Number Theory Project
The Interpretation of the definition
Andre (JianYou) Wang
Joint with
JingYi Xue
Definition
• If f  ax  bxy  cy with measure d  b  4ac  0
and a  0 is called positive definite
Observation 1: c  0
2
Observation 2: 4af   2ax  by   dy 2  0  f  0
2
2
2
That explained partly the etymology of positive .
Definition
• If there exists a linear mapping with integer
coefficient between two different positive
definite polynomials, these two are called
equivalent to each other. i.e. there is a map
x  mX  nY , y  pX  qY , mq  np  1
such that it maps
f g
Explanation of Definition
• Observation 1: this linear map can be
represented as a matrix  m n 

p

q
Observation 2: if the coefficient of f is  a, b, c 
Then the new coefficient has this formula
a '  am  bmp  cp
2
2
b '  2amn  bmq  bnp  2 pqc
c '  an  bnq  cq
2
2
• Observation 3: this definition is actually fitting
into the definition of equivalence relation,
namely, it meets reflex, transitive, and symmetric.
That’s why we can define equivalent class, each
equivalent class is disjoint, in particular, the
element in the same equivalent class has same
measure.
• Observation 4:we may find b  0, c  0, a  0
which further explained the term: positive
polynomials
• Observation 5: the composition of two maps
m n


 p q
r s


u v
is actually the
multiplication of the matrix .
Observation 6: the restriction mp  nq  1 is
the perfect condition to ensure the map has
an inverse which agrees with the law of
inverse in matrix, namely: det X  1
• Claim 1: In every equivalent class, there is a
polynomial with coefficients satisfy the
condition
0bac
• Proof: By well-ordering principle, let a ' be the
minimal value represented by the equivalent
class, and let  a ', b ', c ' be an element in that
class,
2
2
then a  a ' m  b ' mp  c ' p  gcd  m, p   1
• By Bezout identity, we have a proper  q, n 
satisfy the equality mq  np  1 ,which fits our
definition of being a map simultaneously.
Under the map  m n 
 a ', b ', c '   a, b '', c ''

p

q
Under another map  1 x   a, b '', c ''   a, b, c 


0 1
• For this new  a, b, c  there is a proper x
which will enable the tuple to satisfy the claim.
The claim is proven.
• Claim 2: Using Claim we can easily assume that
for each measure d , there are only finitely many
equivalent class, and we can actually compute
the upper bound for coefficient.
d
d  4ac  b  4a  b  3a  a 
3
2
2
2
2
• This suggests that this sort of polynomial has
finite classes.
• Claim 3: The number of the equivalent class
with a fixed measure, is the number of the
tuples  a, b, c  that satisfy
d  b  4ac,0  b  a  c
2
• Proof: Using Claim 1, we know that for every
class there is at least one element satisfy this
condition .All I need to do is to show that two
polynomials in the same class can not be in
the set of tuples together
• If we have f g , then  f    g 
All we’re supposed to check is that
(a f , b f , c f )  (ag , bg , cg )
Without the loss of generality ,we assume a f  a g
By definition, there is a matrix  m n  ,such that,

p

q
a f  ag m 2  bg mp  cg p 2  ag  ag m 2  bg mp  cg p 2
 ag m 2  ag mp  ag p 2  ag ( m  p ) 2  ag mp  ag mp
 mp  1, 0
• Clearly, mp  1 can’t work.
• In the other case, a f  a g , the similar argument
goes to the rest entries.
• Now we prove that there is one to one and onto
relationship between these two sets, which boils
down to the truth, the sizes are equal.
Recap
• These 3 Claims provide a rather efficient way
to determine the equivalent classes of positive
definite quadratic polynomials, in that this
fact can help us classify different class with no
ambiguity
• In light of this, problem 3.2, and the entire
problem 4 is just a simple corollary.
Moreover
• Conjecture 1: if  f    g   the number sets
represented by the former is not the number
set represented by the latter.
• Phenomenon: Different equivalent classes
represents number in a distinct way, some
number sets are disjoint, some are
intertwined, and some are contained in the
other.
• Conjecture 2: If  f  and  g  represents a
single different number , then they represents
infinite many different number.
• We are convinced that these phenomenon are
closely related to the problem 5 and problem
6
• Claim1: If ax 2  bxy  cy 2  kp, (k , p)  1
• then:  d   1, d  b2  4ac
 p
• Pf:
(2ax  by )2  dy 2  4apk
dy 2  (2ax  by )2 (mod p)
• Claim2: If p is an odd prime bigger than 7, then
2
2
 p  1,9(mod 20)
p

x

5
y
•
 20   4   1   5   1   5 
• Pf: "  " :                1
 p 
 p  p  p 
 p  p 
 5  p
If p  1(mod 4),       1, p  1, 4(mod 5)
 p  5
p  1,9(mod 20)
•
•
•
•
•
•
If p  3(mod 4)  p  3,7(mod 20)
cannot be true
" "
 5 
We know  p   1 , so s0 , s0 2  5(mod p )
Consider the set u  vs0 , where 0  u, v  p ,
2
Then the number in the set  p  1  p
So u1  v1s0  u2  v2 s0 (mod p)
(u1  u2 )2  5(v1  v2 )2  0(mod p)
0  (u1  u2 )2  5(v1  v2 )2  6 p
• Let u1  u2  t v1  v2  t '
t
2
2
• If t  t '  5 p ( , t ') is also a solution
5
•
•
•
•
If
If
If
If
t t'
( , ) is also a solution
2 2
t 2  t '2  3 p t 2  2,3(mod5) contradict
t 2  t '2  4 p
contradict
t 2  t '2  p (t , t ') is a solution
t 2  t '2  2 p t 2  2,3(mod5)
2
2
2
n
• Claim3: Any
can be represented by x  5 y
• Any two number can be both represented by
• x 2  5 y 2 , the product also can be represented.
• Pf: The first prop is obvious
•
the second: let A  x12  5 y12 B  x22  5 y22
AB  ( x1 x2  5 y1 y2 ) 2  5( x1 y2
y1 x2 ) 2
• Claim4: If p  3,7(mod 20) , then 2p , 3p can be
represented.
• Pf: the first part of the argument of claim3 is
also valid
• then t 2  t '2  p (claim 2)
t
2
2
• If t  t '  5 p ( , t ') is also a solution(contradict
5
to claim 2)
• If t 2  t '2  4 p ( t , t ' ) is also a solution
2 2
contradiction
• Discover that the two situation both are true,
they coexist. i.e. if u12  5v12  2 p there exist
• u2 2  5v2 2  3 p and vice versa
2
2
u

5
v
• If 1
let u1  2 x1  1, v1  2 y1  1
1  2p
• If 5( y1  x1  1)2  (5 y1  x1  2)2  3 p
2
2
u1 , v1  1(mod 3)
u

5
v
• If 1
1  3p
u1  3x1  1, v  3 y1  1 5( y1  x1 )2  (5 y1  x1  2) 2  2 p
u1  3x1  1, v  3 y1  1 5( y1  x1 )2  (5 y1  x1  2) 2  2 p
u1  3x1  1, v  3 y1  1 5( y1  x1 )2  (5 y1  x1  2) 2  2 p
u1  3x1  1, v  3 y1  1 5( y1  x1 )2  (5 y1  x1  2) 2  2 p
Since 2p,3p one of then has to be true, then the
other one is true
• Claim5: p  3,7(mod 20) pq can be represented
• Pf: Applying claim5, there exist
u12  5v12  2 p
•
u2 2  5v2 2  2 p
(4 numbers are odd)
u1u2  5v1v2 2
u1v2  u2v1 2
pq  (
)  5(
)
2
2
• Claim 6
• If p  3,7  mod 20 ,then 2 p can be
represented, which is a simple corollary from
Claim 5
• Theorem:
Integer n can be represented in the form of
2
2
x  5 y if and only if
 
n  2 5 a p1 p2
2
pm q1q2
Where :
p1 , p2 , , pm  1,9  mod 20 
q1 , q2 , , qn  3,7  mod 20
n    0  mod 2 
qn
 ,   1, 0
• First step:
p1 , p2 ,
pm
can be represented,
a
2
can
be represented, q1 , q2 , , qm , 2 can be
evenly paired, and each pair is in the form, 5 is
in the form.
The product thus can be represneted.
• Second step: prove the converse is also true.
If the converse is not true, it suffices to say
that n can be represented when n  Bp
where
B can be represented, p  3,7, 2  mod 20
• When p  3,7  mod 20 ,3 p is in the form
which means 3  x 2  y 2    u 2  5v 2  s 2  5t 2 
3 x  5 y
2
2
e
2
5f
2
We now apply the
infinite descend to cause contradiction
3 x 2  e 2  mod 5 
x  e  0  mod 5 
• Likewise, the other case can be dealt with in
almost the same way.
Thus, there is a contradiction to our previous
assumption which means the converse is alos
true and the theorem is proven
Problem 6.2
Proposition:
A  2 x 2  7 y 2 , B  2s 2  7t 2  AB  (2 xs  7 yt )2  14( xt
ys)2
• Moreover, by using the method in claim2, we
can give a proof to the conjecture 1:
• If p is prime, p  2,7 then
p  x 2  14 y 2or 2 x 2  7 y 2  p  1,9,15, 23, 25,39(mod 56)
these are primes p  1(mod8)
3 p  x 2  14 y 2  p  3,5,13,19, 27, 45(mod 56)
 14 
 p  1


• Pf: By using
, it’s easy to know the
satisfied primes above.
2
2
x

14
y
 p,2 p, 14 p
• only to focus on
• Here’s the result:
• (call p  1,9,15, 23, 25,39(mod 56) A, p  3,5,13,19, 27, 45(mod 56) B)
1 x 2  14 y 2  p, A
2 x 2  14 y 2  2 p, x  2 x1 , p  2 x12  7 y 2 A
3 x 2  14 y 2  3 p, B
4 x 2  14 y 2  4 p, x  2 x1 , y  2 y1 , p  x12  14 y12 A
5 x  14 y  5 p, x  y (mod 5) B
2
2
2
2
(5u  1) 2  14(5v  1) 2  5 p  (u  14v  3) 2  14(u  v) 2  3 p
(5u 1) 2  14(5v  1) 2  5 p  (u  14v 3) 2  14(u  v ) 2  3 p
(5u  2) 2  14(5v  2) 2  5 p  (u  14v  6) 2  14(u  v ) 2  3 p
(5u 2) 2  14(5v  2) 2  5 p  (u  14v 6) 2  14(u  v )2  3 p
6 x 2  14 y 2  6 p, x  2 x1 ,3 p  2 x12  7 y 2 , x12  y 2 (mod 3) B
2(3u  1) 2  7(3v  1) 2  3 p  (2 x  2 y  3) 2  14( x  y ) 2  3 p
2(3u 1) 2  7(3v  1) 2  3 p  (2 x  2 y 3) 2  14( x  y ) 2  3 p
7 x 2  14 y 2  7 p, x  7 x1 , 2 y 2  7 x12  p A
8 x 2  14 y 2  8 p, x  2 x1 , y  2 y1 , 2 p  x12  14 y12
x1  2 x2 , p  2 x2 2  7 y12 A
9 x 2  14 y 2  9 p A
10 x 2  14 y 2  10 p, x  2 x1 ,5 p  2 x12  7 y 2 ,5 | ( x12  y 2 )
a 5 | x1 ,5 | y, p  5, x1  5 x2 , y  5 y1 , 2 x2 2  7 y12  1, contradiction
b x1  1(mod 5) y  2(mod 5)
5 p  2(5u  1) 2  7(5v  2) 2  (4u  7v  2) 2  14(u  2v  1) 2  3 p
5 p  2(5u 1) 2  7(5v  2) 2  (4u  7v  2) 2  14( u  2v  1) 2  3 p
c x1  2(mod 5) y  1(mod 5)
5 p  2(5u  2) 2  7(5v  1) 2  (4u  7v  3) 2  14(u  2v) 2  3 p
5 p  2(5u 2) 2  7(5v  1) 2  (4u  7v  3) 2  14(u  2v) 2  3 p B
11 x 2  14 y 2  11 p
x 2  1,3, 4,5,9(mod11) contradiction
12 x 2  14 y 2  12 p, x  2 x1 , y  2 y1 ,3 p  x12  14 y12 , B
13 x 2  14 y 2  13 p B
a 13 | x,13 | y, p  13, x12  14 y12  1, x1  1, y1  0
b x 2  1, y 2  1(mod13)
c x 2  3, y 2  3(mod13)
d x 2  4, y 2  4(mod13)
14 x 2  14 y 2  14 p, x  14 x1 , p  14 x12  y 2 , A
• Thank You!