THESIS. Presented in Partial Fulfillment of the Requirements for the Degree Master of Science in the Graduate School of The Ohio State University

The Hasse-Minkowski Theorem in Two and Three Variables THESIS Presented in Partial Fulfillment of the Requirements for the Degree Master of Science in the Graduate School of The Ohio State University By Steven D. Hoehner Graduate Program in Mathematics The Ohio State University 2012 Master's Examination Committee: Professor C. Herbert Clemens, Advisor Professor James Cogdell

Copyright by Steven D. Hoehner 2012

Abstract In this thesis we develop an explanation and proof of the Hasse-Minkowski Theorem for homogeneous quadratic forms in two and three variables using only undergraduate number theory. The goal of this approach is to provide an accessible introduction to this celebrated result of number theory for undergraduates and advanced high school students. Our account of the Hasse-Minkowski Theorem will be expository in nature, providing the reader with the necessary background information to state and prove the theorem. However, some mathematical maturity of the reader is presumed, including familiarity with unique prime factorization, greatest common divisor, least common multiple, the Division Algorithm, the Euclidean Algorithm, the Pigeonhole Principle, and some other basic notions from elementary number theory, set theory, and linear algebra. The remainder of the information necessary for the explanation and proof of the Hasse-Minkowski Theorem is provided in the paper, making it largely selfcontained in this respect. Moreover, throughout the paper we provide the reader with several numerical examples of the various concepts and methods introduced. We now provide the reader with a brief overview of the paper. The first chapter consists of a brief historical account of the study of Diophantine equations in number theory, including Hasse and Minkowski s important contributions to the study of quadratic forms. In Chapter 2 we motivate the discussion of the Hasse-Minkowski Theorem by introducing the type of solutions we seek, and in Chapters 3 and 4 we ii

provide some background information on quadratic forms in two and three variables, respectively. Chapter 5 serves as an introduction to modular arithmetic and some of the results that will be used in the proof of the Hasse-Minkowski Theorem, while in Chapters 6 and 7 we state and prove the Hasse-Minkowski Theorem for forms in two and three variables, respectively. Finally, in Chapter 8 we discuss topics for further study. iii

Dedication In memory of my grandmother Rosemary J. Hoehner. iv

Acknowledgments The completion of this thesis would have been impossible without the support of my committee members, colleagues, family, and friends. For this reason, I am indebted to all of you. I would like to express my deepest gratitude to my advisor, Dr. Herbert Clemens for his encouragement, insight, patience, and guidance in completing the thesis, as well as his support during my two years in the Mathematics for Educators program. I would like to express the same gratitude to Dr. James Cogdell for his guidance with the thesis, as well as for helping me develop an interest and background in number theory. Finally, I would like to thank my girlfriend Jordan Kidder and my fellow colleagues in the Mathematics for Educators program for their help and support during the past two years. v

Vita June 2004... Centennial High School 2008... B.S. Applied Mathematics, Columbia University 2010 to present... Graduate Teaching Associate, Department of Mathematics, The Ohio State University Fields of Study Major Field: Mathematics vi

Table of Contents Abstract... ii Dedication... iv Acknowledgments... v Vita... vi Fields of Study... vi Table of Contents... vii Chapter 1: Introduction... 1 Chapter 2: Motivation... 3 Chapter 3: Binary Quadratic Forms... 5 Chapter 4: Ternary Quadratic Forms... 8 Chapter 5: Modular Arithmetic... 11 Chapter 6: The Hasse-Minkowski Theorem for Binary Forms... 17 Chapter 7: The Hasse-Minkowski Theorem for Ternary Forms... 26 Examples... 42 Chapter 8: Topics for Further Study... 44 References... 46 vii

Chapter 1: Introduction Diophantine equations are polynomial equations with integral coefficients, such as or. The genesis of the study of Diophantine equations is credited to Diophantus, a Greek mathematician who lived from about 200-284 A.D. In his work Arithmetica, Diophantus outlined methods for finding the exact solutions of these equations, and he did so without the use of modern algebraic methods, convenient symbolic notations, or even negative numbers. As a result of his work, the subject dealing with finding exact or approximate solutions to these equations is called Diophantine analysis, and is generally considered a topic of number theory. Following Diophantus work, the Indian mathematician Brahmagupta (ca. 628) was the first to give the general solution of the linear Diophantine equation, where and are integers (LeVeque 27). For this equation to have integral solutions, it must be true that the greatest common divisor of the coefficients and divides. Moreover, Brahmagupta knew that if and are relatively prime then the solutions of this equation exist, and given one solution the remaining solutions are given by and, where is an arbitrary integer. Although the early work on linear Diophantine equations is quite notable, the study of quadratic forms has also played a significant role in number theory. A quadratic form is a homogeneous Diophantine equation of degree 2 (recall that a polynomial is homogeneous of degree if all its monomials have the same total degree ). For 1

example, the quadratic polynomial is homogeneous because its monomials ( and ) all have degree 2. Thus, is a quadratic form. On the other hand, the polynomial is not homogeneous because all of its monomials are of degree 2 except for the term, which has degree 1, so is not a quadratic form. In 1788, the French mathematician Adrien-Marie Legendre gave necessary and sufficient conditions for the equation to have integral solutions, where the coefficients and are nonzero integers, not all of the same sign, whose product is not divisible by any squares. Following Legendre s remarkable solution to the problem of determining the integral solvability of, there was a great deal of interest in the mathematical community to generalize his theorem. One of the first to do this was a young German mathematician named Hermann Minkowski. In his theorem, Minkowski formulated criteria that could be used to determine the existence of integral solutions to homogeneous quadratic forms in any number of variables (although for the purposes of this paper we shall restrict our attention to the two and three variables cases). A few decades later in 1921, another German mathematician named Helmut Hasse provided the proof of a theorem that further generalized Minkowski s result. As a result of their combined effort, today the resulting theorem on the solvability of a homogeneous quadratic equation is called the Hasse-Minkowski theorem. 2

Chapter 2: Motivation In this paper, we provide the reader with an expository account and accessible proof of the Hasse-Minkowski theorem for homogeneous quadratic forms in two and three variables using only undergraduate number theory. Before stating and proving the theorem, we must fix the terminology that will be used throughout the paper and provide some background information on elementary topics in number theory that will be necessary for the proof. We begin with a few definitions and a discussion on the kind of solutions we are looking for. Suppose we have a nontrivial solution in rational numbers. Multiplying each entry of the solution by the least common denominator of the coordinates, we obtain a nontrivial integral solution. For example, the triple is a solution of the equation, and by multiplying each of the three coordinates by, we obtain the integral solution. Note that we can always use this method of using a rational solution to construct an integral solution because we are considering homogeneous forms. In fact, by homogeneity a quadratic form is solvable in the rational numbers if and only if it is solvable in the integers, so we may focus only on the problem of integral solvability. A tuple of numbers for which is called a solution of the quadratic form. The solution is called the trivial solution because it is always solution of regardless of what the coefficients of the polynomial 3

are. Therefore, we seek nontrivial solutions, which are solutions such that for at least one, where or. Moreover, we define a primitive solution to be a nontrivial solution in which the greatest common divisor of the coordinates is. By homogeneity, we see that nontrivial solvability implies primitive solvability, so the two are equivalent. Thus, we may reduce the problem of determining nontrivial solvability to one of determining primitive solvability. 4

Chapter 3: Binary Quadratic Forms The general equation of a binary (i.e. two variable) quadratic form is given by. The number is called the discriminant of. For example, the polynomials and are all examples of binary forms and their discriminants are and, respectively. Consider the binary quadratic form. We attempt to write this equation in the simpler diagonal form by completing the square (this reduction to a diagonal form makes the problem of determining if there are integral solutions much more straightforward). To do this, we first find the terms containing the variable, which are and. Now we have, so, which is the desired form for and. We may also follow this procedure for the general form, where. Finding the terms containing, we write. Thus,, which is the desired diagonal form with 5

and. Since we transformed the form into the form using a linear change of coordinates, the two forms are equivalent. More specifically, we say two forms and are equivalent if there is a integral matrix with such that. From the definition of equivalence we see that given two equivalent forms and, has a nontrivial integral solution if and only if does. Hence the integral solvability of the general form is equivalent to the integral solvability of the diagonal form. Thus, in the two variable case we need only consider equations of the form, where are integers. Note that if at least one of the coefficients in our diagonal form is rational, then by homogeneity of the equation we can clear denominators to obtain an equation with only integral coefficients. We can also assume that since if either of the coefficients is zero, then there are no nontrivial solutions to. Also, we may assume that the greatest common divisor of and is 1. To see this, suppose that. Then dividing both sides of by, we obtain the equivalent integral equation, where Next, we show that we may assume that the coefficients and are square-free, meaning they are not divisible by any squares (for example, the number is square-free, but is not because it is divisible by, which is a square). Suppose first that we have a diagonal form where is not square-free. Then, where and 6

are integers. Then. We may repeat this process for all of the squares that divide, and similarly we may absorb the square factors of into the term, until we have reached a form with and square-free. If solves this last equation, then gives a solution of the original equation. Thus, since the integral solvability of is equivalent to the integral solvability of with squarefree, we may assume from now on that the coefficients are square-free. We will state and prove the Hasse-Minkowski criterion for solvability of binary quadratic forms in Chapter 6. But before concluding this chapter, we make some final remarks regarding the coefficients of our diagonal quadratic form. First, note that if, then, where equality holds if and only if. Thus, there are no nontrivial real solutions to this equation if both and are positive, so there are no nontrivial integral solutions either. Similarly, we see that there are no nontrivial solutions if and are both negative, so for to have a chance of having nontrivial integral solutions, it must be true that and have opposite sign (or equivalently, ). For example, the equation has no nontrivial solutions because both and are positive. The sign condition imposed on the coefficients and allows us to eliminate the blatant offenders that have no nontrivial real solutions, but what about other, less obvious cases such as? 7

Chapter 4: Ternary Quadratic Forms The general equation of a ternary (i.e. three variable) quadratic form is given by. For example, the polynomials,, and are all ternary forms. Like in the two variable case, we would like to reduce the problem of determining the solutions of the general ternary form to that of determining solutions to the diagonal form where and are nonzero integers. Again, we complete the square to write in diagonal form. First, consider the ternary form. We try to find a linear form such that the when the linear form is squared, the resulting terms are. With the proper choice of the linear factor, we see that, so. We now want to finish this process by diagonalizing. We complete the square again, and find that Thus we have, which is 8

now in diagonal form with and. Setting equal to 0, we see that the solvability of this equation is equivalent to the solvability of the integral equation, where the first equality was obtained by absorbing the square factors of the coefficients into the variable terms. We now show that the general ternary form can be diagonalized to a form. We first complete the square on the part of that contains the variable, namely. Now, so we have, where and. Next, we complete the square on the form, and see that, where and. Therefore,, which is now in the desired diagonal form. Since we used a linear change of coordinates to transform the general form into the diagonal form, the general and diagonal forms are equivalent. Therefore, the general form 9

has nontrivial solutions if and only if the diagonal form does, where and are defined as above. After diagonalizing the ternary form, using a similar procedure to that of the two variable case, we may absorb the square factors of the coefficients into the squared variable terms. Hence, we may assume that the coefficients are square-free. Moreover, by homogeneity of the equation, we may assume that the greatest common divisor of the coefficients is. We now show that we can further assume that the coefficients are relatively prime in pairs. To this end, suppose we have the form with and square-free and, and that two of the coefficients, say and share a common prime (note that since and are squarefree and relatively prime, only the first power of divides and ). Then and, so we must have that. Thus, either or and since we must have. Writing and factoring out a we have Dividing both sides of this equality by, we obtain the form with We may repeat this process for the remaining prime factors that and share, until the g.c.d. of the two coefficients is. In a similar fashion, we may eliminate all of the prime factors that and share as well as those that and share, until we have arrived at a form with square-free, pairwise relatively prime coefficients. As we will see in Chapter 7, this representation of the quadratic form is convenient for proving the Hasse-Minkowski theorem for ternary forms. 10

Chapter 5: Modular Arithmetic In order to prove the Hasse-Minkoski theorem in two and three variables, we will need to explore modular arithmetic. Modular arithmetic is a system of arithmetic used on the integers where the numbers start over after they reach a certain value, called the modulus. Perhaps the most familiar use of modular arithmetic occurs in 12-hour clocks. For example, if the current time is 7:00, then 9 hours later it will be 4:00. Usual arithmetic would suggest that the time should be o clock. However, there is no 16 o clock in 12-hour time, because the numbers start over after they have reached 12. From our experience with clocks, we know that because hours have passed since 12:00, the actual time is 4:00. Modular arithmetic is carried out in the same way as clock arithmetic, but with an arbitrary modulus. More specifically, if a nonzero integer divides the difference, where are integers, then we say is congruent to modulo, and we write If does not divide, then we say is not congruent to modulo, and we write. For example,, and. Much of the algebra from usual arithmetic carries over to modular arithmetic, but special care must be taken when performing division. For instance,, but. The following theorem states when we can perform division in modular arithmetic. 11

Theorem 1. Suppose. If and, then To prove Theorem 1, we make use of the following result. Theorem 2. If and are integers, then for some integers and. PROOF. The conclusion of the theorem is obvious if, so assume is nonzero. Consider the set. Both and are in (since and ), as are and. Thus, contains some positive integers, because one of must be positive. Let be the smallest positive integer contained in. We must show that. To see this, note that since, there exist integers and such that. To show, we need to show that is a common divisor of and, and that it is the largest common divisor of the two. By the Division Algorithm, we can write, where. Now, so. If, then we have a contradiction to the minimality of. Thus,, so. Similarly, we see that, so is a common divisor of and. All that remains to be shown is that is the largest divisor of and. Suppose that is a positive common divisor of and. Then since. This completes the proof. PROOF (Theorem 1). If, then and vice versa. Since, by Theorem 2 we can write for some integers and. 12

Hence ; since and, also. That is,. Corollary to Theorem 1. Suppose and is a prime. If, then. PROOF. Since is prime,, so the conclusion follows from Theorem 2. Theorem 3. Let be an integer and be a positive integer. Then if and only if there exists an such that. Moreover, any two such are congruent modulo (in other words, is unique modulo ). PROOF. If, then by Theorem 1 there exist integers and such that. Thus,. Conversely, suppose. Then there exists an integer such that, so. Hence if, and so by Theorem 2 we have. The problem of determining whether there are integral solutions to a homogeneous quadratic form is closely related to the problem of determining whether number is a square modulo. A square modulo is defined as an integer such that (i) and (ii) the congruence has a solution. If the congruence has no solution, then is a non-square modulo. For example, the integers modulo are and. The squares of these numbers are 13

and, respectively. Therefore, the nonzero squares modulo are and, while and are non-squares modulo. Legendre introduced a convenient notation to represent the property of an integer being a square or non-square modulo an odd prime. The Legendre symbol is defined as follows: Moreover, the Legendre symbol has the following properties: I.. Therefore, the product of two squares or non-squares is a square, and the product of a square and a non-square is a non-square. II. III. If, then.. IV.. V.. Property V is the celebrated Law of Quadratic Reciprocity, a very deep result with numerous applications to number theory. For a proof of properties I-V of the Legendre symbol, we refer the interested reader to [3], pages 99, 100, and 102. Next, we provide the reader with the definition of and some properties of the Jacobi symbol, which is a generalization of the Legendre symbol. The Jacobi symbol can be used to determine if an integer is a square modulo a composite number. We define the 14

symbol by defining and if is the prime factorization of some integer, then we define, where is the aforementioned Legendre symbol for. One must be careful when using the Jacobi symbol, however, because does not necessarily imply is a square modulo. To see this, note that it is necessary for for each prime for to be a square modulo. However, if for an even number of, then even though is a non- square modulo. For example, note that the only nonzero square modulo is, so, and using property IV of the Legendre symbol we see that. Hence, by definition of the Jacobi symbol we calculate, but checking values we see that is not a square modulo (the squares modulo are and ). In general, if then no conclusive statement can be given as to whether is a square modulo. In these cases, alternative methods must be used to determine whether is a square modulo. On the other hand, if, then is surely not a square modulo. We will make use of this conclusive property of the Jacobi symbol in Chapter 6 to prove a result that is necessary for the proof of the Hasse-Minkowski theorem for binary forms. Before concluding this chapter, we provide the reader with a few properties of the Jacobi symbol that will be used in the proof of the Hasse-Minkowski Theorem. We omit the details of the proofs here, and refer the interested reader to page 110 in [3]. 15

A.. B. If, then. C.. D.. E. If, then. 16

Chapter 6: The Hasse-Minkowski Theorem for Binary Forms Consider the binary form, where and are square-free, pairwise relatively prime integers of opposite sign. If we have an integral solution to the equation, then for each prime we have a nontrivial solution of the congruence. Therefore, if one can find a prime for which has no solutions modulo, then there are no nontrivial integral solutions to. For instance, by checking cases we see that the congruence has no solution modulo, and therefore the equation has no nontrivial integral solutions. On the other hand, if has a real solution and the congruence holds for some, this does not guarantee the existence of an integral solution to the equation. This leads us naturally to ask the following question: if has a real solution and the congruence is solvable for each prime, does it follow that has an integral solution? In the two variable case, the answer to this question is yes. Theorem (Hasse-Minkowski 1). Let be nonzero, square-free, relatively prime integers of opposite sign. If for each prime the congruence has a solution in integers not both divisible by, then has a nontrivial integral solution. 17

Before proving the theorem, we must prove a few results that we will use in its proof. The first theorem we state and prove is the celebrated Chinese Remainder Theorem. Roughly speaking, the theorem states that it is possible to solve a system of linear congruences with moduli that are pairwise relatively prime. In addition, the Chinese Remainder Theorem provides a solution modulo the product of those moduli. The Chinese Remainder Theorem is one of the most powerful theorems in number theory, and we will use it to prove theorems that will be used in both parts of the proof of the Hasse-Minkowski theorem. Chinese Remainder Theorem. Suppose are positive integers that are relatively prime in pairs, and let be any sequence of integers. Then the congruences have common solutions. Moreover, given any solution that satisfies the above system of congruences, then an integer satisfies the system if and only if where and is some integer. (In other words, the solution is unique modulo.) PROOF. We first show the existence part of the theorem. To this end, let be defined as in the statement of the theorem. Then is an integer for. Since the moduli are relatively prime in pairs, we see that for. Thus, 18

for each there exists an integer such that. Note that if, then, so for all. Now if we let, then using the previous two congruences we see that. Thus, is a solution to the above system of congruences. To show uniqueness, suppose and are two solutions of the system of congruences, so that for. Then for, so is a common multiple of Hence,. Since the are relatively prime in pairs,. Thus, Therefore,, so. This completes the proof. Theorem 4. Suppose is an integer, is a natural number, and let be the prime factorization of. Then is a square modulo if and only if is a square modulo for. PROOF. Suppose first that is a square modulo. Then for some. Since for, this implies for. This proves one direction of the theorem. Conversely, suppose is a square modulo for. Then the equation is solvable for. Since for any, by the Chinese remainder theorem there exists an integer such that 19

for. Thus, for each we have, so by the Chinese Remainder Theorem we have modulo.. In other words, is a square Theorem 5. The congruence is solvable for every prime if and only if for some. (In other words, is a square modulo for every prime if and only if is a square.) PROOF. Suppose first that for some integer. Then the congruence has the solution for every prime. We prove the other direction of Theorem 5 by proving the logically equivalent statement If is not a square, then is not a square modulo for some prime. By Theorem 4, it suffices to show that is not a square modulo for some odd positive integer since then must have some prime factor for which is not a square modulo. We use some of the properties of the Legendre and Jacobi symbols in this proof. If, then for to be a non-square it must contain a prime factor of odd degree. We claim that if and is a non-square, then or where are positive odd integers, is prime, and. In the first case, since is odd, is a non-square. In the second case, if there is an even power of in the factorization of, for to be a non-square there must exist an odd prime factor of that has odd degree. Moreover, the factor ensures is not a square since is odd, and for any positive odd integer (including ) this property does not change if is multiplied by. All that remains to be shown is that any positive non-square integer is 20

of the form or. If is even, then some power of divides. If this power is odd, we are in the first case, and if this power of is even, we are in the second case. Moreover, if is odd, then we are in the second case with. Finally, if then is not a square, and by the previous argument we deduce that all negative numbers are of the form or where and are defined as above. Therefore, the following three cases exhaust all instances in which is not a square. Case I., where and are positive odd integers. Since is odd,. Thus, by the Chinese Remainder Theorem we can choose an integer to satisfy and simultaneously. Now, where the last equality follows from two applications of property A of the Jacobi symbol (see Chapter 5). Now since, for some integer. Thus, using property D of the Jacobi symbol we see. Moreover, by property C we have, where the last equality follows from the fact that is even for any integer. Thus, by property A we see as well, we may write. Therefore,., and since 21

Now by applications of property A, we see that. But since and is even,. Hence, Now since and, by property E of the Jacobi symbol we have. But is odd and is even, so the product is even. Hence,, and since, we have or. Thus in either case we have, so. Now using property B of the Jacobi symbol and the fact that we see. Therefore,, so is not a square modulo. Case II., where and are positive odd integers, is prime, and. Then, so by the Chinese Remainder Theorem we can choose to satisfy and, where is a non-square modulo. Then since, we can write where is some integer. Now we have three applications of property A of the Jacobi symbol., where the last equality follows from By property C of the Jacobi symbol, we have, where the last equality follows from the fact that is even. Also, so we may express both of these equalities compactly as. Thus, from the last equality in the previous paragraph, we see that 22

. By applications of property A, we have, and since we see since is even. Hence,. Now using property A times, we have, so next we calculate. Note that by property E we have, where the final equality follows from the fact that is even. Hence,. Moreover, by property II of the Legendre symbol we have, and since is a non-square modulo we see that. Thus, (since is odd), so we have. By property E, we have since is even. Thus, so and since we have, as desired. Case III.. Pick an integer such that. Using property A, we see. But, so. Hence, Now since, for some integer, so by property C we have. Thus,, so is not a square modulo. This concludes the proof. 23

PROOF (Hasse-Minkowski Theorem 1). Suppose where and are square-free, relatively prime integers such that, and that for all the congruence has a nontrivial solution modulo. If, then we claim that. To see this, we argue by contradiction and suppose. Then, so, and hence either or. Now since we assumed, we must have, so and, which contradicts our assumption that the solution is nontrivial modulo. This contradiction establishes our claim that. Now for each we have, and multiplying both sides by we obtain Since for all, for each we can divide both sides of the congruence by twice to obtain the congruence. Hence, is a square modulo for all. If, then, so is a square modulo for all. Therefore, is a square modulo for all primes, so by Theorem 5 we see for some integer. Now consider the integral pair. Then, so we have found a nontrivial integral solution to the equation. This completes the proof of the Hasse-Minkowski theorem for binary quadratic forms. A homogeneous binary form is locally solvable at if it has a nontrivial solution to the homogeneous congruence modulo. For example, the congruence 24

is not locally solvable at the prime, but is locally solvable at. We say the form is locally solvable if it has a real solutions and is locally solvable at for each prime. The form is globally solvable if it has an integral solution. Clearly, global solvability implies local solvability, so the only interesting question is if local solvability implies global solvability. The Hasse-Minkowski theorem asserts that local solvability is equivalent to global solvability. More specifically, the two variable case of the theorem asserts that if we have a real solution and a solution to the congruence modulo for all primes, then we have an integral solution. For this reason, the Hasse-Minkowski theorem is referred to as a local-to-global theorem. 25

Chapter 7: The Hasse-Minkowski Theorem for Ternary Forms Consider the ternary form, where, and are square-free, pairwise relatively prime integers that do not all have the same sign. In the three variable case, real solvability and solvability of the congruence modulo for every prime is not enough to guarantee the existence of nontrivial integral solutions to. For example, consider the ternary form. We claim that for each prime, the congruence has a nontrivial solution modulo. By inspection we see is a solution to the congruence modulo and is a solution to the congruence modulo. We now show the congruence has a solution for all odd. Assume. Let, and let. We now show all of the members of are distinct. If, then, so we have. Hence, or. But, so. Thus we have, and since we have. Therefore, the elements of lie in distinct residue classes modulo, and by the same argument we see that the elements of lie in distinct residue classes modulo. Since the total number of residue classes is greater than, by the Pigeonhole Principle a member of 26

must be congruent to a member of modulo. In other words, for some pair of integers and. Hence,, so is a solution of the congruence modulo. Since this can be done for any, we have shown that the congruence has a solution for all. To demonstrate that the equation has no nontrivial integral solutions, it suffices to show there exists a prime for which the congruence has no solution. Consider the congruence. Using the fact that the only squares modulo are and, by checking cases we see that this congruence has no nontrivial solution modulo, and hence has no nontrivial integral solutions. By our counterexample, we see that if has a real solution and for each the congruence has a nontrivial solution modulo, then it does not necessarily follow that has an integral solution. However, if we impose stricter requirements for the type of solutions we have modulo, then it does follow that has nontrivial integral solutions. We call a nontrivial integral solution to the congruence -focused if at most one of is divisible by. Theorem (Hasse-Minkowski 2). Let be nonzero, square-free, pairwise relatively prime integers not all of the same sign. If for each odd prime the congruence has a -focused solution in integers, then has a nontrivial integral solution. 27

As mentioned in Chapter 1, Legendre elegantly addressed the problem of determining when the form has nontrivial integral solutions, where and are square-free, pairwise relatively prime integers not all of the same sign. The theorem is remarkable for its succinctness and power, because it shows that only three simple conditions need to be satisfied to ensure the existence of an integral solution. These conditions are provided below in the statement of Legendre s Theorem. Before proving Legendre s Theorem, we must state and prove a few results that will be needed in its proof. Following this, we prove Legendre s Theorem, and then we prove the Hasse-Minkowski theorem for ternary forms by showing it is a consequence of Legendre s Theorem. Theorem (Legendre). Let and be square-free, pairwise relatively prime integers not all of the same sign. Then has a nontrivial integral solution if the following conditions are satisfied: (i) is a square modulo, (ii) is a square modulo, and (iii) is a square modulo. At this point, one should note that although the theorem states these conditions are sufficient, Legendre showed that they are also necessary. For the proof of the Hasse- Minkowski theorem, however, we only need the conditions to be sufficient. 28

Lemma 1. Let be positive real numbers such that an integer. Then any congruence has a nontrivial integral solution such that and. PROOF. Let range over the values over and over where denotes the greatest integer not exceeding. This gives different triples. Now so (since there are only values modulo ) by the Pigeonhole Principle there exist two distinct triples with entries in these ranges such that Thus, we have and as desired. Lemma 2. Suppose that factors into linear factors modulo and modulo. That is, suppose If then also factors into linear factors modulo PROOF. Using the Chinese remainder theorem, we can choose integers to satisfy 29

Therefore, the congruence holds modulo and modulo and hence it also holds modulo Theorem 6. Suppose that let denote the number of ordered pairs of integers such that and and let denote the number of solutions of the congruence Then. PROOF. Consider any nontrivial solution of Of the four nontrivial ordered pairs exactly one of these pairs has positive first coordinate and nonnegative second coordinate. Let denote the number of ordered pairs that satisfy such that and Then so to prove the lemma we will show To do this, we define a function from the set to the set that is one-to-one and onto. This will show that the two sets have the same cardinality, i.e. We accomplish this in three steps. First, we define a function that maps the ordered pairs of to the appropriate residues classes Next, we show that this function is one-to-one, and finally we show that the function is onto. Suppose and are integers such that and We claim that To see this, assume that this is not the case. Then there exists some prime such that and Hence which contradicts the g.c.d. 1 condition. This contradiction establishes the claim. Now since and we can divide both sides of this congruence by twice to obtain (upon 30

further manipulation) Taking we obtain Next, we show that our function from the ordered pairs counted by to the residue classes counted by is one-to-one. Suppose for we have and. We will show that if then. Assume that. Then, and from our assumption we see. Now since, we can divide both sides by to arrive at the congruence. But because for we see that and, whence it follows that and. Thus, and. Since these numbers both lie in the interval and are congruent modulo they are equal. Moreover, since we have. From this we deduce that from the fact that Similarly,. Since for we have. Substituting this into the equation, we conclude, as desired. Hence we have established that the function is one-to-one. The final step in our proof of the lemma is in showing our function is onto. More precisely, given any such that we must show that there exists a primitive integral pair such that Suppose such an has been given, so that. Then there exists an integer such that, which implies. Hence 31

is a binary quadratic form with discriminant. We claim that the form is positive definite, meaning for all integral pairs. To see this, note that for all integral pairs, and since we assumed, we conclude that is positive definite (and ). We now show that is equivalent to a positive definite reduced binary quadratic form. A positive definite form is called reduced if and if, in addition, equality holds in one of the inequalities (i.e. or ) then. To finish the proof that our function is onto, we will need the following theorem. Theorem 7. Each equivalence class of positive definite binary quadratic forms of discriminant contains at least one reduced form. PROOF. Let be an equivalence class of positive definite binary forms of discriminant, and let be an element of such that is minimal amongst the elements of. Then we have, since is equivalent to (use the matrix ). Let. Then we have, where 32

, and. Hence is equivalent to the form. We now claim that. To see this, first note that since is defined to be the greatest integer not exceeding, we have Multiplying the previous inequality by we obtain, whence it follows that.. Hence,, so, which establishes the claim. Since is minimal, by our argument above we see that. The only problem that could occur at this point would be if and. In this case, if we let, then where and. Hence is reduced. This completes the proof of Theorem 7. (Proof of Theorem 6 continued) Next we claim that if is a positive definite reduced form with discriminant, then. To see this, first note that since, so. Since is positive definite, we deduce and. Moreover, since is a reduced form, and. Thus. From this inequality we see 33

, so. Therefore, if is a positive definite reduced form with discriminant, we have, which implies since is an integer. Next, we claim that all positive definite binary forms with discriminant are equivalent. By Theorem 7 it suffices to show that there is exactly one positive definite reduced binary form with discriminant, because then there would be only one equivalence class of positive definite forms with discriminant, meaning all such forms are equivalent. Consider the positive definite reduced form with discriminant. To prove the claim we show that is the only positive definite reduced form with discriminant. To this end, suppose is a positive definite reduced form with discriminant. We want to show. By the inequality at the end of the previous paragraph we have, so. Since is reduced, or, and since is an integer in either case we have or. If, then the discriminant of is, and since we assumed the discriminant of to be we have, which is impossible because is an integer. Hence, which implies. Since the discriminant of is, we have, so and thus. Therefore, is the only positive definite reduced form with discriminant, so by Theorem 7 we deduce that all positive definite forms with discriminant are equivalent. 34

By the argument in the preceding paragraph, since is positive definite and has discriminant, it is equivalent to the reduced positive definite form. Thus by definition of equivalent forms there exists a matrix such that. Hence, and by comparing coefficients we see,, and. Furthermore, since, by Theorem 2 we have. Therefore where the congruence follows since (i.e. ). Now since we have. Thus, so. If and, then it suffices to let since then we have,,, and. Thus in this case we have found a primitive integral pair such that and we are done. On the other hand, if one of the inequalities and does not hold, then we can choose to be one of the primitive integral pairs. To see this, note that by the congruences and we have. Thus in any of these cases we have. Therefore, in all cases we have shown that given any there exists a primitive integral pair such that 35

, which shows the function is onto. Hence, and the proof is complete. Now that we are equipped with the previous results, we are ready to prove Legendre s Theorem. PROOF (Legendre s Theorem). We want to show that if is a square modulo is a square modulo and is a square modulo then has nontrivial integral solutions. We see that this property remains unchanged if we replace the coefficients by their negatives. Since are not all of the same sign, by multiplying both sides of by (if necessary) we may assume that one of them is positive and two of them are negative. Moreover, (if necessary) we may also change variables to assume and For example, given the equation we may multiply both sides by to obtain and changing variables we can write this as Define to be a solution of the congruence (note that exists because of our assumption that is a square modulo ). Since by Theorem 3 there exists a unique integer such that. Then we have which implies. Hence, is a product of linear factors 36

modulo and similarly it is a product of linear factors modulo and modulo Using Lemma 2 twice, we have is a product of linear factors modulo. Thus there exist integers such that. Next, we apply Lemma 1 to the congruence. To this end, we define the positive real numbers and so that, a positive integer. By the lemma, there exists a solution such that. Since are square-free and relatively prime in pairs, we see that is an integer only if it is equal to, and similarly for and. Hence, we have, with equality only if, with equality only if and, with equality only if and. Since and, and unless. Thus, ignoring the special case for now we have. Now since is a solution of, from the congruence we see that it is also a solution of. Using this congruence and the previous inequality, we see that or. In the first case, we have our solution and we are done. In the second case, we may write this equality as. Multiplying both sides by, we have 37

. Thus, we have, so is a solution of. If at least one of is nonzero, then the solution is nontrivial and we are done. On the other hand if, then which implies because is square-free. Hence and, so. Therefore, if, then is a nontrivial solution. We now consider the special case. Since is a square modulo, the congruence is solvable. Hence, where is the number of solutions of the previous congruence. Thus by Lemma 3 we have, so there exists a primitive solution to the equation. From this equation we obtain, so is a solution of. This completes the proof of Legendre s Theorem. ternary forms. Using Legendre s Theorem, we now prove the Hasse-Minkowski theorem for PROOF (Hasse-Minkowski Theorem 2). Let where and are nonzero, square-free, pairwise relatively prime integers not all of the same sign, and for each odd assume that has a -focused solution. 38

First suppose is an odd prime such that and the congruence has a -focused solution. We claim that is a square modulo. Note that by Theorem 4 it suffices to prove is a square modulo for all, so this is what we show. Let be a -focused solution to the congruence. For each we have. If or if, then, which is a square modulo. So suppose. Then we have, and since is -focused, either divides none or exactly one of. Suppose first that divides none of. Then, and since by Theorem 1 we can divide both sides of by twice to obtain. Multiplying both sides of this congruence by we see, so is a square modulo. Now suppose divides exactly one of. If we re done, so suppose divides exactly one of. We assume and (the case for and is similar). Then we have, and since we can divide both sides of this congruence by twice to obtain. Multiplying both sides by, we see, so is a square modulo. Therefore, since has a -focused solution for all, by Theorem 4 is a square modulo. Similarly, since has a -focused solution for all odd and all odd, is a square modulo and is a square modulo. 39

Notably absent from the statement of the theorem is any condition regarding the prime. We now show that no such condition is needed, so that it suffices to consider only the odd. To see this, note that are always squares modulo because they are either odd or even, and so are congruent to or modulo, which are both squares. Therefore, since is automatically a square modulo, by Theorem 4 we need only show is a square modulo for all odd to show is a square modulo. Similarly, to show is a square modulo and is a square modulo, we need only show is a square modulo for all odd and is a square modulo for all odd, respectively. We now show that if the congruence has a focused solution for all odd, then has a nontrivial integral solution. Since the congruence has a -focused solution for all odd, it has a -focused solution for all odd, all odd, and all odd. Hence is a square modulo, is a square modulo, and is a square modulo. Thus we have satisfied the hypotheses of Legendre s Theorem, so has a nontrivial integral solution. This completes the proof of the Hasse-Minkowski Theorem for ternary forms. The Hasse-Minkowski Theorem for ternary forms reduces the problem of determining if has nontrivial integral solutions to one of determining if has a -focused solution for finitely many primes (namely for 40

the odd ). With this in mind, we can now fully state the method for determining if has nontrivial integral solutions outlined in this paper. Given a general homogeneous ternary form, we first diagonalize so that, where are defined as in Chapter 3. If all have the same sign, then has no nontrivial real solutions (and thus no nontrivial integral solutions) and we are done. If are not all of the same sign, then may have rational coefficients, in which case we multiply both sides of the previous equality by the least common denominator of to arrive at an equation with integral coefficients. Having done this, using the process outlined in Chapter 3 we may perform a series of transformations to until we have arrived at the form with integral coefficients that are square-free, pairwise relatively prime, and not all of the same sign. This form is now the same as the one in the statement of the hypothesis of the Hasse-Minkowski Theorem, so we factorize to find all of the odd primes that divide. Now for each of these we must determine if the congruence has a -focused solution. If for each we can find a -focused solution to the congruence, then by the Hasse-Minkowski Theorem the equation has nontrivial integral solutions. 41

Examples We now provide some examples of determining if a given ternary form has nontrivial integral solutions. First consider our earlier example. Then and so. The only odd prime that divides is, so consider the congruence. Then we have, and since and are the only squares modulo we deduce is the only solution to the above congruence, so any nontrivial solution to must be of the form where. Therefore, the congruence has no -focused solution, so by the Hasse-Minkowski Theorem the equation has no nontrivial integral solutions. Next consider. The odd primes that divide the product of the coefficients are and. Hence to show has nontrivial integral solutions we must show has a -focused solution for and. First consider the congruence. Reducing this congruence modulo we have, and by inspection we see is a -focused solution to the congruence. Next, consider the congruence. Since the only squares modulo are and, we see that is a -focused solution to the congruence. For the prime we have, and by checking values we see is a -focused solution to. Finally, we 42

want to determine if there is a -focused solution to the congruence. The squares modulo are and. By checking values, we see is a -focused solution to. Thus we have shown that has a -focused solution for all odd primes that divide the product of the coefficients, so by the Hasse-Minkowski Theorem has a nontrivial integral solution. In the previous examples, we have seen polynomials in which the product of the coefficients is relatively small. If the coefficients of the polynomial are large, then using the above method to determine if has a -focused solution can be a daunting task. We see that as the number of prime factors of grows, so too does the number of congruences we need to find -focused solutions to. Moreover, the coefficients may contain large prime factors, in which case checking values could be an inefficient strategy for determining if has nontrivial integral solutions. In these cases, more efficient approaches and results can be used to determine whether there are nontrivial integral solutions to. For example, one may write a computer program to determine if has a focused solution for all odd or use Legendre s Theorem to determine if there are nontrivial integral solutions to. 43

Chapter 8: Topics for Further Study As mentioned in the Chapter 1, Hasse and Minkowski generalized Legendre s Theorem to include forms of an arbitrary number of variables. The proof of the genral theorem is usually divided into 4 cases: and, where is the number of variables of the quadratic form. Therefore, to prove the general theorem, one must prove the remaining two cases of and. The statement of the general theorem is provided below. Theorem (Hasse-Minkowski 3): Let be nonzero, square-free, pairwise relatively prime integers not all of the same sign. Then the equation has a nontrivial integral solution if and only if for each prime and exponent the congruence has a solution not all divisible by. Like in the two and three variable cases, the statement of the theorem is more general than it first appears. Indeed, much like in the three variable case, given any variable form, we can diagonalize it to an equivalent form using an iterative method of completing the square. Moreover, by homogeneity of the equation we can reduce to the case in which the product of the coefficients is square-free. The proof of the remaining two cases of the theorem is usually treated in a more advanced context, 44