POINTS ON CONICS MODULO p

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1 POINTS ON CONICS MODULO TEAM 2: JONGMIN BAEK, ANAND DEOPURKAR, AND KATHERINE REDFIELD Abstract. We comute the number of integer oints on conics modulo, where is an odd rime. We extend our results to conics on the rojective lane modulo, and then to conics on the rojective lane modulo n, for an integer n. We relate the existence of solutions on the rojective lane modulo n to the existence of rational solutions using the Hasse Minkowski theorem. 1. Introduction Consider the quadratic olynomial q given by (1.1) q(x, y) = ax 2 + by 2 + cxy + dx + ey + f, where a, b, c, d, e and f are integers with no common divisor. We know that the set oints (x, y) R 2 such that q(x, y) = 0 forms an ellise, a arabola, a hyerbola or in degenerate cases a air of lines, a line, a oint or an emty set. We call the set of solutions of q(x, y) = 0 the affine conic determined by q. Although we understand conics in R 2 quite well, it is much more challenging to find oints on conics if we restrict (x, y) to Z 2. Even the question of deciding whether the conic has any oints in Z 2 can be nontrivial. For examle, it is unclear if the olynomial x 2 + 5y 2 3 has any zeros in Z 2, although it has infinitely many zeros in R 2. Similarly, we can consider the homogeneous quadratic (1.2) Q(x, y, z) = ax 2 + by 2 + fz 2 + cxy + dxz + eyz. As we shall see, the zeros of Q in Z 3 are closely related to the zeros of q in Q 2. We call the set of nonzero solutions of Q(x, y, z) = 0, u to constant multiles, the rojective conic determined by Q. Again, the question of the existence of integer oints on rojective conics is difficult to answer. For examle, it is not clear whether x 2 + 5y 2 3z 2 has any integer solutions other than (0, 0, 0). We note that the integer solutions of q(x, y) = 0 or Q(x, y, z) = 0 give us solutions modulo n, where n is a ositive integer, in articular Date: December 12,

2 2 J. BAEK, A. DEOPURKAR, AND K. REDFIELD a rime integer. Since integers modulo a rime form a field, the question of finding zeros modulo is much easier that the question of finding integer zeros. Thus, we may find the zeros of q or Q modulo (and modulo n, if ossible) and from this knowledge, try to extract some information about the integer zeros. In this aer, we focus on finding the zeros of q modulo a rime and the the zeros of Q modulo and modulo an arbitrary ositive integer n. The aer is organized as follows: In Section 2 we find the number of oints on the affine conic q(x, y) 0 (mod ). In Section 3 we find the number of oints on the rojective conic Q(x, y, z) 0 (mod ). In Section 4, we look at the number of oints on the rojective conic Q(x, y, z) 0 (mod n). Finally, in Section 5 we discuss the imlications of the number of solutions of Q(x, y, z) 0 (mod n) to the existence of integer solutions of Q(x, y, z) = Points on Affine Conics modulo Definition 2.1. An affine conic C q Z 2 is the set of solutions to q(x, y) = 0 where q has the form in Equation (1.1). Similarly, we define C q (n) (Z /n Z) 2 to be the set of solutions to q(x, y) 0 (mod n). Given an arbitrary odd rime and a quadratic olynomial q(x, y), we are interested in finding the number of oints in the affine conic C q (). Because there are only a finite number of values in Z / Z, we know that there are only finitely many such oints. To count the number of oints in C q (), we start by examining the equation these oints satisfy: (2.1) q(x, y) = ax 2 + bxy + cy 2 + dx + ey + f 0 (mod ). Now, let X be [ the column ] vector x, y, and let M be the symmetric a b/2 matrix M = with entries in Z / Z. We can then write b/2 c Equation (2.1) in bilinear form as: (2.2) q(x) = X T MX + [ d e ] X + f 0 (mod ). Once we start using the equation s bilinear form, we see that we can change the basis in (Z / Z) 2 to simlify the coefficients inside M: Lemma 2.2. There exists a change of basis A GL 2 () and some S (Z / Z) 2 such that (2.3) q (X) = q(ax + S) X T DX + [ d e ] X + f (mod ) where D is a diagonal matrix. Furthermore, if M is invertible, we can choose our S such that the linear term disaears. In this case, we can

3 rewrite Equation (2.3) to say: POINTS ON CONICS MODULO 3 (2.4) q (X) X T DX + f (mod ). Equivalently, we can write Equation (2.3) and Equation (2.4) in linear form as follows: (2.5) q (x, y) a x 2 + b y 2 + d x + e y + f (mod ) (2.6) q (x, y) a x 2 + b y 2 + f (mod ). Proof. By Theorem A.1 in the aendix, there exists an invertible matrix A and a diagonal matrix D such that D = A T MA. Then, after change of basis via A, letting S be the column vector 0, 0, we see that q (X) (AX + S) T M(AX + S) + [ d (AX) T M(AX) + [ d X T (A T MA)X + [ d X T DX + [ d e ] (AX) + f e ] (AX) + f e ] (AX) + f X T DX + [ d e ] X + f (mod ), as desired. Furthermore, if M is invertible, choosing S = 1 [ ] d 2 (M 1 ) T e instead leads to e ] (AX + S) + f q (X) (AX + S) T M(AX + S) + [ d e ] (AX + S) + f X T (A T MA)X + ( 2S T MA + [ d e ] A ) X + ( S T MS + f ) ( ( X T DX ) [ ] d e M 1 MA + [ d e ] ) A X 2 + ( S T MS + f ) X T DX + ( S T MS + f ) as desired. X T DX + f (mod ), Definition 2.3. Two quadratic olynomials q and q related as in Lemma 2.2 are called identical u to a linear change of coordinates. Definition 2.4. We say that the quadratic olynomial q(x, y) as given by Equation (2.1) is nondegenerate modulo if b 2 4ac 0 (mod ).

4 4 J. BAEK, A. DEOPURKAR, AND K. REDFIELD Possible value of Z q () 3 1, 2, 4, 5 7 1, 6, 8, , 16, 18, , 28, 30, 57 Table 1. Possible values of Z q () in case q(x, y) is nondegenerate modulo, for small odd rimes. Hence if q(x, y) is nondegenerate modulo, we can write its bilinear form as in Equation (2.2) using an invertible matrix M. It is clear from Lemma 2.2 that if q(x, y) is nondegenerate modulo, there exists some (2.7) q (x, y) = a x 2 + b y 2 + f 0 (mod ) which is identical to q(x, y) u to a linear change of coordinates. Corollary 2.5. If two olynomials q(x, y) and q (x, y) are identical u to a linear change of coordinates, there are exaclty as many oints in C q () as there are in C q (). Proof. This fact follows almost directly from Lemma 2.2, because we can write q(ax + S) = q (X) where A is invertible there exists a bijection between the solutions of q and the solutions of q Number of oints in C q () if q is nondegenerate modulo. Let us introduce a simle notation that reresents the number of oints in C q (): Z q () = C q (). We emirically comuted Z q () for small odd rimes and all unique nondegenerate quadratic olynomials q(x, y), u to linear change of coordinates, by writing a rogram that manually tests all oints in (Z / Z) 2. The outut of our rogram can be seen in Table 1. These results indicate that deending on the conic, Z q () can be either 1, 1, + 1 or 2 1. The following theorems exlain the results found by our rogram, and are based on the integral solutions of Equation (2.6). Below the ( exression x ) refers to the Legendre symbol (See Aendix B).

5 POINTS ON CONICS MODULO 5 Theorem 2.6. Given an odd rime and a nondegenerate quadratic olynomial q (x, y) a x 2 + b y (mod ), ( ) a b = +1 = Z q () = 2 1. Proof. We can rewrite the condition q (x, y) 0 (mod ) as (2.8) a x 2 b y 2 (mod ) Note that by hyothesis a, b 0 because q is nondegenerate modulo. If y = 0, then the only solution to this equation is x = 0. If y 0, then there are just as many solutions to Equation (2.8) as there are to the equation: (x/y) 2 b /a (mod ) The fact that a b is a nonzero quadratic residue imlies that b /a is also a nonzero quadratic residue, so for each of the remaining 1 ossible values of y (Z / Z), there are 2 ossible values of x (Z / Z) for which the equality holds, namely, (2.9) x ±y b a (mod ). Altogether, there are 1 + 2( 1) = 2 1 solutions to Equation (2.8), meaning that Z q () = 2 1. Theorem 2.7. Given an odd rime and a nondegenerate quadratic olynomial q (x, y) a x 2 + b y (mod ), ( ) a b = 1 = Z q () = 1. Proof. If a b is a not a quadratic residue, then neither is b /a. Hence, Equation (2.9) is not satisfiable in (Z / Z) 2 unless y = 0. Since setting y = 0 yields only one solution, (x, y) = (0, 0), it must be that Z q () = 1. Theorem 2.8. Given an odd rime and a nondegenerate quadratic olynomial q (x, y) a x 2 + b y 2 + f (mod ) where f 0, ( ) a b = +1 = Z q () = 1. Proof. We can rewrite the condition q (x, y) 0 (mod ) as (2.10) a x 2 ( b )y 2 f (mod ). The fact that a b is a quadratic residue imlies that either b and a are both quadratic residues or both are not. If not, we can simly multily the whole equation by another quadratic nonresidue

6 6 J. BAEK, A. DEOPURKAR, AND K. REDFIELD to make the coefficients of x 2 and y 2 quadratic residues. Therefore, we can assume without loss of generality that a and b are quadratic residues. Let α 2 = a and β 2 = b. Then we can factor Equation (2.10) to say: (2.11) (αx βy)(αx + βy) f (mod ). Since f 0, we can consider the above equation as factoring f into two terms: k 1 and k 2. This can be done in 1 ways for x, y Z / Z. For each such factoring we are simultaneously solving: αx βy k 1 (mod ) αx + βy k 2 (mod ). This gives us a unique solution since (α, β) and (α, β) are indeendent. Thus there are 1 total ways to solve Equation (2.10) in (Z / Z) 2, and Z q () = 1. Theorem 2.9. Given an odd rime and a nondegenerate quadratic olynomial q (x, y) = a x 2 + b y 2 + f where f 0, ( ) a b = 1 = Z q () = + 1. Proof. Starting with the same conditions as the roof for Theorem 2.8, assume without loss of generality that a is a quadratic residue. Then the fact that a b is a nonresidue imlies that b is a nonresidue. Let α 2 = a. We ( have) two cases: Case (1): = +1. f Let φ 2 = f. We can factor Equation (2.10) to say (2.12) (αx φ)(αx + φ) b y 2 (mod ) Assuming that y 0, we are factoring b y 2 into two terms, k 1 and k 2, which can be done in 1 ways as before with Equation (2.11). Once again, each factorization yields a unique solution. If y = 0, then there are two more solutions from αx ±φ (mod ), giving us a total of + 1 solutions. Hence Z q () = + 1. Case (2): ( f ) = 1 First, let us note that there are ( 1)/2 such values for f in Z / Z. Let f 1, f 2 be two of these values, and let q 1(x, y) = a x 2 + b y 2 = f 1

7 POINTS ON CONICS MODULO 7 q 2(x, y) = a x 2 + b y 2 = f 2. For any given solution (x 1, y 1 ) to q 1, we see that we can find a unique f solution ( 2 f x f 1 1, 2 y f 1 1 ) for q 2. Therefore if we can find the total number of solutions to (2.13) q (x, y) = a x 2 + b y 2 C (mod ). in (Z / Z) 2 where C is a nonzero quadratic nonresidue, we can easily comute the number of solutions for a secific C = f because the solutions are uniformly distributed. We know that there are 2 1 ossible nonzero values of C, and by Case (1) we know that if C is a quadratic residue there are exactly + 1 solutions to Equation (2.13). Since there are ( 1)/2 quadratic residues in (Z / Z) we see that there are exactly ( 2 1) ( + 1)( 1) 2 = ( + 1)( 1) 2 combinations of x, y (Z / Z) for which Equation (2.13) roduces a nonzero quadratic nonresidue. Therefore, for each of the ( 1)/2 ossible nonzero, nonresidue values of C = f, Equation (2.13) has exactly (+1)( 1) / ( 1) = + 1 solutions, so Z 2 2 q () = + 1. Theorems 2.6 through 2.9 account for all ossible conditions on a, b, f, hence we have comletely characterized Z q () for any nondegenerate quadratic olynomial q(x, y) modulo Number of oints in C q () if q is degenerate modulo. We next exlored the roerties of degenerate conics. Our findings allowed us to modify the rogram used for Section 2.1 so that we could calculate Z q () when q(x, y) was degenerate modulo. A key asect of our original rogram deended uon the fact that there are a finite number of unique nondegenerate quadratic olynomials modulo u to a linear change in coordinates. We therefore first set out to discover the unique degenerate quadratic olynomials modulo. Lemma If q(x, y) is degenerate modulo, there exists some q of the form (2.14) q (x, y) = a x 2 + e y + f that is identical to q(x, y) u to a linear change of coordinates.

8 8 J. BAEK, A. DEOPURKAR, AND K. REDFIELD Possible values of Z q () 3 0, 3, 6, 9 7 0, 7, 14, , 17, 34, , 29, 58, 841 Table 2. Possible values of Z q () in case q(x, y) is degenerate modulo, for small odd rimes. Proof. From Lemma 2.2, we know that there exists a change of basis A that diagonalizes the bilinear form. Since the quadratic is degenerate, either the first diagonal entry, a, or the second, b, in the diagonalized q(ax) will be zero. Without loss of generality, let b = 0. We have q(ax) = X T DX + [ d e ] X + f. [ ] a 0 Where D =. Adding a vector S to X, where S = 0 0 d, 0, 2a yields the further change of linear coordinates, q (X) = q(a(x + S)) = (X + S) T D(X + S) + [ d e ] (X + S) + f = X T DX + ( 2S T D + [ d e ]) X + ( [ d e ] S + f) = X T DX + ([ d 0 ] + [ d e ]) X + ( [ d e ] S + f) = X T DX + [ 0 e ] X + f, as [ desired. In the cases where a, b = 0, the linear equation q(ax) = d e ] X + f 0 (mod ). We see that by a simle further change of basis, Z q () = Z q () for some q (X) = [ d e ] X + f. In linear form, this can be written as q (x, y) = e x + f as desired. In both cases, we see that q(x) is identical to q (X) u to a linear change of coordinates. By Lemma 2.10 we see that for any quadratic olynomial q(x, y) which is degenerate modulo rime there exists a q (x, y) = a x 2 +e y+ f to which q(x, y) is identical u to a linear change of coordinates. We used this fact along with Corollary 2.5 to modify the earlier version of our rogram to manually comute Z q () for small odd rimes and all degenerate quadratic olynomials q(x, y) unique u to a linear change of coordinates. The modified rogram roduced the results seen in Table 2. These results indicate that given a rime and a quadratic olynomial q which is degenerate modulo, the number of oints Z q () can

9 POINTS ON CONICS MODULO 9 be either 0,, 2 or 2. The following theorems exlain these results and are based uon the integral solutions of Equation (2.14). Theorem Given an odd rime and a degenerate quadratic olynomial q (x, y) a x 2 + e y + f (mod ), e 0 = Z q () =. Proof. Since e 0, we can rearrange the terms in q (x, y) 0 (mod ) to the following: (2.15) y ( f a x 2 )/e (mod ). This indicates that for any value of x, there exists a unique y such that (x, y) satisfies q (x, y) 0 (mod ). Hence there are as many solutions as the ossible values of x. Because x Z / Z, it must then be that Z q () =. Since we have already analyzed the case in which e 0, we now examine the cases in which e = 0. Theorem Given an odd rime and a degenerate quadratic olynomial q (x, y) a x 2 + e y + f (mod ), a = e = f = 0 = Z q () = 2. Proof. All ossible airs in (Z / Z) 2 satisfy q (x, y) 0 (mod ), so Z q () = 2. Theorem Given an odd rime and a degenerate quadratic olynomial q (x, y) a x 2 + e y + f (mod ), a = e = 0, f 0 = Z q () = 0. Proof. In this case, q (x, y) 0 (mod P ) reads f 0 (mod ), which is not true by hyothesis. Hence, no air in (Z / Z) 2 satisfies the equation, so Z q () = 0. Theorem Given an odd rime and a degenerate quadratic olynomial q (x, y) a x 2 + e y + f (mod ), a, f 0, e = 0 = Z q () = 2 or 0. Proof. If a 0 and e = 0, we can rearrange the terms in q (x, y) 0 (mod ) to the following: (2.16) ( ) x 2 f /a (mod ) Case (1): = +1. a f

10 10 J. BAEK, A. DEOPURKAR, AND K. REDFIELD In this case, the right-hand side of Equation (2.16) is a nonzero quadratic residue, so the equation has two solutions. Because y is free, there are a total of 2 oints in C q (). Thus Z q () = 2. Case (2): ( a f ) = 1. If a f is a nonresidue, Equation (2.16) has no solutions, as the left-hand side is a residue while the right-hand side is not. Therefore Z q () = 0. Theorem Given an odd rime and a degenerate quadratic olynomial q (x, y) a x 2 + e y + f (mod ), a 0, e = f = 0 = Z q () =. Proof. Now the equation q (x, y) 0 (mod ) reads a x 2 0 (mod ), whose solutions are (0, y) for all y Z / Z. Hence Z q () =. Theorems 2.11 through 2.15 account for all ossible conditions on a, e, f, hence we have comletely characterized Z q () in terms of q(x, y) in case it is degenerate modulo. 3. Points on Projective Conics modulo We are now interested in rational oints on conics. To study rational oints, we will make use of the rojective lane, which is defined as follows: Definition 3.1. Let F be a field. The rojective lane over F, denoted by P 2 (F ), is the set of equivalence classes in { a, b, c F 3 a, b, c 0, 0, 0 } where scalar multiles are identified. relation is given by a, b, c F 3 0, λ F 0, In other words, the similarity a, b, c λa, λb, λc. If we let F = Q, then members of P 2 (Q) are triles of rational numbers unique u to scalar multilication. By scaling aroriately, we can treat them as triles of integers unique u to scalar multilication. Then, they corresond naturally to ordered air of rational numbers, as shown in the next roosition. Proosition 3.2. There is a bijective corresondence between the set of oints x, y, z in P 2 (Q) with z 0 and the set of rational oints in Q 2, given by x, y, z x/z, y/z.

11 POINTS ON CONICS MODULO 11 Proof. The corresondence above is injective. If x, y, z = x, y, z in P 2 (Q), then it must be that x = λx, y = λy, z = λz, for some λ. Then λx x /z, y /z = λz, λy λz = x/z, y/z. It is also surjective: if q 1, q 2 Q 2 is a rational oint, the rojective line (q 1 z, q 2 z, z), where z is the least common multile of the denominators of q 1, q 2, will ma to (q 1, q 2 ). We can now relate the number of rational oints on (2.1) to oints on the rojective lane. We begin by homogenizing (2.1): (3.1) Q(x, y, z) = ax 2 + bxy + cy 2 + dxz + eyz + fz 2 = 0. Using Proosition 3.2, we see that counting solutions to (3.1) in P 2 (Q) tells us something about the number of nontrivial solutions to (2.1) in Q 2. (To be exact, they are equal once rojective oints with z = 0 are omitted, according to Proosition 3.2.) Therefore, the number of solutions to (3.1) in P 2 (Q) is of interest to us. For the remainder of the section, denotes an odd rime, and we will use P 2 as a shorthand for P 2 (F ). Note that Proosition 3.2 is still relevant because P 2 can be interreted as P 2 (Q) modulo. More formally, we take the aroach in Section 2 and attemt to characterize the number of solutions in P 2 to an arbitrary quadratic olynomial Q(x, y, z). In treating Equation (3.1), we can once again emloy symmetric bilinear forms: a b/2 d/2 (3.2) X T b/2 c e/2 X 0 (mod ). d/2 e/2 f Here X denotes the column vector x, y, z. We will be using Q to denote the matrix of the symmetric bilinear form as well. In that notation, the equation reads X T QX 0 (mod ). Definition 3.3. A rojective conic C Q () is the set of solutions to Q(x, y, z) 0 (mod ) in P 2 where Q has the form in (3.1). We denote by Z Q () the cardinality of C Q (). Our goal, then, is to comute values of Z Q () for different symmetric bilinear forms Q.

12 12 J. BAEK, A. DEOPURKAR, AND K. REDFIELD 3.1. Counting equivalent classes of symmetric bilinear forms. As in (Z / Z) 2, it is not necessary to count the number of solutions to all symmetric bilinear forms in P 2. We exloit the fact that the numbers of solutions to two symmetric bilinear forms Q and Q are the same if the two forms are similar under the equivalence relation defined in Section A of the aendix, i.e. there exists an A GL 3 () such that Q = A T QA. To see why, realize that the condition Q = A T QA is the same as saying Q (X) = Q(AX). Thus A is simly a change of basis, and the number of solutions is reserved. By Theorem A.1, all symmetric bilinear forms are similar to diagonal bilinear forms. In turn, we can rove that there are very few equivalence classes within diagonal bilinear forms. Lemma 3.4. Given an odd rime, there exists a quadratic nonresidue η that can be written as the sum of two quadratic residues. Proof. Suose the contrary that all sums of two quadratic residues are quadratic residues (or zero). This imlies that quadratic residues are closed under addition. Note that Z / Z is generated by 1, which is a quadratic residue. So all nonzero elements of Z / Z would be residues, which is false. Lemma 3.5. Two diagonal bilinear forms D and E over P 2 are similar if the following two conditions hold: rk (D) = rk (E), where rk denotes the rank modulo, and [# of quadratic residues in entries of D] [# of quadratic residues in entries of E] (mod 2). Proof. We begin by noting that D is similar to any diagonal matrix with its entries ermuted, because we can ick row-switching elementary matrices as the change of basis. Also, icking an arbitrary nonsingular diagonal matrix as the change of basis has the effect of searately multilying each entry by a square. Comosing these oerations, we see that D is similar to D = I i ηi j, 0 where i, j corresond to the numbers of residues and nonresidues along the diagonal of D, resectively. Note that i + j = rk (D). In case j 2, we erform the following additional oeration: let η be the nonresidue written as the sum σ σ 2 2, which exists by Lemma

13 POINTS ON CONICS MODULO Take the identity [ matrix ] I 3 and relace a 2-by-2 square along the diagonal by 1 σ1 σ 2. Here the 2-by-2 square in I η σ 2 σ 3 is chosen to 1 corresond to a 2-by-2 subsquare of ηi j sitting inside D. Denote the resulting matrix by J. We see that J T D J looks exactly like D, excet two η s along the diagonal are relaced by 1 s. To check this, it suffices to examine only the rows corresonding to the nonidentity 2-by-2 subsquare of J: [ ] T [ ] [ ] 1 σ1 σ 2 η 0 1 σ1 σ 2 = 1 [ ] [ ] σ1 η σ 2 η σ1 σ 2 η σ 2 σ 1 0 η η σ 2 σ 1 η 2 σ 2 η σ 1 η σ 2 σ 1 = 1 [ ] η 2 0 η 2 0 η 2 = I 2. Alying this oeration as necessary, we can show that D is similar to I i ηi j where j {0, 1}. In summary, the canonical form of 0 the equivalence relation deends only on the rank of D and the arity of the number of residues along the diagonal of D. Therefore, if the remise of the lemma holds, then D and E have the same canonical form, which indicates that they are similar. Corollary 3.6. A symmetric bilinear form Q over P 2 is equivalent to one of the following: I 3, , η 0, , , where η is a quadratic nonresidue modulo. Proof. We list all ossible simlified forms obtained in the roof of Lemma 3.5, droing scalar multiles. We caution the reader that if the rank is odd, there is exactly one canonical form instead of two: the two matrices (one for each arity of the number of residues along the diagonal) are not similar, but when treated as bilinear forms over the rojective lane, they are identified under scalar multilication. For instance, I 3 η η η When the rank is even, scaling does not fli the arity, so the two canonical forms are reserved.

14 14 J. BAEK, A. DEOPURKAR, AND K. REDFIELD Corollary 3.6 tells us that there are at most five distinct equivalence classes. We ask if there are exactly five equivalence classes: Theorem 3.7. The diagonal forms in Corollary 3.6 are distinct under the equivalence relation. Proof. Let D be a diagonal form given in Corollary 3.6. First note that for any invertible matrix modulo A, both A and A T have full ranks. Therefore, the rank of A T DA is equal to the rank of D. Hence the rank is invariant under the equivalence relation. Next, the quadratic residuosity of the roduct of the nonzero diagonal entries in D is invariant as well. Suose that D E, where D and E are two different forms in Corollary 3.6 with the same rank k. We have A T DA = E for some invertible matrix A. Now we take the submatrices of A, D, E corresonding to the k-by-k to-left corner, and denote them by Ã, D, Ẽ resectively. They are related as follows: Ã T DÃ = Ẽ. Note that because Ẽ is nonsingular, Ã must also be nonsingular. Also, ( ) ( ) ( ) det Ẽ det = ÃT DÃ det = Ã 2 ( det D ) ( det = D ). Since the determinants of D and Ẽ are simly the roducts of the nonzero diagonal entries of D and E, resectively, our claim of invariance is roven. Insecting the forms in Corollary 3.6, we see that any two diagonal matrices differ in either their ranks, or in the quadratic residuosity of the roduct of nonzero diagonal entries. Therefore, they cannot be similar. Remark: We did not use the fact that we are oerating in P 2 rather than in a higher-dimensional rojective sace P n, which consists of lines in (Z / Z) n+1. Therefore, we exect that the number of equivalent symmetric bilinear forms over P n to be similarly determined. We conjecture it to be 3n/ This number is obtained by counting one equivalence class for each odd rank, two for each even rank, and one for the all-zero form.

15 POINTS ON CONICS MODULO 15 Possible values of Z Q () 3 1, 4, 7, , 6, 11, , 8, 15, , 12, 23, , 14, 27, 183 Table 3. Possible values of Z Q (), for small odd rimes Number of oints in C q (). Using Matlab, we comuted Z Q () for small odd rimes, and for several olynomials Q reresenting each of the five equivalence classes in Theorem 3.7. (In fact, it would suffice to only try the five canonical forms in Corollary 3.6.) The results are given in Table 3. We observe emirically that Z Q () {1, + 1, 2 + 1, } for all symmetric bilinear form Q. The following theorems exlicitly characterize the ossible values of Z Q () we observe. The simlest number to account for is 2 ++1, which is the number of all elements in P 2. Theorem 3.8. Q 0 (mod ) Z Q () = Proof. If Q is the zero matrix modulo, all elements of the form X = x, y, z satisfies Equation (3.2). Since P 2 = , it follows that Z Q () = Conversely, if all elements of P 2 satisfies the equation, then for all X, Y P 2, we have X T QY (X T QY + Y T QX)/2 (X + Y ) T Q(X + Y )/2 0. Fixing X, we get (X T Q) is a trivial linear ma on GL 3 (). But since this holds for all X, it must be that Q 0. Theorem 3.9. rk (Q) = 1 = Z Q () = + 1. Proof. If Q has rank 1, we know from Corollary 3.6 that Q must be equivalent to after a change of basis. A solution x, y, z to this diagonal form obeys the equation x 2 0. So x 0, while y, z are free. Counting triles in (Z / Z) 3 yields 2 solutions. Subtracting the

16 16 J. BAEK, A. DEOPURKAR, AND K. REDFIELD all-zero solution and identifying scalar multiles, we obtain Z Q () = + 1. Theorem rk (Q) = 2 = Z Q () = 1 or Proof. Using the same logic as in Theorem 3.9, the symmetric bilinear form Q must be similar to either or η 0 where η is a nonresidue. In either case, the corresonding olynomial has the form x 2 + b y 2 0 (mod ), where b 0 (mod ). Rearranging the terms, we have (3.3) b (y/x) 2, or x y 0. If b is a quadratic residue modulo, then there exists β such that β 2 b. Then Equation (3.3) is equivalent to y/x ±β. Counting solutions in (Z / Z) 3 yields (2 1). Subtracting the all-zero solution and dividing by 1 to account for scaling, we get Z Q () = If b is not a quadratic residue, b (y/x) 2 has no solution, in which case we only count solutions with x y 0. Then there is exactly one solution in P 2, namely 0, 0, 1. So Z Q () = 1. Theorem rk (Q) = 3 = Z Q () = + 1. Proof. Because Q has rank 3, Q is similar to I 3. So we can count solutions to x 2 + y 2 + z 2 0. Rearranging the terms yields (3.4) x 2 + y 2 z 2 (mod ). In comarison, consider solutions to (3.5) x 2 + y 2 ηz 2 (mod ), where η is a nonresidue modulo. We remark that exactly one of 1 and η is a quadratic residue modulo, since their roduct is a nonresidue. Let K 1 and K 2 be the number of nonzero solutions to (3.4) and (3.5), resectively, in (Z / Z) 3. Note that a nonzero air (x, y) satisfies exactly one of the two equations, deending on the quadratic residuosity of x 2 + y 2. Therefore, K 1 + K 2 is equal to the number of ossible nonzero airs (x, y): K 1 + K 2 = 2( 2 1).

17 POINTS ON CONICS MODULO 17 The factor of 2 arises from that there are two ossible values of z for each equation. Now let β be such that β 2 = 1 or β 2 = η, deending on which one is the quadratic residue. Then, one of the two equations above factors as (3.6) (x βz)(x + βz) y 2. Solving (3.6) is equivalent to simultaneously solving x βz k 1 x + βz k 2 where k 1 k 2 y 2. If y 0, there are 1 ways to factor y 2 into k 1 k 2, and each system of equation yields a unique solution. Lastly, there are 1 ways to choose y. Counting all these yields ( 1) 2 solutions. If y 0, we require x = ±βz, which has 2( 1) solutions. The total, then, is ( 1) 2 + 2( 1) = 2 1. This tells us that either K 1 or K 2 is 2 1. However, since they add u to 2( 2 1), they must both be 2 1. Then, in either case, we obtain = + 1 solutions to Equation (3.4) in P2. In summary, we conclude the following about the number of solutions to Equation (3.1) on the rojective lane: Theorem The number of solutions to Equation (3.1) in P 2 is , if rk (Q) = 0, + 1, if rk (Q) = 1, Z Q () = 1 or 2 + 1, if rk (Q) = 2, + 1, if rk (Q) = Relation to C q (). The numbers we observe differ somewhat from the numbers of integral solutions to q(x, y) modulo. This can be exlained by the fact that we are counting oints at infinity, or oints on the rojective lane with z = 0. After a linear change of coordinates, these corresond to an arbitrary 2-dimensional subsace V (Z / Z) 3. Hence, the number of solutions on P 2 that actually gives rise to a rational oint are {X V {0} X Z T QX 0 (mod )} Q (). 1

18 18 J. BAEK, A. DEOPURKAR, AND K. REDFIELD 3.4. Imlication of factoring. We can ask how the condition of Q factoring over F relates to the number of solutions. If Q factors nontrivially, then for some nonzero v = v 1, v 2, v 3 and w = w 1, w 2, w 3, Q(x, y, z) (v 1 x + v 2 y + v 3 z)(w 1 x + w 2 y + w 3 z) (v T X) T (w T X) X T (vw T )X (mod ). If we treat Q as a symmetric bilinear form, then for all X, X, X T QX ((X + X ) T Q(X + X ) X T QX X T QX )/2 (X T (vw T )X + X T (vw T )X)/2 ( ) vw X T T + wv T X (mod ). 2 Thus Q vwt +wv T 2 (mod ). A solution to Q here satisfies either v T X 0 or w T X 0. It is easy to see that v T X 0 has + 1 solutions in P 2, which are simly lines orthogonal to v. Similarly, w T X 0 has + 1 solutions. If we further assume that v, w are indeendent, there exists exactly one line orthogonal to both v and w. Hence, the total count of solutions is ( + 1) + ( + 1) 1 = From Theorem 3.12, it follows that the rank of Q must be 2. If v and w are deendent, we can write w = kv, so Q = 1 2 kvvt. Since the rowsace of 1 2 kvvt is sanned by {v}, its rank is 1, and there are + 1 solutions total. 4. Points on rojective conics modulo n Let a,b,c be nonzero integers such that gcd(a, b, c) = 1 and n be a ositive integer. In this section, we look at the solutions of the equation (4.1) ax 2 + by 2 + cz 2 0 (mod n). Observe that Equation (4.1) is trivially satisfied by (0, 0, 0). However, there may be other trivial solutions. For examle, if n = 2, where is a rime, then (,, ) is a nonzero trivial solution of Equation (4.1). Somehow we need to disregard these trivial solutions. We realize that we must look at solutions (x, y, z) where x, y and z do not share a common factor with n. This motivates the following definition.

19 POINTS ON CONICS MODULO 19 Definition 4.1. Let n be a ositive integer. We define the set A 3 n of rimitive triles modulo n as follows: A 3 n := (Z /n Z) 3 d>1 d n d (Z /n Z) 3. A trile (x, y, z) Z 3 gives a trile in (Z /n Z) 3 if we interret x, y, z modulo n. Also, observe that (x, y, z) reresents a trile in A 3 n if and only if gcd(x, y, z, n) = 1. Before we look at the solutions of Equation (4.1) in A 3 n, let us examine the structure of A 3 n for different n. Let n 1 and n 2 be ositive integers and set n = n 1 n 2. We can interret an element of Z /n Z as an element of Z /n 1 Z. In other words, we have a natural surjection from Z /n Z onto Z /n 1 Z. It is clear that this surjection alied to each coordinate yields a surjection π : A 3 n A 3 n 1. Proosition 4.2. Let n 1 and n 2 be relatively rime ositive integers and set n = n 1 n 2. Let π i : A 3 n A 3 n i be the natural surjections for i = 1, 2. Then we have a bijection π 1 π 2 : A 3 n A 3 n 1 A 3 n 2. Proof. Given triles (x i, y i, z i ) A 3 n i for i = 1, 2, there is a unique element (x, y, z) Z /n Z such that π i (x, y, z) = (x i, y i, z i ) by the Chinese remainder theorem. Since (x i, y i, z i ) are rimitive tules modulo n i for i = 1, 2, it is easy to see that (x, y, z) is a rimitive trile modulo n. Note that if (x, y, z) A 3 n is a solution of Equation (4.1), then so is (λx, λy, λz) for any λ in (Z /n Z). Hence, it is reasonable to count the number of solutions of Equation (4.1) u to multilication by a unit of Z /n Z. To make this idea recise, we define a relation on A 3 n by letting (x, y, z) (λx, λy, λz) for all λ (Z /n Z). Clearly, is an equivalence relation. Definition 4.3. Let n be a ositive integer. The set of equivalence classes of A 3 n under is called the rojective lane modulo n and is denoted by P 2 n. For an element (x, y, z) A 3 n, we denote its equivalence class in P 2 n by (x, y, z). Note that every trile in A 3 n has φ(n) elements in its equivalence class. Therefore, φ(n) P 2 n = A 3 n. Also, if n is rime then the new definition of the rojective lane agrees with the revious definition.

20 20 J. BAEK, A. DEOPURKAR, AND K. REDFIELD Let n 1, n 2 be ositive integers and set n = n 1 n 2. Recall that we have a surjection π : A 3 n A 3 n 1. Note that if two elements of A 3 n are equivalent, then so are their images. Hence the rojection π gives a rojection π : P 2 n P 2 n 1. We have a result analogous to Proosition 4.2. Proosition 4.4. Let n 1 and n 2 be relatively rime ositive integers and set n = n 1 n 2. Consider the rojections π i : P 2 n P 2 n i for i = 1, 2. We have a bijection π 1 π 2 : P 2 n P 2 n 1 P 2 n 2. Proof. By Proosition 4.2, we have a bijection π 1 π 2 : A 3 n A 3 n 1 A 3 n 2. Let X, Y be two triles in A 3 n. Observe that X Y in A 3 n if and only if π i (X) π i (Y ) in A 3 n i for i = 1, 2. Hence π 1 π 2 is a bijection. We now have the necessary setu to discuss the solutions of Equation (4.1). As the reader may have guessed, some asects of the discussion are not limited to equations like Equation (4.1), but can be extended to any homogeneous equation. In articular, if Q is a homogeneous olynomial in three variables with integer coefficients and (x, y, z) A 3 n is a zero of Q modulo n, then all elements of A 3 n equivalent to (x, y, z) are also zeros of Q modulo n. We introduce some notation that extends the notation of the revious section. Definition 4.5. Let Q be a homogeneous olynomial in three variables with integer coefficients and n a ositive integer. Define C Q (n) := {(x, y, z) A 3 n Q(x, y, z) 0 C Q (n) := {(x, y, z) P 2 n Q(x, y, z) 0 Z Q (n) := C Q (n), Z Q (n) := C Q (n). (mod n)}, (mod n)}, Let P : A 3 n P 2 n be the ma We clearly have the following: Proosition 4.6. (1) P 1 ( C Q (n)) = C Q (n), (2) Z Q (n) = φ(n) Z Q (n). P : (x, y, z) (x, y, z).

21 POINTS ON CONICS MODULO 21 We can rove the analogue of Proosition 4.2 and Proosition 4.4 for the zero sets C Q and C Q. Proosition 4.7. Let n 1 and n 2 be relatively rime ositive integers and set n = n 1 n 2. Let π i : A 3 n A 3 n i and π i : P 2 n P 2 n i be the natural rojections for i = 1, 2. Then, we have the commutative diagram: π 1 π 2 C Q (n) CQ (n 1 ) C Q (n 2 ), C Q (n) eπ 1 eπ 2 CQ (n 1 ) C Q (n 2 ) where the mas π 1 π 2 and π 1 π 2 are bijective. Proof. We know that we have a commutative diagram A 3 n π 1 π 2 A 3 n1 A 3 n 2, P 2 eπ 1 eπ 2 n P 2 n1 P 2 n 2 where π 1 π 2 and π 1 π 2 are bijections. Since gcd(n 1, n 2 ) = 1, we have Q(x, y, z) 0 (mod n) if and only if Q(x, y, z) 0 (mod n i ) for i = 1, 2. Hence we get a bijection π 1 π 2 : C Q (n) C Q (n 1 ) C Q (n 2 ). This, along with Proosition 4.6 imlies that π 1 π 2 gives a bijection π 1 π 2 : P 2 n P 2 n 1 P 2 n 2. As a direct corollary, we have the following: Proosition 4.8. Let n 1 and n 2 be relatively rime ositive integers. Then, (1) Z Q (n 1 n 2 ) = Z Q (n 1 )Z Q (n 2 ), (2) Z Q (n 1 n 2 ) = Z Q (n 1 ) Z Q (n 2 ). For the rest of the section, we fix Q to be the quadratic form given by Q(x, y, z) = ax 2 + by 2 + cz 2, where a, b, c are nonzero integers such that gcd(a, b, c) = 1. Recall that we are interested in the numbers Z Q (n). Thanks to Proosition 4.8, we can comute the numbers Z Q if we know them for owers of rimes. In

22 22 J. BAEK, A. DEOPURKAR, AND K. REDFIELD other words, if n = α 1 1 α k k, where i are distinct rimes and α i > 0, then Z Q (n) = Z Q ( α 1 1 ) Z Q ( α k k ). Table 4 shows the values Z Q for some forms Q and the first 8 owers of x 2 + y 2 + z x 2 + 2y 2 z x 2 + y 2 81z x 2 + 8y 2 5z Table 4. ZQ (3 k ) for some Q. The integer 2 often behaves differently in situations involving quadratic equations. Table 5 shows the values Z Q for some forms Q and the first 8 owers of x 2 + y 2 + z x 2 + 2y 2 z x 2 + y 2 81z x 2 + 8y 2 5z Table 5. ZQ (2 k ) for some Q and k Observe in Table 4 and Table 5 that the numbers Z Q ( k ) for k = 1, 2,... eventually form a geometric series with ratio. This is not an accident. Theorem 4.9. Let be a rime and Q(x, y, z) = ax 2 +by 2 +cz 2, where a, b, c are nonzero ositive integers. Let N be a ositive integer such that N+1 does not divide any of a, b or c. Let N 0 = N + 1 if is odd and N 0 = N + 3 if = 2. Then, for n N 0 we have Z Q ( n+1 ) = Z Q ( n ). Proof. Since we know that Z Q ( t ) = φ( t ) Z Q ( t ) (Proosition 4.6) and φ( t ) = t 1 ( 1), we may rove Z Q ( n+1 ) = 2 Z Q ( n ). Recall that interreting elements of Z / n+1 Z modulo n gives us a surjection π : C Q ( n+1 ) C Q ( n ). Let α, β and γ be the highest owers of dividing 2a, 2b and 2c resectively. For (x, y, z) C Q ( n ) such that x, define A 1 (x, y, z) := {(x + i n α, y, z) 0 i α 1}.

23 POINTS ON CONICS MODULO 23 Notice that A 1 (x, y, z) = α. It is easy to rove that for n N 0, we have a(x + i n α ) 2 ax 2 (mod n ). Therefore, we see that A(x, y, z) C Q ( n ). Similarly, for an element (x, y, z) C Q ( n ) such that y, we define A 2 (x, y, z) := {(x, y + j n β, z) 0 j β 1}, and for (x, y, z) C Q ( n ) such that z, A 3 (x, y, z) := {(x, y, z + k n γ ) 0 n γ 1}. See that all A i (x, y, z) are subsets of C Q ( n ). Also, if A i (x 1, y 1, z 1 ) and A i (x 2, y 2, z 2 ) have an element in common, then A i (x 1, y 1, z 1 ) = A i (x 2, y 2, z 2 ). Also, if (x, y, z) is an element of C Q ( n ), then at least one of x, y or z is not divisible by, because (x, y, z) is a rimitive trile. Thus, we see that we can write C Q ( n ) as a disjoint union of sets S r, where each S r has the form A i (x, y, z) for some i {1, 2, 3} and (x, y, z) C Q ( n ). As a consequence, the roosition is roved if we rove that the number of elements in the reimage of A i (x, y, z) in C Q ( n+1 ) is 2 A i (x, y, z). Without loss of generality, we may rove this assertion for A 1 (x, y, z). Choose integers x, y, z such that x and (x, y, z) C Q ( n ). Elements in A 3 n+1 that ma into A 1(x, y, z) under π have the form (x + i n α, y + j n, z + k n ), where 0 i α+1 1, 0 j 1 and 0 k 1. However, (x + i n α, y + j n, z + k n ) lies in C Q ( n+1 ) if and only if (4.2) a(x + i n α ) 2 + b(y + j n ) 2 + c(z + k n ) 2 0 (mod n+1 ). Set D := Q(x, y, z)/ n, which is an integer since (x, y, z) C Q ( n ). Set a = 2ax/ α, b = 2by and c = 2cz. Note that a is an integer not divisible by by our choice of α. Since n N 0, the terms ai 2 2(n α), bj 2 2n and ck 2 2n are all divisible by n+1. Hence, Equation (4.2) is equivalent to (4.3) D + ia + jb + kc 0 (mod ). Recall that 0 i α+1 1 and 0 j, k 1. Since gcd(, a ) = 1, we see that we can choose i in α ways for every choice of j and k. Therefore, A i (x, y, z) has α+2 reimages in C Q ( n+1 ). Recall that A i (x, y, z) = α, and hence the roof is comlete. Theorem 4.10 and Theorem 3.12 give us an exlicit formula for Z Q ( n ) if is an odd rime and Q is nondegenerate modulo.

24 24 J. BAEK, A. DEOPURKAR, AND K. REDFIELD Theorem Let be an odd rime and Q(x, y, z) = ax 2 +by 2 +cz 2 be such that abc. Then, Z Q ( n ) = n 1 ( + 1). Proof. If abc, then we may take N = 0 in the statement of Theorem 4.9. Then we have, Z Q ( n ) = n 1 ZQ (). Since none of a, b, c is divisible by, we conclude from Theorem 3.12 that Z Q () = + 1. In articular, note that if Q is nondegenerate modulo, then it has zeros modulo k for all k N. The number 2 behaves a bit differently. We have the following result. Theorem Let Q(x, y, z) = ax 2 + by 2 + cz 2 be a quadratic form where a, b, c are odd integers. Then Z Q (2 n ) = 2 n 3 ZQ (8) for n 3. Proof. This follows from Theorem 4.9 since we can take N = 1. In articular, note that if Q is nondegenerate modulo 2, then it has zeros modulo 2 k for all k N if it has zeros modulo Integer oints on rojective conics Let a, b, c Z be nonzero integers such that gcd(a, b, c) = 1. Define a quadratic form Q by Q(x, y, z) = ax 2 + by 2 + cz 2. In the revious sections, we looked at the zeros of Q modulo n for different ositive integers n. In this section, we see if zeros of Q modulo n for all ositive integers n give any information about the integer zeros of Q. Let (x, y, z) Z 3 be such that Q(x, y, z) = 0 but (x, y, z) (0, 0, 0). Since Q is homogeneous, we can assume that gcd(x, y, z) = 1. See that (x, y, z) gives us a zero of Q modulo n, for all ositive integers n. In other words, (x, y, z) is an element of CQ (n) for all ositive integers n. Furthermore, (x, y, z) is also a real zero of Q. Hence, if Q has a nontrivial integer zero, then it must have a nontrivial zero modulo n for all n and also have a nontrivial real zero. One can ask if having a nontrivial real zero and a nontrivial zero modulo n for all n is sufficient to guarantee a nontrivial integer zero of Q. The celebrated Hasse Minkowski theorem asserts that this is indeed the case.

25 POINTS ON CONICS MODULO 25 Theorem 5.1 (Hasse Minkowski). [1,.2] 1 Let Q(x, y, z) = ax 2 + by 2 + cz 2, where a,b,c are nonzero integers. Then Q has a nontrivial zero in Z 3 if and only if it has a nontrivial zero in R 3 and a zero in P 2 n for all ositive integers n. In fact, it turns out that the statement is true even if we dro the hyothesis of Q having real zeros ([2, Ch IV, 4, Corollary 3]). In other words, we have the following: Theorem 5.2. Q has a nontrivial zero in Z 3 if and only if it has a zero in P 2 n for all ositive integers n. By Proosition 4.8, we see that Q has zeros in P 2 n for all ositive integers n if and only if it has zeros in P 2 for all rimes and ositive k integers k. Theorem 3.12 shows that Q has zeros in P 2 for all rimes. The two examles show that given an odd rime and k 2, we can construct a form Q that has zeros in P 2 for n < k but fails to have a n zero in P 2. k Examle 5.3. Let k 2 be even and an odd rime. Choose α Z such that α is a quadratic nonresidue modulo. Consider the form Q(x, y, z) = x 2 αy 2 k 1 z 2. Clearly (0, 0, 1) gives a rimitive zero of Q in P 2 for n < k. We claim n that Q has no zero in P 2. k Let x, y, z Z be such that Q(x, y, z) 0 (mod k ). Let u be the highest ower of dividing x 2 αy 2, and v the highest ower of dividing k 1 z 2. Since α is a nonresidue modulo, it follows that u is even. On the other hand, v is odd since k 1 is odd. Therefore, for Q(x, y, z) 0 (mod k ), we must have u k and v k. Hence divides z and x 2 αy 2. Again, since α is a nonresidue, divides x and y. Thus, (x, y, z) is not a rimitive trile modulo. It follows that Q has no zeros in P 2. k Examle 5.4. Let k > 2 be odd and be an odd rime. Choose α Z such that α is a nonresidue modulo, as before. Consider the form Q(x, y, z) = x 2 k 2 y 2 α k 1 z 2. Again, (0, 0, 1) gives a zero of Q in P 2 n no zero in P 2. k for n < k. We show that Q has 1 This is not the most common way to formulate the theorem. It is best hrased in terms of -adic numbers. See [2, Ch IV, 3].

26 26 J. BAEK, A. DEOPURKAR, AND K. REDFIELD Let x, y, z Z be such that Q(x, y, z) 0 (mod k ). It follows that k 2 x 2. Since k 2 is odd, we see that k 1 x 2. This, in turn, imlies that y 2, and hence 2 y 2. Thus, we have Letting x 2 = k 1 x 2 1, we get Q(x, y, z) x 2 α k 1 z 2 0 (mod k ). x 2 1 αz 2 0 (mod ). Since α is a nonresidue, we see that z and x 1. Thus, divides x,y and z, which shows that (x, y, z) is not a rimitive trile. It follows that Q has no zeros in P 2 k. Finally, the next examle shows that given any k 2, we can find an odd rime and a form Q that satisfies the following: (1) Q has zeros in P 2 qn for all rimes q and all n N. (2) Q has zeros in P 2 n for all n < k, (3) Q does not have a zero in P 2. k Examle 5.5. Let k 2 and be a rime congruent to 7 modulo 8. Consider the form Q(x, y, z) = x 2 + k 2 y 2 + k 1 z 2. Note that if k is even, then (1, 0, 1) gives a zero of Q in P 2 8. If k is odd, then (1, 1, 0) gives a zero of Q in P 2 8. By Theorem 4.11, we conclude that Q has zeros in P 2 2n for all n N. Next, Q is nondegenerate modulo all rimes q. By Theorem 4.10, we see that Q has zeros in P 2 qn for all odd rimes q and n N. Clearly, (0, 0, 1) gives a zero of Q in P 2 for n < k. We claim that n Q has no zero in P 2. Let x, y, z Z be such that Q(x, y, z) 0 k (mod k ). We see that k 1 x 2 + k 2 y 2 and k 2 x 2. We have two cases: Case (1) k is odd. In this case, k 2 > 0 is odd. Since k 2 x 2, we conclude that k 1 x 2, and hence y. Thus, we get 2 y, and hence Q(x, y, z) x 2 + k 1 z 2 (mod k ). Letting x 2 = k 1 x 2 1, we see that we have x z 2 0 (mod ). Since 3 (mod 4), we see that z and x 1. In articular, divides x, y and z. Hence (x, y, z) is not a rimitive trile.

27 POINTS ON CONICS MODULO 27 Case (2) k is even. Writing x 2 = k 2 x 2 1, the equation Q(x, y, z) 0 (mod k ) gives x y 2 + z 2 0 (mod 2 ). In articular, x y 2. Since 3 (mod 4), we see that x 1 and y. Thus, 2 x y 2, which imlies that z. Hence, divides x, y and z. Hence (x, y, z) is not a rimitive trile. In any case, we see that we do not get a zero of Q in P 2 k. Aendix A. Equivalence of Symmetric Matrices mod Let be a rime, and denote by S the set of n-by-n symmetric matrices over F. Consider the oeration on S that takes M S to A T MA, where A is an n-by-n invertible matrix. The oeration gives rise to a natural relation over S : for all M 1, M 2 S, where the relation M 1 M 2 holds if and only if there exists an invertible matrix A such that M 1 A T M 2 A (mod ). We note that is in fact an equivalence relation it is reflexive, symmetric and transitive. We show in the theorem below that a symmetric matrix modulo must be similar to a diagonal matrix: Theorem A.1. For any M S, there exists a diagonal matrix D and an invertible matrix A GL n () such that D A T MA (mod ) Proof. We will exlicitly consturct a diagonal matrix D satisfying the theorem. Consider M as a bilinear form. We begin by inductively showing that for each ositive integer k n, there exists a linearly indeendent set of vectors {v1,..., vk } such that (A.1) i j {1,..., k}, (v i ) T M(v j ) 0 (mod ). For k = 1, the condition in (A.1) is vacuous, so we can ick an arbitrary vector v1. For the inductive case, assume that we already have a linearly indeendent set of vectors {v1,..., vk } satisfying (A.1). By extending the set as a basis, ick a vector v k+1 such that v1,..., vk, v k+1 are linearly indeendent. Then we let k vk+1 v k+1 vi (vi ) T M(v k+1 ) (vi )T M(vi ) (mod ). i=1

28 28 J. BAEK, A. DEOPURKAR, AND K. REDFIELD First, for all j k + 1, we see that (v j ) T M(v k+1) (v j ) T M(v k+1 ) k i=1 (v j ) T M(v i ) (v i ) T M(v k+1 ) (v i )T M(v i ) (v j ) T M(v k+1 ) (v j ) T M(v j ) (v j ) T M(v k+1 ) (v j )T M(v j ) 0 (mod ), as desired. By symmetry of M, we also obtain (vk+1 )T M(vj ) 0 (mod ). Therefore, (A.1) still holds. In addition, we see that the vectors v1,..., vk+1 are linearly indeendent, so the inductive hyothesis is satisfied. The induction above yields a basis {v1,..., vn} satisfying (A.1). Now let A be the matrix formed by taking v1,..., vn as column vectors. A is nonsingular because the columns are linearly indeendent. Pick D = A T MA (mod ). Note that the desired condition in the theorem is trivially satisfied. It remains to check that D is diagonal. From our choice of A, we see that D ij = (vi ) T M(vj ), which is zero when i j. Thus we have successfully shown that M is similar to a diagonal matrix. Aendix B. Quadratic residues Quadratic residues are useful in treatments of quadratic equations modulo, where is a rime. Definition B.1. A number q Z / Z is called a quadratic residue modulo (or residue for short) if there exists x 0 (mod ) such that x 2 q (mod ). Zero is never a quadratic residue, since x 2 0 imlies x 0. It is also easy to see that there are exactly ( 1)/2 residues. The condition of a number being a quadratic residue is called the quadratic residuosity, and is exressed commonly by the Legendre symbol: ( ) a := 0, if a 0 (mod ), +1, if a is a quadratic residue modulo, 1, otherwise. An imortant roerty of Legendre symbols is that they are multilicative, given a fixed rime :

29 POINTS ON CONICS MODULO 29 Proosition B.2. For all a, b Z / Z, ( ) ( ) ( ) a b ab (B.1) =. ( ) ( ) ( Proof. If a or b is zero, then both and a b ) ab are zero, so (B.1) is satisfied. So we may restrict our attention to cases where a, b 0 (mod ). If a and b are both residues, then there exist α, β Z / Z such that α 2 = a and β 2 = b. Then ab is also a residue because (αβ) 2 = ab. Then both sides of (B.1) is 1, so it is satisfied. If a is a residue with α 2 = a and b is not, then it cannot be that ab is a residue. Suose the contrary, i.e. there exists δ Z / Z such that δ 2 = ab. Then b = ( δ 2, α) which contradicts the assumtion. The same holds when a is not a residue and b is. In either case, both sides of (B.1) is -1. The only remaining case is when a, b are both nonresidues. We can use the following counting argument: as a, b vary in {1,..., 1} each, the roduct ab takes each value in {1,..., 1} exactly 1 times. Since half of {1,..., 1} are residues, ab is a residue for exactly ( 1) 2 /2 airs of values of (a, b). We have already accounted for ( 1) 2 /4 airs where both a, b are residues. The remaining ( 1) 2 /4 airs then must come from airs where both a, b are nonresidues. But there are exactly ( 1) 2 /4 such airs, so it must be the roduct of two nonresidue is always a residue, satisfying (B.1). We have examined all ossible cases for quadratic residuosity of a and b, and in all cases, (B.1) holds. Aendix C. Primary Authors Section 2 was rimarily authored by Katherine Redfield; Section 3 and the aendices (Section A and B) were rimarily authored by Jongmin Baek; Section 1, Sections 4 and Section 5 were rimarily authored by Anand Deourkar. References [1] W. Aitken, F. Lemmermeyer: Counterexamles to the Hasse Princile: An Elementary Introduction, htt://ublic.csusm.edu/aitken_html/m372/ diohantine.df [2] J.-P. Serre: A Course in Arithmetic, Sringer-Verlag New York, 1973