Shawn GodinShawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 11 / 110 Rings and Fields A ring is a collection of elements, R, along with two binary operators, and , that

Quadratics

Shawn Godin

Cairine Wilson S.SOrleans, ON

[email protected]

October 14, 2017

Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 1 / 110

Binary Quadratic Form

A form is a homogeneous polynomial, that is a polynomial where eachterm has the same degree.

Specifically, a binary quadratic form is a homogeneous polynomial in twovariables of degree 2, that is a polynomial of the form

f (x , y) = ax2 + bxy + cy2.


Number Systems: In the Beginning

Natural numbers, N = {1, 2, 3, . . . }Whole numbers, W = {0, 1, 2, 3, . . . }

closed under addition (i.e. if x , y ∈ N then (x + y) ∈ N),not closed under subtraction (for example 2− 5 6∈ N),closed under multiplication,

not closed under division


Number Systems: Linear Equations ax + b = 0

N – closed under + and ×, not under − and ÷W – closed under + and ×, not under − and ÷Integers, Z = {. . . ,−2,−1, 0, 1, 2, . . . }

closed under addition,

closed under subtraction,

closed under multiplication,

not closed under division

not all equations ax + b = 0, with a, b ∈ Z have solutions in Z


Groups

A group is a collection of elements, G , along with a binary operator, ⊕,that satisfy the following conditions:

G is closed under ⊕ (i.e. if x , y ∈ G then x ⊕ y ∈ G ),

⊕ is associative, that is for x , y , z ∈ G , x ⊕ (y ⊕ z) = (x ⊕ y)⊕ z ,

there exists a element,e, called the identity such that for any x ∈ G ,e ⊕ x = x ⊕ e = x ,

each x ∈ G has an inverse, denoted x−1, that satisfiesx ⊕ x−1 = x−1 ⊕ x = e.

A group in which ⊕ is also commutative, that is for all x , y ∈ G we havex ⊕ y = y ⊕ x , is called an Abelian group.


Some Abelian Groups

The following are all Abelian groups:

If G is any of the sets: Z, Q, R, or C with regular addition. Theidentity is 0 and the inverse of an element x is its negative −x .

If G is any of the sets: Q \ {0}, R \ {0}, or C \ {0} with regularmultiplication, ×. The identity is 1 and the inverse of an element x isits reciprocal 1x .

If G is the integers modulo n, Zn, with addition modulo n. Theidentity is 0 and the inverse of an element is its additive inversemodulo n.

If G is Zp \ {0}, for some prime p, with multiplication modulo p. Theidentity is 1 and the inverse of an element is its multiplicative inversemodulo n.


A Non-Abelian Group: The Symmetries of an EquilateralTriangle

An equilateral triangle has 6 symmetries: counterclockwise rotationthrough 120◦ (r1) or 240

◦ (r2), reflection in an axis of symmetry (`1), (`2),or (`3) and do nothing (or rotate through 360

◦) (e).

`1

`2 `3


Composition of Symmetries

Transformations can be combined using composition. a ◦ b means to dotransformation b then transformation a. Composing any two symmetriesresults in another symmetry. For example `1 ◦ `2 yields

`1

`2 `3

−→̀2

`1

`2 `3

−→̀1

`1

`2 `3

which is the same as r1

`1

`2 `3

−→r1

`1

`2 `3



Yet when we calculate `2 ◦ `1 we get

`1

`2 `3

−→̀1

`1

`2 `3

−→̀2

`1

`2 `3

which is the same as r2

`1

`2 `3

−→r2

`1

`2 `3

Thus `1 ◦ `2 = r1 6= r2 = `2 ◦ `1, so composition is not commutative.



Using G = {e, r1, r2, `1, `2, `3} and ⊕ = ◦ forms an non-Abelian groupcalled the dihedral group of order 6, D6.

◦ e r1 r2 `1 `2 `3e e r1 r2 `1 `2 `3r1 r1 r2 e `3 `1 `2r2 r2 e r1 `2 `3 `1`1 `1 `2 `3 e r1 r2`2 `2 `3 `1 r2 e r1`3 `3 `1 `2 r1 r2 e

Table: Table of composition of symmetries



N – closed under + and ×, not for − and ÷W – closed under + and ×, not for − and ÷Z – closed under +, − and ×, not ÷, (Z,+) is a groupRational numbers, Q =

{ab |a, b ∈ Z, b 6= 0

}closed under addition,



closed under division

all equations ax + b = 0, with a, b ∈ Q have solutions in Q(Q,+) and (Q \ {0},×) are groups


Rings and Fields

A ring is a collection of elements, R, along with two binary operators, ⊕and �, that satisfy the following conditions:

R is closed under both ⊕ and �,(R,⊕) is an Abelian group,� is associative,the distributive laws hold, that is for all x , y ∈ R we have

(x ⊕ y)� z = (x � z) + (y � z)

andx � (y ⊕ z) = (x � y)⊕ (x � z)

A ring is called commutative if � is also commutative. A ring is said tohave an identity (or contain a 1) if there is an element 1 ∈ R such that1× a = a× 1 = a for all a ∈ R.A field is a commutative ring with identity in which all non-zero elementshave a multiplicative inverse.Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 12 / 110


N – closed under + and ×, not for − and ÷W – closed under + and ×, not for − and ÷Z – closed under +, − and ×, not ÷, (Z,+) is a group, (Z,+,×) is a ringRational numbers, Q =

{ab |a, b ∈ Z, b 6= 0

}closed under addition,



closed under division

all equations ax + b = 0, with a, b ∈ Q have solutions in Q(Q,+) and (Q \ {0},×) are groups, (Q,+,×) is a ring, Q is a field


Measurement: The Square

s

A = s2


The Perfect Squares

1

11

21

31

41

51

61

71

81

91

2

12

22

32

42

52

62

72

82

92

3

13

23

33

43

53

63

73

83

93

4

14

24

34

44

54

64

74

84

94

5

15

25

35

45

55

65

75

85

95

6

16

26

36

46

56

66

76

86

96

7

17

27

37

47

57

67

77

87

97

8

18

28

38

48

58

68

78

88

98

9

19

29

39

49

59

69

79

89

99

10

20

30

40

50

60

70

80

90

100

1 4 9

16

25

36

49

64

81

100


Consecutive Squares

02 + 1 = 1 = 12

12 + 3 = 4 = 22

22 + 5 = 9 = 32

32 + 7 = 16 = 42

42 + 9 = 25 = 52


Squares as Sums of Odd Numbers

Thus

1 = 12

1 + 3 = 22

1 + 3 + 5 = 32

....

1 + 3 + 5 + · · ·+ (2n − 1) = n2

....

Note

n2 − (n − 1)2 = n2 − (n2 − 2n + 1)= 2n − 1


The Geoboard Problem

How many different areas of squares are possible on an 11× 11 pingeoboard?

16



1, 4, 9, 16, 25, 36, 49, 64, 81, 100



2, 8, 18, 32, 50 = 2× 1, 2× 4, 2× 9, 2× 16, 2× 25



A square with area of 13 square units.




c

a

b


The Pythagorean Theorem

If ABC is a right angled triangle with legs a and b, and hypotenuse c

C

A

Ba

bc

thena2 + b2 = c2


Visual Proof of the Pythagorean Theorem

a2

b2c2


Measurement: The Square Revisted

s

A = s2


Measurement: The Square Revisited

s =√A

A


What About√

2?

1

1

√2


Continued Fraction Proof of Irrationality of√

2

A little algebraic manipulation yields

√2 = 1 + (−1 +

√2)

= 1 + (−1 +√

2)

(1 +√

2

1 +√

2

)= 1 +

1

1 +√

2



2

Now we can substitute our expression into itself

√2 = 1 +

1

1 +√

2

= 1 +1

1 + 1 + 11+√

2



2

and again . . .

√2 = 1 +

1

1 +√

2

= 1 +1

2 + 11+√

2

= 1 +1

2 + 11+1+ 1

1+√

2



2

√2 = 1 +

1

2 + 12+ 1

2+ 1

2+ 12+···

The convergents are

1

1,

3

2,

7

5,

17

12,

41

29,

99

70,

239

169,

577

408,

1393

985, · · ·

Note that

√2 = 1.41421 . . .

99

70= 1.41428 . . .

141

100= 1.41


Hurwitz’s Theorem

For every irrational number α there are infinitely many relatively primeintegers m and n such that ∣∣∣α− m

n

∣∣∣ < 1√5 n2

.

The convergents of the continued fraction expansion of α satisfy Hurwitz’stheorem.


Number Systems: Polynomial Equations

N – closed under + and ×, not for − and ÷W – closed under + and ×, not for − and ÷Z – closed under +, − and ×, not ÷; (Z,+) is a group, (Z,+,×) is a ringQ – closed under +, −, ×, and ÷; (Q,+) and (Q \ {0},×) are groups,(Q,+,×) is a ring, Q is a field. Some convergent sequences have limitoutside Q. Some polynomials not solvable.Real numbers, R

closed under addition,



closed under division,

(R,+) and (R \ {0},×) are groups, (R,+,×) is a ring, R is a field,all convergent sequences in R has limit in R,all equations ax + b = 0, with a, b ∈ R have solutions in R,many polynomials (not all) “unsolvable” in Q, are solvable in R.


The Quadratic Polynomial f (x) = ax2 + bx + c

Consider the polynomial function

f (x) = ax2 + bx + c, a, b, c ∈ R, a 6= 0

then it is well known that the equation f (x) = 0 has solutions

x =−b ±

√b2 − 4ac

2a.


The Discriminant

The discriminant, Dn, of the degree n polynomial function

f (x) = anxn + an−1x

n−1 + · · ·+ a2x2 + a1x + a0, ai ∈ R

is a function of the coefficients Dn(a0, a1, . . . , an) such that

Dn(a0, a1, . . . , an) = 0 if and only if f has at least one multiple root,

if Dn(a0, a1, . . . , an) < 0 then f has some non-real roots,

if f has n distinct real roots then Dn(a0, a1, . . . , an) > 0.


The Discriminant of a Quadratic Polynomial

In particular, for the quadratic polynomial

f (x) = ax2 + bx + c, a, b, c ∈ R, a 6= 0

the discriminant is D = b2 − 4ac, where

if D > 0 then f has two distinct real roots,

if D = 0 then f has a repeated real root,

if D < 0 then f has no real roots,

if D is a perfect square, then f has two distinct rational roots and fcan be factored into two linear factors with rational or integercoefficients.


The Polynomial f (x) = x2 + 1

Consider the polynomial f (x) = x2 + 1, its roots are the solution to theequation

x2 + 1 = 0

x2 = −1

for which there are no real roots.

Note: a = 1, b = 0, c = 1 so D = 02 − 4(1)(1) = −4.

Thus there are degree n polynomials with real coefficients that do nothave n real roots (counting multiplicities).


The Complex Numbers C

If we define a number number i , the imaginary unit, such that

i2 = −1

then we can define a new number system

C = {a + bi |a, b ∈ R}

called the complex numbers.


Number Systems: Polynomial Equationsanx

n + an−1xn−1 + · · ·+ a2x2 + a1x + a0 = 0

N,W – closed under + and ×, not for − and ÷Z – closed under +, − and ×, not ÷; (Z,+) is a group, (Z,+,×) is a ringQ – is a field; some convergent sequences have limit outside Q; somepolynomials not solvable.R – is a field; all convergent sequences have limit in R; some polynomialsnot solvable.Complex numbers, C

is a field,

all convergent sequences in C has limit in C,all polynomial equations anx

n + an−1xn−1 + · · ·+ a2x2 + a1x + a0 = 0,

with ai ∈ C have n solutions in C (counting multiplicities).


The Graph of a Quadratic Function

The graph with equation y = ax2 + bx + c is a parabola

x

y

y = ax2 + bx + c


Conic Sections

Consider the double cone sliced by various planes.


Conic Sections


The Circle

A circle is the locus of points that are a fixed distance, called the radiusof the circle, from a fixed point called the centre of the circle.

radius

centre


The Ellipse

An ellipse is the locus of points such that the sum of the distances to twofixed points, called the foci (singular focus), is a constant.

major axis

minor axis

P PF1 + PF2 = constant

focusF1

focusF2


The Parabola

A parabola is a locus of points such that the distance from a point on theparabola to a fixed point, called the focus, is equal to the distance to afixed line, called the directrix.

PF = PD

directrix D

focusF

P


The Hyperbola

A hyperbola is the locus of points such that the difference of thedistances to two fixed points, called the foci, is a constant.

major axis

minor axisP

focus

F1

focus

F2

|PF1 − PF2| = constantasymptote asymptote


Equations of Conic Sections

The equationAx2 + Bxy + Cy2 + Dx + Ey + F = 0

describes a (possibly degenerate) conic section.

The discriminant D = B2 − 4AC tells us the conic is

an ellipse if D < 0 (and a circle if A = C and B = 0),

a parabola if D = 0,

a hyperbola if D > 0.


Binary Quadratic Form

A form is a homogeneous polynomial, that is a polynomial where eachterm has the same degree.

Specifically, a binary quadratic form is a homogeneous polynomial in twovariables of degree 2, that is a polynomial of the form

f (x , y) = ax2 + bxy + cy2.


The Discriminant of a Binary Quadratic Form

Multiplying the binary quadratic form

f (x , y) = ax2 + bxy + cy2

by 4a and completing the square yields

4af (x , y) = 4a2x2 + 4abxy + 4acy2

= (2ax)2 + 2(2a)(by) + (by)2 − (by)2 + 4acy2

= (2ax + by)2 − (b2 − 4ac)y2

= (2ax + by)2 −∆y2

where ∆ = b2 − 4ac is called the discriminant.


Properties of the Discriminant of a Binary Quadratic Form

Since∆ = b2 − 4ac

we have

∆ ≡ b2 − 4ac (mod 4)≡ b2 (mod 4)

and hence ∆ ≡ 0, 1 (mod 4).


Existence of a Form with a Given Discriminant

If ∆ ≡ 0 (mod 4) then ∆4 is an integer, and

x2 −(

∆

4

)y2

is a binary quadratic form with discriminant ∆.

Similarly, if ∆ ≡ 1 (mod 4) then ∆−14 is an integer, and

x2 + xy −(

∆− 14

)y2

is a binary quadratic form with discriminant ∆.

Hence, for every ∆ ≡ 0, 1 (mod 4) there exists at least one binaryquadratic form with discriminant ∆.


Existence of a Form with a Given Discriminant: Examples

Some binary quadratic forms with given discriminant:

Case 1: ∆ ≡ 0 (mod 4)

if ∆ = 20: x2 −(

20

4

)y2 = x2 − 5y2,

if ∆ = −44: x2 −(−44

4

)y2 = x2 + 11y2,

Case 2: ∆ ≡ 1 (mod 4)

if ∆ = 5: x2 + xy −(

5− 14

)y2 = x2 + xy − y2,

if ∆ = −11: x2 + xy −(−11− 1

4

)y2 = x2 + xy + 3y2.


Representation of n by a Binary Quadratic Form

We say that a binary quadratic form

f (x , y) = ax2 + bxy + cy2

represents an integer n, if there exists integers x0 and y0 such that

f (x0, y0) = n.

If gcd(x0, y0) = 1 then the representation is called proper, otherwise it iscalled improper.


Representation Problems

The following representation problems are of interest:

Which integers do the form f represent?

Which forms represent the integer n?

How many ways does the form f represent the integer n?


Types of Binary Quadratic Forms

A binary quadratic form f (x , y) = ax2 + bxy + cy2 can be one of threetypes.

Indefinite if f takes on both positive and negative values. Thishappens when ∆ > 0.

Semi-definite if f (x , y) ≥ 0 (positive semi-definite) or f (x , y) ≤ 0(negative semi-definite) for all integer values of x and y . Thishappens when ∆ ≤ 0.

Definite if it is semi-definite and the only solution to f (x , y) = 0 isx = y = 0. This happens when ∆ < 0 and thus a and c have thesame sign. Thus we can have positive definite (if a, c > 0) ornegative definite (if a, c < 0) forms.


Improper Representation

Suppose that n is represented by (x0, y0) with gcd(x0, y0) = d > 1, thenx0 = dX and y0 = dY for some integers X and Y with gcd(X ,Y ) = 1.Thus

f (x0, y0) = n

ax20 + bx0y0 + cy20 = n

a(dX )2 + b(dX )(dY ) + c(dY )2 = n

d2(aX 2 + bXY + cY 2) = n

which implies that d2 | n, and f properly represents nd2

.


Example of Proper and Improper Representation

Consider the binary quadratic form

f (x , y) = x2 + y2

then x = 7, y = 1 is a proper representation of 50 since

f (7, 1) = 72 + 12 = 50

and gcd(1, 7) = 1, yet x = y = 5 is an improper representation of 50 since

f (5, 5) = 52 + 52 = 50

and gcd(5, 5) = 5 = d > 1. Hence d2 = 25 | 50, so x = y = 55 = 1 is aproper representation of 5025 = 2 as

f (1, 1) = 12 + 12 = 2.


Solution Set to x2 + y 2 = 50

x

y



x

y



x

y

x2 + y2 = 2


Forms Representing 0

If ∆ is a perfect square, or 0, then√

∆ is a positive integer and

4af (x , y) = (2ax + (b +√

∆)y)(2ax + (b −√

∆)y).

Thus our form is factorable, and so f (x , y) = 0 has many solutions.

If ∆ is a not perfect square, nor 0, then the only solution to f (x , y) = 0 isx = y = 0.


Examples of Forms Representing 0

If ∆ = 16 = 42, then

f (x , y) = x2 −(

16

4

)y2 = x2 − 4y2

has the given discriminant and hence

f (x , y) = (x + 2y)(x − 2y)

so any solution to x + 2y = 0 or x − 2y = 0 satisfies f (x , y) = 0, that is

f (±2k, k) = 0, ∀k ∈ Z.


Solution Set to x2 − 4y 2 = 0

x

y


More on Forms Representing 0

If we want to find all integer solutions to f (x , y) = x2 − 4y2 = 21 thenfactoring yields

(x + 2y)(x − 2y) = 21.

Since x , y ∈ Z, then (x + 2y), (x − 2y) ∈ Z, so (x + 2y) | 21 and(x − 2y) | 21. Each pair of factors of 21 yields a system of equationswhich yield a solution to the original equation. For example, using3× 7 = 21 gives

x + 2y = 3 (1)

x − 2y = 7 (2)

which has solution x = 5, y = −1. The full solution set is

(x , y) ∈ {(±5,±1), (±11,±5)}.


Solution Set to x2 − 4y 2 = 21

x

y


Equivalence of Binary Quadratic Forms

Consider the formf (x , y) = 7x2 + 3y2

which represents 103 four ways as

f (±2,±5) = 103.Consider the new form g defined by

g(x , y) = f (2x + y , x + y)

= 7(2x + y)2 + 3(x + y)2

= 31x2 + 34xy + 10y2.

Solving the system

2x + y = 2

x + y = 5

yields x = −3, y = 8, which impliesf (2, 5) = g(−3, 8) = 103


Equivalence of Binary Quadratic Forms

Looking at all the representations of 103 we get

f (2, 5) = g(−3, 8) = 103 f (2,−5) = g(7,−12) = 103f (−2, 5) = g(−7, 12) = 103 f (−2,−5) = g(3,−8) = 103

x

y


Linear Transformation of a Binary Quadratic Form

Starting with the form

f (x , y) = ax2 + bxy + cy2

if we define a new form

f ′(x , y) = f (αx + βy , γx + δy) = a′x2 + b′xy + c ′y2

then

a′ = aα2 + bαγ + cγ2

b′ = b(αδ + βγ) + 2(aαβ + cγδ)

c ′ = aβ2 + bβδ + cδ2.


Linear Transformation of a Binary Quadratic Form

The discriminant of the new form will be

∆′ = b′2 − 4a′c ′

= (αδ − βγ)2(b2 − 4ac)= (αδ − βγ)2∆

so that if(αδ − βγ)2 = 1

then∆′ = ∆.


Equivalent Forms

If two forms, f and g , are related by a transformation of the same typewith αδ − βγ = +1, then the forms are called properly equivalent andwe write

f ∼ g .

If two forms are equivalent, they have the same discriminant and theyrepresent the same integers.From our example

7x2 + 3y2 ∼ 31x2 + 34xy + 10y2.


Reduced Positive Definite Forms

A positive definite form

f (x , y) = ax2 + bxy + cy2, a, c > 0, b2 − 4ac < 0

is called reduced if

−a < b ≤ a ≤ c , with b ≥ 0 if c = a.

For example

7x2 + 3y2 and 31x2 + 34xy + 10y2

are unreduced forms but3x2 + 7y2

is reduced.


The Reduction Algorithm

If f (x , y) = ax2 + bxy + cy2 is a positive definite form then we can find aninteger δ such that

| − b + 2cδ| ≤ c

thenax2 + bxy + cy2 ∼ a′x2 + b′xy + c ′y2

where |b′| ≤ a′ and

a′ = c

b′ = −b + 2cδc ′ = a− bδ + cδ2.

If a′ ≤ c ′ you are done, if not repeat the process.


Example: Reducing 31x2 + 34xy + 10y 2

To reduce 31x2 + 34xy + 10y2, we need a δ such that

| − 34 + 2(10)δ| ≤ 10

which is satisfied by δ = 2, thus we get

a′ = c = 10

b′ = −b + 2cδ = −34 + 2(10)(2) = 6c ′ = a− bδ + cδ2 = 31− 34(2) + 10(2)2 = 3

so31x2 + 34xy + 10y2 ∼ 10x2 + 6xy + 3y2

which is unreduced. If we perform the process one more time we get thereduced form

31x2 + 34xy + 10y2 ∼ 10x2 + 6xy + 3y2 ∼ 3x2 + 7y2.


The Class Number

For each discriminant ∆ < 0 there are a number of classes of equivalentforms. Each class contains a unique reduced form. The number of classesfor a given discriminant ∆ < 0 is called the class number, h(∆).

For example, h(−84) = 4 so there are 4 equivalence classes of forms withdiscriminant −84. The reduced forms in the classes are

x2 + 21y2, 2x2 + 2xy + 11y2, 3x2 + 7y2, 5x2 + 4xy + 5y2

Each class will represent its own set of numbers.

The classes form an Abelian group called the class group where the groupoperation is called composition.


Numbers Represented by the Form f (x , y) = x2 + y 2

1

11

21

31

41

51

61

71

81

91

2

12

22

32

42

52

62

72

82

92

3

13

23

33

43

53

63

73

83

93

4

14

24

34

44

54

64

74

84

94

5

15

25

35

45

55

65

75

85

95

6

16

26

36

46

56

66

76

86

96

7

17

27

37

47

57

67

77

87

97

8

18

28

38

48

58

68

78

88

98

9

19

29

39

49

59

69

79

89

99

10

20

30

40

50

60

70

80

90

100

1 4 9

16

25

36

49

64

81

100



1

11

21

31

41

51

61

71

81

91

2

12

22

32

42

52

62

72

82

92

3

13

23

33

43

53

63

73

83

93

4

14

24

34

44

54

64

74

84

94

5

15

25

35

45

55

65

75

85

95

6

16

26

36

46

56

66

76

86

96

7

17

27

37

47

57

67

77

87

97

8

18

28

38

48

58

68

78

88

98

9

19

29

39

49

59

69

79

89

99

10

20

30

40

50

60

70

80

90

100

1 4 9

16

25

36

49

64

81

100

2 5 10

17

26

37

50

65

82



1

11

21

31

41

51

61

71

81

91

2

12

22

32

42

52

62

72

82

92

3

13

23

33

43

53

63

73

83

93

4

14

24

34

44

54

64

74

84

94

5

15

25

35

45

55

65

75

85

95

6

16

26

36

46

56

66

76

86

96

7

17

27

37

47

57

67

77

87

97

8

18

28

38

48

58

68

78

88

98

9

19

29

39

49

59

69

79

89

99

10

20

30

40

50

60

70

80

90

100

1 4 9

16

25

36

49

64

81

100

2 5 10

17

26

37

50

65

82

8

13 20

29

40

53

68

85



1

11

21

31

41

51

61

71

81

91

2

12

22

32

42

52

62

72

82

92

3

13

23

33

43

53

63

73

83

93

4

14

24

34

44

54

64

74

84

94

5

15

25

35

45

55

65

75

85

95

6

16

26

36

46

56

66

76

86

96

7

17

27

37

47

57

67

77

87

97

8

18

28

38

48

58

68

78

88

98

9

19

29

39

49

59

69

79

89

99

10

20

30

40

50

60

70

80

90

100

1 4 9

16

25

36

49

64

81

100

2 5 10

17

26

37

50

65

82

8

13 20

29

40

53

68

85

18

34

45

58

73

90



1

11

21

31

41

51

61

71

81

91

2

12

22

32

42

52

62

72

82

92

3

13

23

33

43

53

63

73

83

93

4

14

24

34

44

54

64

74

84

94

5

15

25

35

45

55

65

75

85

95

6

16

26

36

46

56

66

76

86

96

7

17

27

37

47

57

67

77

87

97

8

18

28

38

48

58

68

78

88

98

9

19

29

39

49

59

69

79

89

99

10

20

30

40

50

60

70

80

90

100

1 4 9

16

25

36

49

64

81

100

2 5 10

17

26

37

50

65

82

8

13 20

29

40

53

68

85

18

34

45

58

73

90

32

41

52

65

80

97

61

74

89

72

85

98



1 37 73

5 41 77

9 45 81

13 49 85

17 53 89

21 57 93

25 61 97

29 65 101

33 69 105

2 38 74

6 42 78

10 46 82

14 50 86

18 54 90

22 58 94

26 62 98

30 66 102

34 70 106

3 39 75

7 43 79

11 47 83

15 51 87

19 55 91

23 59 95

27 63 99

31 67 103

35 71 107

4 40 76

8 44 80

12 48 84

16 52 88

20 56 92

24 60 96

28 64 100

32 68 104

36 72 108

1 2 4

5 8

9 10

13 16

17 18 20

25 26

29 32

34 36

37 40

41

45

49 50 52

53

58

61 64

65 68

72

73 74

80

81 82

85

89 90

97 98 100

101 104

106


Sums of Squares Modulo 4

hin n2 (mod 4)

0 01 12 03 1

m2 + n2 (mod 4)m\n 0 1 2 3

0 0 1 0 11 1 2 1 22 0 1 0 13 1 2 1 2

.


Writing n as a Sum of Two Squares

Diophantus–Brahmagupta–Fibonacci identity:

(a2 + b2)(c2 + d2) = (ac − bd)2 + (ad + bc)2

Theorem: If p ≡ 1 (mod 4) is a prime, then there exists positive integersa and b such that a2 + b2 = p.

Theorem (Fermat): If n is factored into primes as

n = 2α∏i

pβii

∏j

qγjj

where pi and qj are primes with pi ≡ 1 (mod 4) and qj ≡ 3 (mod 4), forall i and j , then n can be expressed as a sum of two squares if and only ifγj is even for all j .


Examples of D-B-F Identity

1 37 73

5 41 77

9 45 81

13 49 85

17 53 89

21 57 93

25 61 97

29 65 101

33 69 105

2 38 74

6 42 78

10 46 82

14 50 86

18 54 90

22 58 94

26 62 98

30 66 102

34 70 106

3 39 75

7 43 79

11 47 83

15 51 87

19 55 91

23 59 95

27 63 99

31 67 103

35 71 107

4 40 76

8 44 80

12 48 84

16 52 88

20 56 92

24 60 96

28 64 100

32 68 104

36 72 108

1 2 4

5 8

9 10

13 16

17 18 20

25 26

29 32

34 36

37 40

41

45

49 50 52

53

58

61 64

65 68

72

73 74

80

81 82

85

89 90

97 98 100

101 104

106



1 37 73

5 41 77

9 45 81

13 49 85

17 53 89

21 57 93

25 61 97

29 65 101

33 69 105

2 38 74

6 42 78

10 46 82

14 50 86

18 54 90

22 58 94

26 62 98

30 66 102

34 70 106

3 39 75

7 43 79

11 47 83

15 51 87

19 55 91

23 59 95

27 63 99

31 67 103

35 71 107

4 40 76

8 44 80

12 48 84

16 52 88

20 56 92

24 60 96

28 64 100

32 68 104

36 72 108

1 2 4

5 8

9 10

13 16

17 18 20

25 26

29 32

34 36

37 40

41

45

49 50 52

53

58

61 64

65 68

72

73 74

80

81 82

85

89 90

97 98 100

101 104

106




(a2 + b2)(c2 + d2) = (ac − bd)2 + (ad + bc)2



n = 2α∏i

pβii

∏j

qγjj



Primes p ≡ 1 (mod 4)

1 37 73

5 41 77

9 45 81

13 49 85

17 53 89

21 57 93

25 61 97

29 65 101

33 69 105

2 38 74

6 42 78

10 46 82

14 50 86

18 54 90

22 58 94

26 62 98

30 66 102

34 70 106

3 39 75

7 43 79

11 47 83

15 51 87

19 55 91

23 59 95

27 63 99

31 67 103

35 71 107

4 40 76

8 44 80

12 48 84

16 52 88

20 56 92

24 60 96

28 64 100

32 68 104

36 72 108

1 2 4

5 8

9 10

13 16

17 18 20

25 26

29 32

34 36

37 40

41

45

49 50 52

53

58

61 64

65 68

72

73 74

80

81 82

85

89 90

97 98 100

101 104

106




(a2 + b2)(c2 + d2) = (ac − bd)2 + (ad + bc)2



n = 2α∏i

pβii

∏j

qγjj



Sum of Two Squares Example

n = 2× 13× 32 = 234Since

2 = 12 + 12, 13 = 22 + 32, 9 = 02 + 32

we have

26 = 2× 13 = (12 + 12)(22 + 32)= (1× 2− 1× 3)2 + (1× 3 + 1× 2)2

= (−1)2 + 52

= 12 + 52

so

234 = 26× 9 = (12 + 52)(02 + 32)= (1× 0− 5× 3)2 + (1× 3 + 5× 0)2

= 152 + 32


A Curious Result

Suppose α = a + ib and β = c + id are two complex numbers(a, b, c, d ∈ R), then

α× β = (a + ib)(c + id)= ac + iad + ibc + (i2)bd

= (ac − bd) + i(ad + bc)


(a2 + b2)(c2 + d2) = (ac − bd)2 + (ad + bc)2


Modulus of a Complex Number

Recall for a complex number z = x + iy , x , y ∈ R, the modulus of z , |z |,satisfies

|z |2 = zz̄ = (x + iy)(x − iy) = x2 + y2

or|z | =

√x2 + y2.

<

=

z = x + iy


A Curious Result Revisited

Suppose α, β ∈ C with α = a + ib and β = c + id , then

αβ = (ac − bd) + i(ad + bc)

Thus

|α|2 = a2 + b2,|β|2 = c2 + d2,

|α× β|2 = (ac − bd)2 + (ad + bc)2,

so the Diophantus–Brahmagupta–Fibonacci identity tells us

|αβ|2 = |α|2|β|2

which, since |z | ≥ 0, is equivalent to

|αβ| = |α||β|.


A Curious Identity

We can write(x21 + x

22 )(y

21 + y

22 ) = z

21 + z

22

where

z1 = x1y1 − x2y2z2 = x1y2 + x2y1

as a statement of|X ||Y | = |XY |

where X ,Y ∈ C with X = x1 + ix2 and Y = y1 + iy2.


Another Curious Identity

We can also write

(x21 + x22 + x

23 + x

24 )(y

21 + y

22 + y

23 + y

24 ) = z

21 + z

22 + z

23 + z

24

where

z1 = x1y1 − x2y2 − x3y3 − x4y4z2 = x1y2 + x2y1 + x3y4 − x4y3z3 = x1y3 + x3y1 − x2y4 + x4y2z4 = x1y4 + x4y1 + x2y3 − x3y2


More on Sums of Squares

Sums of Three Squares: Every positive integer n can be written in theform

n = a2 + b2 + c2, a, b, c ∈ Z

except for those n of the form n = 4a(8b + 7) where a and b arenon-negative integers.

Sums of Four Squares: Every positive integer n can be written in theform

n = a2 + b2 + c2 + d2, a, b, c , d ∈ Z.


Quaternions

If we define three distinct new numbers, i , j , and k , that satisfy

i2 = −1 j2 = −1 k2 = −1ij = k jk = i ki = j

then ifq = a + bi + cj + dk

we call q a quaternion and the set of all quaternions is denoted H.

Using the definitions of i , j , and k , we find that

ji = −k = −ij kj = −i = −jkik = −j = −ki ijk = −1

so multiplication of quaternions is not commutative.


Yet Another Curious Identity

We can also write

(x21 + x22 + · · ·+ x28 )(y21 + y22 + · · ·+ y28 ) = z21 + z22 + · · ·+ z28

where

z1 = x1y1 − x2y2 − x3y3 − x4y4 − x5y5 − x6y6 − x7y7 − x8y8,z2 = x1y2 + x2y1 + x3y4 − x4y3 + x5y6 − x6y5 − x7y8 + x8y7,z3 = x1y3 + x3y1 − x2y4 + x4y2 + x5y7 − x7y5 + x6y8 − x8y6,z4 = x1y4 + x4y1 + x2y3 − x3y2 + x5y8 − x8y5 − x6y7 + x7y6,z5 = x1y5 + x5y1 − x2y6 + x6y2 − x3y7 + x7y3 − x4y8 + x8y4,z6 = x1y6 + x6y1 + x2y5 − x5y2 − x3y8 + x8y3 + x4y7 − x7y4,z7 = x1y7 + x7y1 + x2y8 − x8y2 + x3y5 − x5y3 − x4y6 + x6y4,z8 = x1y8 + x8y1 − x2y7 + x7y2 + x3y6 − x6y3 + x4y5 − x5y4.


Vector Spaces

A set V is said to be a vector space over a field F if (V ,+) is an Abeliangroup and for each a ∈ F and v ∈ V there is an element av ∈ V suchthat:

a(u + v) = au + av ,

(a + b)v = av + bv ,

a(bv) = (ab)v ,

1v = v ,

for all a, b ∈ F and for all u, v ∈ V , where 1 ∈ F is the multiplicativeidentity.

If v ∈ V , then v is called a vector.

If a ∈ F , then a is called a scalar.


Normed Algebras

A ring R is called an algebra over a field F if R is a vector space over Fand

(au)× (bv) = (ab)(u × v)

for all scalars a, b,∈ F and all vectors u, v ∈ R, where × representsmultiplication within the ring.

A norm, ‖ · ‖, of a vector space V over a field F , is a function‖ · ‖ : V → R such that:‖0‖ = 0,‖v‖ > 0 for all v 6= 0 ∈ V ,‖av‖ = |a|‖v‖ for all a ∈ F and for all v ∈ V ,‖u + v‖ ≤ ‖u‖+ ‖v‖.

An algebra with a norm is called a normed algebra.


Examples of Normed Algebras

The complex numbers C with ‖z‖ = |z | for all z ∈ C,

Three dimensional Euclidean vectors R3 with the cross product withthe Euclidean norm ‖(x , y , z)‖ =

√x2 + y2 + z2,

The quaternions H with ‖a + bi + cj + dk‖ =√a2 + b2 + c2 + d2.

The octonions O with

‖a0 + a1i1 + · · ·+ a7i7‖ =√a20 + a

21 + · · ·+ a27


Adding Something and Losing Something

The real numbers R as a normed algebra, is an ordered set where × iscommutative and associative.

The complex numbers C as a normed algebra, is a non-ordered set where× is commutative and associative.

The quaternions H as a normed algebra, is a non-ordered set where × isnon-commutative but is associative.

The octonions O as a normed algebra, is a non-ordered set where × isnon-commutative and non-associative.



How many different areas of squares are possible on an 11× 11 pingeoboard?

16


The End


Documents

Shawn GodinShawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 11 / 110 Rings and Fields A ring is a collection of elements, R, along with two binary operators, and , that