√ Continued Fraction Proof of Irrationality of 2 A little algebraic manipulation yields √ √ 2 = 1 + ( − 1 + 2) √ � � √ 1 + 2 = 1 + ( − 1 + 2) √ 1 + 2 1 = 1 + √ 1 + 2 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 31 / 110
√ Continued Fraction Proof of Irrationality of 2 Now we can substitute our expression into itself √ 1 2 = 1 + √ 1 + 2 1 = 1 + 1 1 + 1 + √ 1+ 2 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 32 / 110
√ Continued Fraction Proof of Irrationality of 2 and again . . . √ 1 2 = 1 + √ 1 + 2 1 = 1 + 1 2 + √ 1+ 2 1 = 1 + 1 2 + 1 1+1+ √ 1+ 2 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 33 / 110
√ Continued Fraction Proof of Irrationality of 2 √ 1 2 = 1 + 1 2 + 1 2+ 1 2+ 1 2+ 2+ ··· 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 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 34 / 110
Hurwitz’s Theorem For every irrational number α there are infinitely many relatively prime integers m and n such that 1 � α − m � � √ � < 5 n 2 . � � n The convergents of the continued fraction expansion of α satisfy Hurwitz’s theorem. Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 35 / 110
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 ring Q – closed under +, − , × , and ÷ ; ( Q , +) and ( Q \ { 0 } , × ) are groups, ( Q , + , × ) is a ring, Q is a field. Some convergent sequences have limit outside Q . Some polynomials not solvable. Real numbers , R closed under addition, closed under subtraction, closed under multiplication, 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 . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 36 / 110
The Quadratic Polynomial f ( x ) = ax 2 + bx + c Consider the polynomial function f ( x ) = ax 2 + bx + c , a , b , c ∈ R , a � = 0 then it is well known that the equation f ( x ) = 0 has solutions √ b 2 − 4 ac x = − b ± . 2 a Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 37 / 110
The Discriminant The discriminant , D n , of the degree n polynomial function f ( x ) = a n x n + a n − 1 x n − 1 + · · · + a 2 x 2 + a 1 x + a 0 , a i ∈ R is a function of the coefficients D n ( a 0 , a 1 , . . . , a n ) such that D n ( a 0 , a 1 , . . . , a n ) = 0 if and only if f has at least one multiple root, if D n ( a 0 , a 1 , . . . , a n ) < 0 then f has some non-real roots, if f has n distinct real roots then D n ( a 0 , a 1 , . . . , a n ) > 0. Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 38 / 110
The Discriminant of a Quadratic Polynomial In particular, for the quadratic polynomial f ( x ) = ax 2 + bx + c , a , b , c ∈ R , a � = 0 the discriminant is D = b 2 − 4 ac , 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 f can be factored into two linear factors with rational or integer coefficients. Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 39 / 110
The Polynomial f ( x ) = x 2 + 1 Consider the polynomial f ( x ) = x 2 + 1, its roots are the solution to the equation x 2 + 1 = 0 x 2 = − 1 for which there are no real roots. Note: a = 1, b = 0, c = 1 so D = 0 2 − 4(1)(1) = − 4. Thus there are degree n polynomials with real coefficients that do not have n real roots (counting multiplicities). Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 40 / 110
The Complex Numbers C If we define a number number i , the imaginary unit , such that i 2 = − 1 then we can define a new number system C = { a + bi | a , b ∈ R } called the complex numbers . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 41 / 110
Number Systems: Polynomial Equations a n x n + a n − 1 x n − 1 + · · · + a 2 x 2 + a 1 x + a 0 = 0 N , W – closed under + and × , not for − and ÷ Z – closed under +, − and × , not ÷ ; ( Z , +) is a group, ( Z , + , × ) is a ring Q – is a field; some convergent sequences have limit outside Q ; some polynomials not solvable. R – is a field; all convergent sequences have limit in R ; some polynomials not solvable. Complex numbers , C is a field, all convergent sequences in C has limit in C , all polynomial equations a n x n + a n − 1 x n − 1 + · · · + a 2 x 2 + a 1 x + a 0 = 0, with a i ∈ C have n solutions in C (counting multiplicities). Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 42 / 110
The Graph of a Quadratic Function The graph with equation y = ax 2 + bx + c is a parabola y y = ax 2 + bx + c x Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 43 / 110
Conic Sections Consider the double cone sliced by various planes. Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 44 / 110
Conic Sections Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 45 / 110
The Circle A circle is the locus of points that are a fixed distance, called the radius of the circle, from a fixed point called the centre of the circle. radius centre Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 46 / 110
The Ellipse An ellipse is the locus of points such that the sum of the distances to two fixed points, called the foci (singular focus ), is a constant. P PF 1 + PF 2 = constant minor axis focus focus major axis F 1 F 2 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 47 / 110
The Parabola A parabola is a locus of points such that the distance from a point on the parabola to a fixed point, called the focus, is equal to the distance to a fixed line, called the directrix . PF = PD F focus P directrix D Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 48 / 110
The Hyperbola A hyperbola is the locus of points such that the difference of the distances to two fixed points, called the foci, is a constant. asymptote asymptote | PF 1 − PF 2 | = constant minor axis P F 1 F 2 focus focus major axis Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 49 / 110
Equations of Conic Sections The equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 describes a (possibly degenerate) conic section. The discriminant D = B 2 − 4 AC 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. Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 50 / 110
Binary Quadratic Form A form is a homogeneous polynomial, that is a polynomial where each term has the same degree. Specifically, a binary quadratic form is a homogeneous polynomial in two variables of degree 2, that is a polynomial of the form f ( x , y ) = ax 2 + bxy + cy 2 . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 51 / 110
The Discriminant of a Binary Quadratic Form Multiplying the binary quadratic form f ( x , y ) = ax 2 + bxy + cy 2 by 4 a and completing the square yields 4 af ( x , y ) = 4 a 2 x 2 + 4 abxy + 4 acy 2 = (2 ax ) 2 + 2(2 a )( by ) + ( by ) 2 − ( by ) 2 + 4 acy 2 = (2 ax + by ) 2 − ( b 2 − 4 ac ) y 2 = (2 ax + by ) 2 − ∆ y 2 where ∆ = b 2 − 4 ac is called the discriminant . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 52 / 110
Properties of the Discriminant of a Binary Quadratic Form Since ∆ = b 2 − 4 ac we have ∆ ≡ b 2 − 4 ac (mod 4) ≡ b 2 (mod 4) and hence ∆ ≡ 0 , 1 (mod 4). Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 53 / 110
Existence of a Form with a Given Discriminant If ∆ ≡ 0 (mod 4) then ∆ 4 is an integer, and � ∆ � x 2 − y 2 4 is a binary quadratic form with discriminant ∆. Similarly, if ∆ ≡ 1 (mod 4) then ∆ − 1 is an integer, and 4 � ∆ − 1 � x 2 + xy − y 2 4 is a binary quadratic form with discriminant ∆. Hence, for every ∆ ≡ 0 , 1 (mod 4) there exists at least one binary quadratic form with discriminant ∆. Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 54 / 110
Existence of a Form with a Given Discriminant: Examples Some binary quadratic forms with given discriminant: Case 1: ∆ ≡ 0 (mod 4) � 20 � if ∆ = 20: x 2 − y 2 = x 2 − 5 y 2 , 4 � − 44 � if ∆ = − 44: x 2 − y 2 = x 2 + 11 y 2 , 4 Case 2: ∆ ≡ 1 (mod 4) � 5 − 1 � if ∆ = 5: x 2 + xy − y 2 = x 2 + xy − y 2 , 4 � − 11 − 1 � if ∆ = − 11: x 2 + xy − y 2 = x 2 + xy + 3 y 2 . 4 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 55 / 110
Representation of n by a Binary Quadratic Form We say that a binary quadratic form f ( x , y ) = ax 2 + bxy + cy 2 represents an integer n , if there exists integers x 0 and y 0 such that f ( x 0 , y 0 ) = n . If gcd( x 0 , y 0 ) = 1 then the representation is called proper , otherwise it is called improper . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 56 / 110
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 ? Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 57 / 110
Types of Binary Quadratic Forms A binary quadratic form f ( x , y ) = ax 2 + bxy + cy 2 can be one of three types. Indefinite if f takes on both positive and negative values. This happens 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 . This happens when ∆ ≤ 0. Definite if it is semi-definite and the only solution to f ( x , y ) = 0 is x = y = 0. This happens when ∆ < 0 and thus a and c have the same sign. Thus we can have positive definite (if a , c > 0) or negative definite (if a , c < 0) forms. Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 58 / 110
Improper Representation Suppose that n is represented by ( x 0 , y 0 ) with gcd( x 0 , y 0 ) = d > 1, then x 0 = dX and y 0 = dY for some integers X and Y with gcd( X , Y ) = 1. Thus f ( x 0 , y 0 ) = n ax 2 0 + bx 0 y 0 + cy 2 0 = n a ( dX ) 2 + b ( dX )( dY ) + c ( dY ) 2 = n d 2 ( aX 2 + bXY + cY 2 ) = n which implies that d 2 | n , and f properly represents n d 2 . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 59 / 110
Example of Proper and Improper Representation Consider the binary quadratic form f ( x , y ) = x 2 + y 2 then x = 7, y = 1 is a proper representation of 50 since f (7 , 1) = 7 2 + 1 2 = 50 and gcd(1 , 7) = 1, yet x = y = 5 is an improper representation of 50 since f (5 , 5) = 5 2 + 5 2 = 50 and gcd(5 , 5) = 5 = d > 1. Hence d 2 = 25 | 50, so x = y = 5 5 = 1 is a proper representation of 50 25 = 2 as f (1 , 1) = 1 2 + 1 2 = 2 . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 60 / 110
Solution Set to x 2 + y 2 = 50 y x Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 61 / 110
Solution Set to x 2 + y 2 = 50 y x Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 62 / 110
Solution Set to x 2 + y 2 = 50 y x Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 63 / 110
Solution Set to x 2 + y 2 = 50 y x Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 64 / 110
Solution Set to x 2 + y 2 = 50 y x Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 65 / 110
Solution Set to x 2 + y 2 = 50 y x 2 + y 2 = 2 x Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 66 / 110
Forms Representing 0 √ If ∆ is a perfect square, or 0, then ∆ is a positive integer and √ √ 4 af ( x , y ) = (2 ax + ( b + ∆) y )(2 ax + ( 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 is x = y = 0. Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 67 / 110
Examples of Forms Representing 0 If ∆ = 16 = 4 2 , then � 16 � f ( x , y ) = x 2 − y 2 = x 2 − 4 y 2 4 has the given discriminant and hence f ( x , y ) = ( x + 2 y )( x − 2 y ) so any solution to x + 2 y = 0 or x − 2 y = 0 satisfies f ( x , y ) = 0, that is f ( ± 2 k , k ) = 0 , ∀ k ∈ Z . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 68 / 110
Solution Set to x 2 − 4 y 2 = 0 y x Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 69 / 110
More on Forms Representing 0 If we want to find all integer solutions to f ( x , y ) = x 2 − 4 y 2 = 21 then factoring yields ( x + 2 y )( x − 2 y ) = 21 . Since x , y ∈ Z , then ( x + 2 y ) , ( x − 2 y ) ∈ Z , so ( x + 2 y ) | 21 and ( x − 2 y ) | 21. Each pair of factors of 21 yields a system of equations which yield a solution to the original equation. For example, using 3 × 7 = 21 gives x + 2 y = 3 (1) x − 2 y = 7 (2) which has solution x = 5, y = − 1. The full solution set is ( x , y ) ∈ { ( ± 5 , ± 1) , ( ± 11 , ± 5) } . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 70 / 110
Solution Set to x 2 − 4 y 2 = 21 y x Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 71 / 110
Equivalence of Binary Quadratic Forms Consider the form f ( x , y ) = 7 x 2 + 3 y 2 which represents 103 four ways as f ( ± 2 , ± 5) = 103 . Consider the new form g defined by g ( x , y ) = f (2 x + y , x + y ) = 7(2 x + y ) 2 + 3( x + y ) 2 = 31 x 2 + 34 xy + 10 y 2 . Solving the system 2 x + y = 2 x + y = 5 yields x = − 3, y = 8, which implies f (2 , 5) = g ( − 3 , 8) = 103 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 72 / 110
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) = 103 f ( − 2 , 5) = g ( − 7 , 12) = 103 f ( − 2 , − 5) = g (3 , − 8) = 103 y x Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 73 / 110
Linear Transformation of a Binary Quadratic Form Starting with the form f ( x , y ) = ax 2 + bxy + cy 2 if we define a new form f ′ ( x , y ) = f ( α x + β y , γ x + δ y ) = a ′ x 2 + b ′ xy + c ′ y 2 then a ′ = a α 2 + b αγ + c γ 2 b ′ = b ( αδ + βγ ) + 2( a αβ + c γδ ) c ′ = a β 2 + b βδ + c δ 2 . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 74 / 110
Linear Transformation of a Binary Quadratic Form The discriminant of the new form will be ∆ ′ = b ′ 2 − 4 a ′ c ′ = ( αδ − βγ ) 2 ( b 2 − 4 ac ) = ( αδ − βγ ) 2 ∆ so that if ( αδ − βγ ) 2 = 1 then ∆ ′ = ∆ . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 75 / 110
Equivalent Forms If two forms, f and g , are related by a transformation of the same type with αδ − βγ = +1, then the forms are called properly equivalent and we write f ∼ g . If two forms are equivalent, they have the same discriminant and they represent the same integers. From our example 7 x 2 + 3 y 2 ∼ 31 x 2 + 34 xy + 10 y 2 . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 76 / 110
Reduced Positive Definite Forms A positive definite form f ( x , y ) = ax 2 + bxy + cy 2 , b 2 − 4 ac < 0 a , c > 0 , is called reduced if − a < b ≤ a ≤ c , with b ≥ 0 if c = a . For example 7 x 2 + 3 y 2 31 x 2 + 34 xy + 10 y 2 and are unreduced forms but 3 x 2 + 7 y 2 is reduced. Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 77 / 110
The Reduction Algorithm If f ( x , y ) = ax 2 + bxy + cy 2 is a positive definite form then we can find an integer δ such that | − b + 2 c δ | ≤ c then ax 2 + bxy + cy 2 ∼ a ′ x 2 + b ′ xy + c ′ y 2 where | b ′ | ≤ a ′ and a ′ = c b ′ = − b + 2 c δ c ′ = a − b δ + c δ 2 . If a ′ ≤ c ′ you are done, if not repeat the process. Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 78 / 110
Example: Reducing 31 x 2 + 34 xy + 10 y 2 To reduce 31 x 2 + 34 xy + 10 y 2 , 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 + 2 c δ = − 34 + 2(10)(2) = 6 c ′ = a − b δ + c δ 2 = 31 − 34(2) + 10(2) 2 = 3 so 31 x 2 + 34 xy + 10 y 2 ∼ 10 x 2 + 6 xy + 3 y 2 which is unreduced. If we perform the process one more time we get the reduced form 31 x 2 + 34 xy + 10 y 2 ∼ 10 x 2 + 6 xy + 3 y 2 ∼ 3 x 2 + 7 y 2 . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 79 / 110
The Class Number For each discriminant ∆ < 0 there are a number of classes of equivalent forms. Each class contains a unique reduced form. The number of classes for a given discriminant ∆ < 0 is called the class number , h (∆). For example, h ( − 84) = 4 so there are 4 equivalence classes of forms with discriminant − 84. The reduced forms in the classes are x 2 + 21 y 2 , 2 x 2 + 2 xy + 11 y 2 , 3 x 2 + 7 y 2 , 5 x 2 + 4 xy + 5 y 2 Each class will represent its own set of numbers. The classes form an Abelian group called the class group where the group operation is called composition. Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 80 / 110
Numbers Represented by the Form f ( x , y ) = x 2 + y 2 91 92 93 94 95 96 97 98 99 100 100 81 81 82 83 84 85 86 87 88 89 90 71 72 73 74 75 76 77 78 79 80 61 62 63 64 64 65 66 67 68 69 70 51 52 53 54 55 56 57 58 59 60 41 42 43 44 45 46 47 48 49 50 49 31 32 33 34 35 36 36 37 38 39 40 21 22 23 24 25 25 26 27 28 29 30 11 12 13 14 15 16 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 1 4 9 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 81 / 110
Numbers Represented by the Form f ( x , y ) = x 2 + y 2 91 92 93 94 95 96 97 98 99 100 100 81 81 82 82 83 84 85 86 87 88 89 90 71 72 73 74 75 76 77 78 79 80 61 62 63 64 64 65 65 66 67 68 69 70 51 52 53 54 55 56 57 58 59 60 41 42 43 44 45 46 47 48 49 50 49 50 31 32 33 34 35 36 36 37 37 38 39 40 21 22 23 24 25 25 26 26 27 28 29 30 11 12 13 14 15 16 16 17 17 18 19 20 1 2 3 4 5 6 7 8 9 10 1 2 4 5 9 10 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 82 / 110
Numbers Represented by the Form f ( x , y ) = x 2 + y 2 91 92 93 94 95 96 97 98 99 100 100 81 81 82 82 83 84 85 85 86 87 88 89 90 71 72 73 74 75 76 77 78 79 80 61 62 63 64 64 65 65 66 67 68 68 69 70 51 52 53 53 54 55 56 57 58 59 60 41 42 43 44 45 46 47 48 49 50 49 50 31 32 33 34 35 36 36 37 37 38 39 40 40 21 22 23 24 25 25 26 26 27 28 29 29 30 11 12 13 13 14 15 16 16 17 17 18 19 20 20 1 2 3 4 5 6 7 8 9 10 1 2 4 5 8 9 10 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 83 / 110
Numbers Represented by the Form f ( x , y ) = x 2 + y 2 91 92 93 94 95 96 97 98 99 100 100 81 81 82 82 83 84 85 85 86 87 88 89 90 90 71 72 73 73 74 75 76 77 78 79 80 61 62 63 64 64 65 65 66 67 68 68 69 70 51 52 53 53 54 55 56 57 58 58 59 60 41 42 43 44 45 46 47 48 49 50 45 49 50 31 32 33 34 34 35 36 36 37 37 38 39 40 40 21 22 23 24 25 25 26 26 27 28 29 29 30 11 12 13 13 14 15 16 16 17 17 18 18 19 20 20 1 2 3 4 5 6 7 8 9 10 1 2 4 5 8 9 10 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 84 / 110
Numbers Represented by the Form f ( x , y ) = x 2 + y 2 91 92 93 94 95 96 97 97 98 98 99 100 100 81 81 82 82 83 84 85 85 85 86 87 88 89 89 90 90 71 72 72 73 73 74 74 75 76 77 78 79 80 80 61 61 62 63 64 64 65 65 65 66 67 68 68 69 70 51 52 52 53 53 54 55 56 57 58 58 59 60 41 42 43 44 45 46 47 48 49 50 41 45 49 50 31 32 32 33 34 34 35 36 36 37 37 38 39 40 40 21 22 23 24 25 25 26 26 27 28 29 29 30 11 12 13 13 14 15 16 16 17 17 18 18 19 20 20 1 2 3 4 5 6 7 8 9 10 1 2 4 5 8 9 10 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 85 / 110
Numbers Represented by the Form f ( x , y ) = x 2 + y 2 1 2 3 4 37 38 39 40 73 74 75 76 1 2 4 37 40 73 74 5 5 6 7 8 8 41 41 42 43 44 77 78 79 80 80 9 9 10 10 11 12 45 45 46 47 48 81 81 82 82 83 84 13 13 14 15 16 16 49 49 50 50 51 52 52 85 85 86 87 88 17 18 19 20 53 54 55 56 89 90 91 92 17 18 20 53 89 90 21 22 23 24 57 58 58 59 60 93 94 95 96 25 25 26 26 27 28 61 61 62 63 64 64 97 97 98 98 99 100 100 29 29 30 31 32 32 65 65 66 67 68 68 101 101 102 103 104 104 33 34 35 36 69 70 71 72 105 106 107 108 34 36 72 106 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 86 / 110
Sums of Squares Modulo 4 m 2 + n 2 (mod 4) hi n 2 (mod 4) n m \ n 0 1 2 3 0 0 0 0 1 0 1 . 1 1 1 1 2 1 2 2 0 2 0 1 0 1 3 1 3 1 2 1 2 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 87 / 110
Writing n as a Sum of Two Squares Diophantus–Brahmagupta–Fibonacci identity: ( a 2 + b 2 )( c 2 + d 2 ) = ( ac − bd ) 2 + ( ad + bc ) 2 Theorem: If p ≡ 1 (mod 4) is a prime, then there exists positive integers a and b such that a 2 + b 2 = p . Theorem (Fermat): If n is factored into primes as q γ j n = 2 α � p β i � i j i j where p i and q j are primes with p i ≡ 1 (mod 4) and q j ≡ 3 (mod 4), for all i and j , then n can be expressed as a sum of two squares if and only if γ j is even for all j . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 88 / 110
Examples of D-B-F Identity 1 1 2 2 3 4 4 37 37 38 39 40 40 73 73 74 74 75 76 5 5 6 7 8 8 41 41 42 43 44 77 78 79 80 80 9 9 10 10 11 12 45 45 46 47 48 81 81 82 82 83 84 13 13 14 15 16 16 49 49 50 50 51 52 52 85 85 86 87 88 17 17 18 18 19 20 20 53 53 54 55 56 89 89 90 90 91 92 21 22 23 24 57 58 58 59 60 93 94 95 96 25 25 26 26 27 28 61 61 62 63 64 64 97 97 98 98 99 100 100 29 30 31 32 65 66 67 68 101 102 103 104 29 32 65 68 101 104 33 34 34 35 36 36 69 70 71 72 72 105 106 106 107 108 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 89 / 110
Examples of D-B-F Identity 1 1 2 2 3 4 4 37 37 38 39 40 40 73 73 74 74 75 76 5 5 6 7 8 8 41 41 42 43 44 77 78 79 80 80 9 9 10 10 11 12 45 45 46 47 48 81 81 82 82 83 84 13 13 14 15 16 16 49 49 50 50 51 52 52 85 85 86 87 88 17 17 18 18 19 20 20 53 53 54 55 56 89 89 90 90 91 92 21 22 23 24 57 58 58 59 60 93 94 95 96 25 25 26 26 27 28 61 61 62 63 64 64 97 97 98 98 99 100 100 29 30 31 32 65 66 67 68 101 102 103 104 29 32 65 68 101 104 33 34 34 35 36 36 69 70 71 72 72 105 106 106 107 108 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 90 / 110
Examples of D-B-F Identity 1 1 2 2 3 4 4 37 37 38 39 40 40 73 73 74 74 75 76 5 5 6 7 8 8 41 41 42 43 44 77 78 79 80 80 9 9 10 10 11 12 45 45 46 47 48 81 81 82 82 83 84 13 13 14 15 16 16 49 49 50 50 51 52 52 85 85 86 87 88 17 17 18 18 19 20 20 53 53 54 55 56 89 89 90 90 91 92 21 22 23 24 57 58 58 59 60 93 94 95 96 25 25 26 26 27 28 61 61 62 63 64 64 97 97 98 98 99 100 100 29 30 31 32 65 66 67 68 101 102 103 104 29 32 65 68 101 104 33 34 34 35 36 36 69 70 71 72 72 105 106 106 107 108 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 91 / 110
Writing n as a Sum of Two Squares Diophantus–Brahmagupta–Fibonacci identity: ( a 2 + b 2 )( c 2 + d 2 ) = ( ac − bd ) 2 + ( ad + bc ) 2 Theorem: If p ≡ 1 (mod 4) is a prime, then there exists positive integers a and b such that a 2 + b 2 = p . Theorem (Fermat): If n is factored into primes as q γ j n = 2 α � p β i � i j i j where p i and q j are primes with p i ≡ 1 (mod 4) and q j ≡ 3 (mod 4), for all i and j , then n can be expressed as a sum of two squares if and only if γ j is even for all j . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 92 / 110
Primes p ≡ 1 (mod 4) 1 1 2 2 3 4 4 37 37 38 39 40 40 73 73 74 74 75 76 5 5 6 7 8 8 41 41 42 43 44 77 78 79 80 80 9 9 10 10 11 12 45 45 46 47 48 81 81 82 82 83 84 13 13 14 15 16 16 49 49 50 50 51 52 52 85 85 86 87 88 17 17 18 18 19 20 20 53 53 54 55 56 89 89 90 90 91 92 21 22 23 24 57 58 58 59 60 93 94 95 96 25 26 27 28 61 62 63 64 97 98 99 100 25 26 61 64 97 98 100 29 29 30 31 32 32 65 65 66 67 68 68 101 101 102 103 104 104 33 34 34 35 36 36 69 70 71 72 72 105 106 106 107 108 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 93 / 110
Writing n as a Sum of Two Squares Diophantus–Brahmagupta–Fibonacci identity: ( a 2 + b 2 )( c 2 + d 2 ) = ( ac − bd ) 2 + ( ad + bc ) 2 Theorem: If p ≡ 1 (mod 4) is a prime, then there exists positive integers a and b such that a 2 + b 2 = p . Theorem (Fermat): If n is factored into primes as q γ j n = 2 α � p β i � i j i j where p i and q j are primes with p i ≡ 1 (mod 4) and q j ≡ 3 (mod 4), for all i and j , then n can be expressed as a sum of two squares if and only if γ j is even for all j . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 94 / 110
Sum of Two Squares Example n = 2 × 13 × 3 2 = 234 Since 2 = 1 2 + 1 2 , 13 = 2 2 + 3 2 , 9 = 0 2 + 3 2 we have 26 = 2 × 13 = (1 2 + 1 2 )(2 2 + 3 2 ) = (1 × 2 − 1 × 3) 2 + (1 × 3 + 1 × 2) 2 = ( − 1) 2 + 5 2 = 1 2 + 5 2 so 234 = 26 × 9 = (1 2 + 5 2 )(0 2 + 3 2 ) = (1 × 0 − 5 × 3) 2 + (1 × 3 + 5 × 0) 2 = 15 2 + 3 2 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 95 / 110
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 + ( i 2 ) bd = ( ac − bd ) + i ( ad + bc ) Diophantus–Brahmagupta–Fibonacci identity: ( a 2 + b 2 )( c 2 + d 2 ) = ( ac − bd ) 2 + ( ad + bc ) 2 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 96 / 110
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 + ( i 2 ) bd = ( ac − bd ) + i ( ad + bc ) Diophantus–Brahmagupta–Fibonacci identity: ( a 2 + b 2 )( c 2 + d 2 ) = ( ac − bd ) 2 + ( ad + bc ) 2 Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 97 / 110
Modulus of a Complex Number Recall for a complex number z = x + iy , x , y ∈ R , the modulus of z , | z | , satisfies | z | 2 = z ¯ z = ( x + iy )( x − iy ) = x 2 + y 2 or x 2 + y 2 . � | z | = ℑ z = x + iy ℜ Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 98 / 110
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 = a 2 + b 2 , | β | 2 = c 2 + d 2 , | 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 | . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 99 / 110
A Curious Identity We can write ( x 2 1 + x 2 2 )( y 2 1 + y 2 2 ) = z 2 1 + z 2 2 where z 1 = x 1 y 1 − x 2 y 2 z 2 = x 1 y 2 + x 2 y 1 as a statement of | X || Y | = | XY | where X , Y ∈ C with X = x 1 + ix 2 and Y = y 1 + iy 2 . Shawn Godin (Cairine Wilson S.S.) Quadratics October 14, 2017 100 / 110
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