6 Binary Quadratic forms

Transcription

1 6 Binary Quadratic forms 6.1 Fermat-Euler Theorem A binary quadratic form is an exression of the form f(x,y) = ax 2 +bxy +cy 2 where a,b,c Z. Reresentation of an integer by a binary quadratic form has attracted attention of many mathematicians in the ast. For examle, Fermat observed that theorem 6.1 An odd rime can be reresented by a sum of two squares if and only if 1 (mod 4). We have seen two roofs of Theorem 6.1 in the last Chater. We now give Euler s roof of this result using Fermat s method of descent. If 3 (mod 4) then it cannot be a sum of two squares since a sum of two squares is either congruent to 1,2, or 0 modulo 4. Suose 1 (mod 4) then ( ) 1 = 1 imlies that x 2 1 (mod ) is solvable. Let u be an integer with 1 such that u 1 2 u 2 1 (mod ). (6.1) This imlies that u 2 +1 = k (6.2) 1 We use the remainders from ( 1)/2 to ( 1)/2 instead of 0 to 1.

2 6.1 Fermat-Euler Theorem 69 for some ositive integer k. Note that by (6.1), u 2 +1 < 2, and hence 0 < k <. Now since (u 1), we conclude that if S := {m Z + x 2 +y 2 = m, xy,m <.}, then by (6.2) and our discussion so far, we conclude that k S and hence S is a non-emty subset of ositive integers. By the least integer axiom, S contains a minimal ositive m 0 satisfying the conditions. If m 0 = 1 then we are done. Suose m 0 > 1. We first note that m 0 cannot divide both x and y. If this were the case, then m 0 would divide. Since m 0 1 and m 0 <, this is not ossible. Therefore, if we write x = x 1 +m 0 r and y = y 1 +m 0 s with x 1 < m 0 /2 and y 1 < m 0 /2, x 1 and y 1 cannot be both zero. This imlies that x 2 1 +y 2 1 > 0. Now, 0 < x 2 1 +y 2 1 m 2 0/2 < m 2 0 (6.3) by our choice of the range of x 1 and y 1. Hence, with m 1 < m 0 by (6.3). Next, since we find that x 2 1 +y 2 1 = m 0 ( 2rx 2sy +m 0 r 2 +m 0 s 2 ) = m 0 m 1 (x iy)(x 1 +iy 1 ) = (xx 1 +yy 1 +i(xy 1 x 1 y), m 2 0 m 1 = (x 2 +y 2 )(x 2 1 +y2 1 ) = (xx 1 +yy 1 ) 2 +(xy 1 x 1 y) 2. Now, observe that X = xx 1 +yy 1 = x(x m 0 r)+y(y m 0 s) = m 0 w, for some integer w. Similarly, Y = xy 1 x 1 y = m 0 v. Therefore, This imlies that m 2 0 m 1 = X 2 +Y 2 = m 2 0 (w2 +v 2 ). w 2 +v 2 = m 1. (6.4) Now, if (wv), then w or v. Without loss of generality, we may assume u.

3 70 Binary Quadratic forms Now, (6.4) imlies that v 2, or v. Again from (6.4), we find that m 1, which is imossible because 0 < m 1 <. Hence, (wv). The above discussion imlies that m 1 S. But this contradicts the minimality of m 0 in S. This imlies that m 0 = 1 in the first lace and we are done. From Theorem 6.1, we see ( that ) rimes of the form x 2 +y 2 can be determined 1 by the Legendre condition = 1. It turns out rimes of the form x 2 +2y 2 ( ) 2 and x 2 + 3y 2 are determined comletely by the conditions = 1 and ( ) 3 = 1, resectively. A natural question to ask is therefore, can all rimes ( ) n of the form x 2 +ny 2 determined by a single condition = 1? The answer is no. For examle, one can show that if and only if = x 2 +5y 2 ( ) 1 = 1 and ( ) 5 = 1. So two conditions are needed. In fact if we only have one condition ( ) 5 = 1, then we can only conclude either = x 2 +5y 2 or = 2x 2 +2xy +3y 2. Thebinaryquadraticformsx 2 +5y 2 and2x 2 +2xy+3y 2 bothhavediscriminant b 2 4ac = 20. Furthermore these binary quadratic forms are genuinely different. In the following sections, we will exlain the meaning of binary quadratic forms being genuinely different. 6.2 Reresentations of integers by binary quadratic forms definition 6.2 Given a binary quadratic form ax 2 +bxy+cy 2 the discriminant (denoted usually by d or ) of the quadratic form is defined to be the number b 2 4ac.

4 6.2 Reresentations of integers by binary quadratic forms 71 theorem 6.3 Let d be an integer. There exists at least one binary quadratic form with integral coefficients and discriminant d, if and only if d 0 or 1 (mod 4). Let f(x,y) = ax 2 +bxy+cy 2 be a binary quadratic form of discriminant d. Since b 2 0 or 1 (mod 4), d = b 2 4ac 0 or 1 (mod 4). Suose d 0 (mod 4). Then the form x 2 d 4 y2 has discriminant d. If d 1 (mod 4), then ( ) d 1 x 2 +xy y 2 4 has discriminant d. definition 6.4 A form f(x,y) is called indefinite if it takes on both ositive and negative values. The form is called called ositive semidefinite (or negative semidefinite) if f(x,y) 0 ( or f(x,y) 0) for all integers x,y. A semidefinite form is called definite if in addition the only integers x,y for which f(x,y) = 0 are x = 0,y = 0. The form f(x,y) = x 2 2y 2 is indefinite. The form f(x,y) = (x y) 2 is semidefinite and the form f(x,y) = x 2 +y 2 is ositive definite form. examle 6.1 Let d be the discriminant of the form f(x,y) = ax 2 +bxy +cy 2. If d < 0 and a > 0, show that f(x,y) is ositive definite. Solutions. We see that ( f(x,y) = a x+ by ) 2 ( + 4ac b2 2a 4a 2 y 2 = a x+ by ) 2 d 2a 4a y2. This immediately shows that f(x, y) is ositive definite. definition 6.5 We say that a quadratic form f(x, y) reresents an integer n if there exists integers k,l such that f(k,l) = n. We say that the reresentation is roer if (k,l) = 1; otherwise, it is imroer. In the first section, we have seen that if 1 (mod 4) then is roerly reresented by x 2 +y 2. Our aim in this chater is to continue our investigation of integers roerly reresented by a articular quadratic forms.

5 72 Binary Quadratic forms theorem 6.6 Let n > 0 and d be given integers with n 0. There exists a binary quadratic form of discriminant d that reresents n roerly if and only if the congruence has a solution. Suose b is a solution of x 2 d (mod 4n) x 2 d (mod 4n). Then b 2 = d+4nc, for some integer c. The form f(x,y) = nx 2 +bxy +cy 2 has discriminant d and roerly reresents n since f(1,0) = n. Next,suosef(k,l) = nand(k,l) = 1.Weclaimthatwecanfindaquadratic form g(x,y) that takes the form nx 2 +Bxy +Cy 2. In order to rove this, we rewrite a general binary quadratic form in the form ( ) ( )( ) a b/2 x x y. b/2 c y We note the discriminant may be defined as 4det(M), where ( ) a b/2 M =. b/2 c So in general if F(x,y) = f(αx + γy,βx + δy), then the discriminant of F is given by ( ) ( ) 2 a b/2 α β 4 det det. b/2 c γ δ Suose f(x,y) = ax 2 +bxy +cy 2. Since (k,l) = 1, there exist integers u,v such that ku lv = 1. We let g(x,y) = f(kx+vy,lx+uy). Then g(x,y) = nx 2 +Bxy +Cy 2. Since det ( ) k l = 1, v u by the above discussion, we find that the discriminant of g(x,y) is given by B 2 4nC = d.

6 6.2 Reresentations of integers by binary quadratic forms 73 This immediately imlies that has a solution. x 2 d (mod 4n) corollary 6.7 Suose d 0 or 1 (mod 4). If is an odd rime, then there is a binary quadratic form of discriminant d that reresents if and only if ( ) d = 1. By Theorem 6.6, we conclude that x 2 d (mod 4) is solvable if can be reresented by a binary quadratic form of discriminant d. This imlies that x 2 d (mod ) is solvable, or Conversely, if ( ) d = 1 then ( ) d = 1. x 2 d (mod ) is solvable. Note that d 0 or 1 (mod 4), then or resectively. Hence, x 2 d 0 (mod 4) x 2 d 1 (mod 4) x 2 d (mod 4) is solvable. By Chinese Remainder Theorem, we conclude that x 2 d (mod 4) is solvable and by Theorem 6.6, is reresented by a binary quadratic form of discriminant d.

7 74 Binary Quadratic forms 6.3 Equivalence of binary quadratic forms We have seen in the roof of Theorem 6.6 that ( matrices ) lay a role in the study α β of quadratic forms. In fact, we observe that is 1, then the discriminant γ δ of g(x,y) that is obtained from f(x,y) under the transformation ( ) ( ) ( ) α β x y x y γ δ is equal to the discriminant of f(x,y). This observation also shows that an integer n can be roerly reresented by f(x,y) if and only if it can be roerly reresentedbyg(x,y).inotherwords,the formsf(x,y)andg(x,y)are similar. This leads us to the next definition. definition 6.8 We say that two quadratic forms f(x,y) = ax 2 +bxy +cy 2 and g(x,y) = Ax 2 +Bxy+Cy 2 are related and write f g if there are integers α,β,γ,δ Z with αδ βγ = 1, such that g(x,y) = f(αx+γy,βx+δy). theorem 6.9 The relation on the set of binary quadratic forms is an equivalence relation. Since f(x,y) = f(x,y), f f. Suose f g, then we can write g(x,y) = Ax 2 +Bxy +Cy 2 = ( x y )( )( ) A B/2 x B/2 C y = ( x y )( )( α γ a b/2 β δ b/2 c )( α β γ δ )( x y Comaring the reresentations above(check that the (1, 2)-entry and(2, 1)-entry of the roduct of three matrices in the last exression are the same), we conclude that ( ) A B/2 = B/2 C ( α γ β δ With the above reresentation, we see that ( ) a b/2 = b/2 c ( α γ β δ )( a b/2 b/2 c ) 1( A B/2 B/2 C and hence g f. Lastly, to check transitivity, we suose that h(x,y) = A x 2 +B xy +C y 2. )( α β γ δ ) ( ( α γ β δ ). ) 1 ) T ).

8 6.4 Reduced forms 75 ( ) ( α β α β Suose f g via and g h via γ δ γ δ ( A B ) ( /2 α γ )( )( α γ a b/2 B /2 C = β δ β δ b/2 c ). Then )( α β γ δ )( α β γ δ ). This comutation shows that f h. Remark 6.10 If f g, then g(x,y) = f(αx+γy,βx+δy). Note that since αδ βγ = 1, the discriminant of f and g are the same and so the relation is an equivalence relation on the set of binary quadratic forms with the same discriminant. From now on, we will concentrate on the set of ositive definite binary quadratic forms (the discriminant is negative and that its absolute value is not a erfect square). We will also introduce the notion of reduced forms and show that in each equivalence class under, there exists one and only one reduced form which serves as a reresentative of that class. We will then show that given a fixed discriminant, there are finitely many equivalence classes and this number is equal to the number of reduced forms. 6.4 Reduced forms definition 6.11 Let f(x,y) = ax 2 +bxy+cy 2 be a ositive definite binary quadratic form whose discriminant d < 0 with ac 0. We call f reduced if a < b a < c or if 0 b a = c. We will first show the following: theorem 6.12 Let d be a given negative integer.each equivalence class of ositive definite binary quadratic forms of discriminant d contains at least one reduced form. Suose f(x,y) = ax 2 +bxy +cy 2.

9 76 Binary Quadratic forms Since ac 0, a 0 and c 0. Let F := {A Z + Ax 2 +Bxy +Cy 2 f(x,y)}. Note that F is non-emty and because a 0 and f f. By the least integer axiom, there exists a smallest ositive integer u in the set F. Let g(x,y) = ux 2 +vxy +wy 2 such that g(x,y) f(x,y). Suose the second coefficient of g, namely, v does not lie in ( u, u ]. Then by Division Algorithm, there exists a unique integer m such that u < 2um+v u. The form g(x+my,y) = ux 2 +(2um+v)xy +w y 2 will then have the middle coefficient lying in ( u, u ]. Note that if w < u, then by alying ( ) 0 1 S =, 1 0 we arrive at an equivalent form w x 2 (2um+v)xy +uy 2, which contradicts the minimality of u. Hence, w u. If w = u and 2um+v = 0 then u = 1 since ux 2 +uy 2 has to be rimitive. If 2um+v > 0, then is reduced. If 2um+v < 0, then since ux 2 +(2um+v)xy +uy 2 ux 2 +(2um+v)xy +uy 2 ux 2 (2um+v)xy +uy 2, and ux 2 (2um +v)xy + uy 2 is reduced, we observe that f is equivalent to a reduced form. lemma 6.13 Let f(x,y) = ax 2 +bxy+cy 2 be a reduced ositive definite form. If for some air of integers x and y we have (x,y) = 1 and f(x,y) c, then f(x,y) = a or c and the oint (x,y) is one of the six oints ±(1,0),±(0,1),±(1, 1). Moreover, the number of roer reresentations of a by f is 2 if a < c, 4 if 0 b < a = c, and 6 if a = b = c.

10 6.4 Reduced forms 77 Suose that y 2. Then 4af(x,y) = (2ax+by) 2 dy 2 dy 2 4d = 16ac 4b 2 > 8ac 4b 2 4a 2 4b 2 +4ac 4ac. Thus, f(x,y) > c if y 2. This imlies that if f(x,y) c then y 1. Now, suose that y = 1 and x 2. Then Therefore, 2ax+by 2ax by 4a b 3a. 4af(x,y) = (2ax+by) 2 dy 2 9a 2 dy 2 = 9a 2 d = 9a 2 +4ac b 2 > a 2 b 2 +4ac 4ac. Thus, f(x,±1) > c. Next, if y = 0 then since (x,y) = 1, we conclude that x = ±1. Collecting what we obtain so far, we conclude that if f(x,y) c, then y 1 and x 1.Therearealtogethereightoints((0,0)isexcludedsincethegreatest common divisor of0and 0 is not 1) (x,y) satisfying the aboveinequalities. These are ±(1,0)±(0,1),±(1, 1) and ±(1,1). Next, since b > a, we find that f(±1,±1) = a+b+c > c. Hence, (1,1) and ( 1, 1) are excluded as well. We have thus arrived at the six ossible solutions. The last assertion follows on observing that f(1,0) = a,f(0,1) = c and f(1, 1) = a b+c. We observe that from the above comutations, a and c are the smallest ositive integer and second smallest ositive integer reresented by f resectively. theorem 6.14 Let f(x,y) = ax 2 +bxy+cy 2 and g(x,y) = Ax 2 +Bxy+Cy 2 be reduced ositive definite quadratic forms. If f g then f = g. Fromthe reviouslemma,wefindthat aisthesmallestintegerthat isroerly reresented by f(x,y). Since f is equivalent to g, we may write f(x,y) = g(αx+γy,βx+δy).

11 78 Binary Quadratic forms Now, two solutions to f(x,y) = a are (±1,0). Thus, g(±α,±β) = a. Since αδ βγ = 1, we conclude that (α,β) = 1 and the reresentation of a by g(x,y) is roer. Therefore a A since A is the smallest ositive integer reresented by g. Interchanging the roles of f and g and using similar arguments, we conclude that A a. Hence, a = A. Suose a < c. Then f(x,y) = a has only two solutions, namely, (±1,0). If C = A = a, then by revious lemma, g(x,y) = a would have more than 2 solutions with (x,y) = 1. Hence, we conclude that C > a. Observe that c is the second smallest integer roerly reresented by f(x,y) and the solutionsf(x,y) = care(0,±1).thisimlies that g(x,y) roerlyreresents c. Hence c C since C is the second smallest integer roerly reresented by g(x,y). Interchangingthe roles of f and g, we conclude that C c and hence, C = c. To show that b = B suose g(x,y) = f(α x+γ y,β x+δ y). Note that g(x,y) = ax 2 +Bxy +cy 2 = (aα 2 +bα β +cβ 2 )x 2 +B xy +(aγ 2 +bγ δ +cδ 2 )y 2. Comaring the coefficients of x 2, we find that f(α,β ) = a. The only two solutions to the above are (±1,0). Therefore, α = ±1 and β = 0. Now, by considering solutions to f(x,y) = c we conclude that (γ,δ ) = (0,±1). We then conclude that g(x,y) = f( x, y) or f(x,y) and b = B. Suose a = c. Then f(x,y) = a has at least 4 roer reresentations by f. Therefore the same is true for g. Hence, C = a = c. Now, B 2 4AC = b 2 4ac and the fact that b 0,B 0 yields b = B and our roof is comlete. theorem 6.15 Let f be a reduced ositive definite binary quadratic form whose discriminant d is not a erfect square. Then 0 < a d/3. The number of reduced forms of a given nonsquare discriminant d is finite. We see that d = 4ac b 2 4a 2 a 2 = 3a 2.

12 6.4 Reduced forms 79 Hence, a d/3. Now a < b a and b is determined by the equation b 2 4ac = d. This imlies that there are finitely many reduced forms. There is a formula to comute the number of reduced binary quadratic form of discriminant d < 0. For examle if d 1 (mod 4) and d > 3, then the number is 1 d For examle, if d = 11, we have (n,d)=1 1<n< d ( ) n n. d 1 { } = 1. 11