e �b � 1

2

(mod 2), it follows

that �b is a quadrati residue modulo 2p = bn� 1.

19

Corollary 4.7 If n � 1 or 2 (mod 4), then n is a sum of three squares.

If n � 3 (mod 4), then n = (n � 1) + 1

2

, where n � 1 is a sum of three

squares, so that n is a sum of four squares.

If n � 0 (mod 4), then n = 4

a

(4b+ r), r = 1; 2; 3. Sin e 4

a

= (2

a

)

2

and

4b+ r must be a sum of 4 squares, learly n is a sum of 4 squares.

20