Hilbert s Tenth Problem

Transcription

1 Hilbert s Tenth Problem Nicole Bowen BS, Mathematics An Undergraduate Honors Thesis Submitted in Partial Fulfillment of the Requirements for the Degree of Bachelor of Science at the University of Connecticut May 2014 i

2 Copyright by Nicole Bowen May 2014 ii

3 APPROVAL PAGE Bachelor of Science Honors Thesis Hilbert s Tenth Problem Presented by Nicole Bowen, BS Math Honors Major Advisor Honors Thesis Advisor William Abikoff David Reed Solomon University of Connecticut May 2014 iii

4 ACKNOWLEDGMENTS Many thanks to Professor Solomon for helping me through all the details of Hilbert s Tenth Problem, and for undertstanding that things always take longer than epected iv

5 Hilbert s Tenth Problem Nicole Bowen, BS University of Connecticut, May 2014 ABSTRACT In 1900, David Hilbert posed 23 questions to the mathematics community, with focuses in geometry, algebra, number theory, and more In his tenth problem, Hilbert focused on Diophantine equations, asking for a general process to determine whether or not a Diophantine equation with integer coefficients has integer solutions Seventy years later, Yuri Matiyasevich and his colleagues showed that such a process does not eist, with a proof that has had many applications for modern computability theory In this thesis, we give a background on Diophantine equations and computability theory, followed by an in-depth eplanation of the unsolvability of Hilbert s Tenth Problem v

6 Contents I Introduction 1 II Background: Diophantine Equations and Sets 3 Ch 1 Key Definitions and Concepts 4 11 Diophantine Equations 4 12 Simplifying Hilbert s Problem 4 13 Diophantine Sets 6 14 Diophantine Functions, Relations, and Properties 7 15 Unions and Intersections of Diophantine Sets 9 Ch 2 More Eamples More Diophantine Relations More Diophantine Functions Another Diophantine Function:Eponentiation Overview of the Proof Eamining the Sequence α b S is a Diophantine Set S is a Diophantine Set The set {< a, b, c > a = b c } is Diophantine 44 III Background: Computability Theory 56 Ch 3 Key Concepts and Definitions Register Machines Computable and Computably Enumberable Sets 59 vi

7 IV The Proof 61 Ch 4 Comparison of Diophantine and CE Sets Further Eamination of Register Machines Preparation for Determining Diophantine Conditions Determining the Diophantine Conditions 65 Ch 5 Hilbert s Tenth Problem is Unsolvable 83 Bibliography 84 vii

8 Part I Introduction 1

9 2 In countless areas of mathematics, finding solutions to equations is a necessity Often, such solutions are sought over the real numbers However, in some cases, one may attempt to find solutions that are more restricted, perhaps to certain subsets of the reals For eample, say we are dealing with an equation whose and y variables represent the number of horses and sheep on a farm In this eample, a solution of say = 2 and y = 16 is not so meaningful, so we may want to restrict our solutions to the natural numbers Such eamples were the focus for a Greek mathematician named Diophantus (200 s AD), who was concerned with finding only natural or positive rational number solutions to equations, as he considered other values to be nonsensical Thus, equations whose solutions are restricted to the natural numbers, integers, or rationals are often called Diophantine equations Even today, many Diophantine equations remain difficult to solve In particular, the methods for finding solutions vary depending on the number of variables and the degree of an equation As either of these increase, it becomes more and more difficult to solve the equation, to the point where we still do not have complete methods to do so One particular difficulty when attempting to solve Diophantine equations is that we may not be sure that a solution even eists, and thus any attempt to find one may be in vain In 1900, David Hilbert asked for a method to help solve this dilemma in what came to be known as Hilbert s tenth problem In particular, the problem was given as follows: 10 DETERMINATION OF THE SOLVABILITY OF A DIOPHANTINE EQUATION Given a diophantine equation with any number of unknown quantities and with rational integral numerical coefficients: To devise a process according to which it can be determined by a finite number of operations whether the equation is solvable in rational integers In 1970, a Russian mathematician named Yuri Matiyasevich found that such a process does not eist, with the help of his colleagues Martin Davis, Julia Robinson, and Hilary Putnam His proof, which has had many applications in modern computability theory, is described in detail in this thesis

10 Part II Background: Diophantine Equations and Sets 3

11 Chapter 1 Key Definitions and Concepts 11 Diophantine Equations In order to understand Hilbert s tenth problem, we must first know what a Diophantine equation is In general, a Diophantine equation is classified not only by its form, but also by the range of its unknowns, which are often restricted to the rationals, integers, or natural numbers For our purposes, we will define a Diophantine equation based on the specifications that Hilbert used in the statement of his problem When Hilbert states rational integers, he was refering to the integers, so we can define a Diophantine equation as follows: Definition 111 A Diophantine equation is an equation of the form D( 1,, m ) = 0 where D is a polynomial with integer coefficients, and where solutions for 1,, m are restricted to the integers 12 Simplifying Hilbert s Problem As previously stated, when Hilbert posed his tenth problem, he wanted a method for determining whether or not integer solutions eist for any Diophantine equation 4

12 5 In this section, we will see that it is equivalent to work in the natural numbers, which for our purposes will include 0 It is important to note that for a particular Diophantine equation, the problem of deciding whether or not it has integer solutions is a different problem from deciding whether or not it has natural number solutions However, in terms of deciding whether or not a general process eists for checking the solvability of an equation, it is equivalent to work with the natural numbers In other words, we will see in this section that there is a general process for determining whether or not Diophantine equations have natural number solutions if and only if there is a general process for determining whether or not Diophantine equations have integer solutions First, we will show that if there is no general process for checking the eistance of natural number solutions, there there is no general process for checking the eistance of integer solutions Note that we can take a system of Diophantine equations and compress it into a single Diophantine equation In particular, the system D 1 ( 1,, m ) = 0 D k ( 1,, m ) = 0 has an integer solution 1,, m if and only if the Diophantine equation D 2 1( 1,, m ) + + D 2 k( 1,, m ) = 0 has an integer solution 1,, m Now, let D( 1,, m ) = 0 be any arbitrary Diophantine equation, and let E( 1,, m, y 1,1,, y m,4 ) = 0 be the Diophantine equation formed by compressing the system D( 1,, m ) = 0 1 = y 2 1,1 + y 2 1,2 + y 2 1,3 + y 2 1,4 m = y 2 m,1 + y 2 m,2 + y 2 m,3 + y 2 m,4 If D( 1,, m ) = 0 has a solution in the natural numbers, then E( 1,, m, y 1,1,, y m,4 ) = 0 must have a solution in the integers, since any natural number can be written as the sum of four squares Likewise, if E( 1,, m, y 1,1,, y m,4 ) = 0 has a solution in the integers, then D( 1,, m ) = 0 must have a solution in the natural numbers Thus, any arbitrary Diophantine equation D( 1,, m ) = 0 has natural number solutions if and only if E( 1,, m, y 1,1,, y m,4 ) = 0 has an integer solu-

13 6 tion So, if we were to find that there is no method of checking whether or not D( 1,, m ) = 0 has natural number solutions, then there is no way of checking whether or not E( 1,, m, y 1,1,, y m,4 ) = 0 has integer solutions This means that we cannot check the eistance of integer solutions for any arbitrary Diophantine equation,as E is one such equation Net, we will show that if there is a general process for checking natural number solutions, then there is a general process for checking integer solutions Suppose that an arbitrary Diophantine equatio D( 1,, m ) = 0 has a solution in the integers Note that any integer k can be written as the difference of two natural numbers a k and b k Thus, the Diophantine equation D(a 1 b 1,, a m b m ) = 0 must have a solution for a 1,, a m, b 1,, b m in the natural numbers Likewise, if D(a 1 b 1,, a m b m ) = 0 has a solution for a 1,, a m, b 1,, b m in the natural numbers, then 1 = a 1 b 1,, m = a m b m is a solution to D( 1,, m ) = 0 in the integers Thus, any arbitrary Diophantine equation D( 1,, m ) = 0 is solvable in the integers if and only if the Diophantine equation D(a 1 b 1,, a m b m ) = 0 is solvable in the natural numbers So, if we were to find that there is a method of checking whether or not any Diophantine equation if solvable in the natural numbers, then we would know that there is a method of finding whether or not D(a 1 b 1,, a m b m ) = 0 is solvable in the natural numbers And if we can check the eistance of solutions for D(a 1 b 1,, a m b m ) = 0 in the natural numbers, then we can check the eistance of integer solutions for any arbitrary D( 1,, m ) = 0 With this, the question of the solvability of Hilbert s problem in the integers is reducible to the question of its solvability in the natural numbers In general, this will make our work in proving that Hilbert s tenth problem is unsolvable easier, as it allows us to work within the natural numbers only For the remainder of this thesis, all lowercase variables can be assumed to be natural numbers, unless otherwise stated 13 Diophantine Sets With the definition of a Diophantine equation in hand, we can define another important object of study, called a Diophantine set Definition 131 Let S be a set of n-tuples of natural numbers Then S is called a Diophantine set if there eists some Diophantine equation D(a 1,, a n, 1,, m ), with parameters a 1,, a n and unknowns 1,, m, such that a 1,, a n S 1,, m [D(a 1,, a n, 1,, m ) = 0] In order to understand this definition, it is perhaps best to look at some eamples

14 7 Eample 132 Let S = {3}, and consider the equation D(a, ) = ( + 1)(a 3) First, suppose a S Then a = 3, so D(a, ) = ( + 1)(a 3) = 0 for all Instead, suppose a / S Then a 3, so a 3 0 Then, since N, D(a, ) = (+1)(a 3) 0 for all Therefore, we can conclude that S is a Diophantine set, since we have found a Diophantine equation D(a, ) such that a S [D(a, ) = 0] Eample 133 Let S = {3, 5} Then consider the Diophantine equation D(a, ) = ( + 1)(a 3)(a 5) By the same reasoning described in the previous eample, a S [D(a, ) = 0] Therefore, S is a Diophantine set 14 Diophantine Functions, Relations, and Properties In the previous section, we found that the sets S = {3} and S = {3, 5} were Diophantine In some cases, rather than showing that a set of particular n-tuples is Diophantine, we may want to show that a set of n-tuples with certain properties is Diophantine, as in the net eample Eample 141 Let S = { a, b, c a + b = c}, and let D(a, b, c, ) = ( + 1)(a + b c) Then a, b, c S [D(a, b, c, ) = 0], so S is a Diophantine set In this eample, rather than concluding that the set S = { a, b, c a + b = c} is a Diophantine set, we might instead state that addition is a Diophantine function Definition 142 A function of natural numbers is a Diophantine f unction when its set of solutions is a Diophantine set In the net eample, we show that multiplication is a Diophantine function Eample 143 Let S = { a, b, c ab = c}, and let D(a, b, c, ) = ( + 1)(ab c) Then a, b, c S [D(a, b, c, ) = 0], so S is a Diophantine set

15 8 Similarly, we can show that many relations are Diophantine Definition 144 A relation between n natural numbers is a Diophantine relation when the set of all n-tuples for which the relation holds is a Diophantine set In the following eample, we show that the less than (and likewise greater than) relation is Diophantine Eample 145 Let S = { a, b a < b}, and let D(a, b, ) = (b a) ( + 1) Then [D(a, b, ) = (b a) ( + 1) = 0] [b a = + 1] a < b Therefore, S is a Diophantine set Similarly, the relation less than or equal to (and likewise greater than or equal to) is Diophantine Eample 146 Let S = { a, b a b} Then consider the equation D(a, b, ) = (b a) Then [D(a, b, ) = (b a) = 0] [b a = ] a b Therefore, S is a Diophantine set The relation of divisibility is also Diophantine Eample 147 Let S = { a, b a b}, and let D(a, b, ) = a b Then [D(a, b, ) = a b = 0] [a = b] a b Therefore, S is a Diophantine set Further, we can show that certain properties are Diophantine Definition 148 A property of natural numbers is a Diophantine property when the set of numbers for which this property holds is Diophantine In the following eample, we see that the property is an even number is Diophantine Eample 149 Let S be the set of all even numbers, and let D(a, ) = 2 a Then a S [D(a, ) = 0], so S is a Diophantine set

16 9 We will also note that for any Diophantine function, relation, or property X(a 1,, a n ), the condition k+1,, n [X(a 1,, a k, k+1,, n )] is also Diophantine In other words, if the set is a Diophantine set, then the set { a 1,, a n X(a 1,, a n )} { a 1,, a k k+1,, n [X(a 1,, a k, k+1,, n )]} is also a Diophantine set For eample, suppose there is some Diophantine relation R Then there must eist a Diophantine equation D(a 1,, a n, 1,, m ) such that Then it follows that R(a 1,, a n ) 1,, m [D(a 1,, a n, 1,, m ) = 0] 1,, m+k [D(a 1,, a n k, 1,, m+k ) = 0] 1,, k [R(a 1,, a n k, 1,, k )] Thus the set is a Diophantine set as well { a 1,, a n k 1,, k [R(a 1,, a n k, 1,, k )]} 15 Unions and Intersections of Diophantine Sets In this section, we will see that the union and intersection of Diophantine sets is also Diophantine Proposition 151 The union of two Diophantine sets of n-tuples is Diophantine Proof Suppose that S 1 and S 2 are two Diophantine sets of n-tuples Then there must be Diophantine equations D 1 and D 2 such that a 1,, a n S 1 1,, m [D 1 (a 1,, a n, 1,, m ) = 0] and a 1,, a n S 2 y 1,, y l [D 2 (a 1,, a n, y 1,, y l ) = 0]

17 10 Then consider the equation D 3 (a 1,, a n, 1,, m, y 1,, y l ) = D 1 (a 1,, a n, 1,, m ) D 2 (a 1,, a n, y 1,, y l ) Then 1,, m, y 1,, y l [D 3 (a 1,, a n, 1,, m, y 1,, y l ) = 0] 1,, m [D 1 (a 1,, a n, 1,, m ) = 0] or y 1,, y m [D 2 (a 1,, a n, y 1,, y l ) = 0] a 1,, a n S 1 or a 1,, a n S 2 a 1,, a n S 1 S 2 Therefore, S 1 S 2 is a Diophantine set Further note that this proposition can be interpreted in terms of Diophantine functions, relations, and properties using the conjunction or Proposition 152 The intersection of two Diophantine sets of n-tuples is Diophantine Proof Suppose that S 1 and S 2 are two Diophantine sets of n-tuples Then there must be Diophantine equations D 1 and D 2 such that a 1,, a n S 1 1,, m [D 1 (a 1,, a n, 1,, m ) = 0] and a 1,, a n S 2 y 1,, y l [D 2 (a 1,, a n, y 1,, y l ) = 0] Then consider the equation D 3 (a 1,, a n, 1,, m, y 1,, y l ) = D 2 1(a 1,, a n, 1,, m ) + D 2 2(a 1,, a n, y 1,, y l )

18 11 Then 1,, m, y 1,, y l [D 3 (a 1,, a n, 1,, m, y 1,, y l ) = 0] 1,, m [D 1 (a 1,, a n, 1,, m ) = 0] and y 1,, y m [D 2 (a 1,, a n, y 1,, y l ) = 0] a 1,, a n S 1 and a 1,, a n S 2 a 1,, a n S 1 S 2 Therefore, S 1 S 2 is a Diophantine set Further note that this proposition can be interpreted in terms of Diophantine functions, relations, and properties using the conjunction and Up until now, in order to show that a set is Diophantine, we have had to find a particular Diophantine equation that fits Defintion 131 However, with the results of this section, we can show that a set is Diophantine by representing it as the union or intersection of other Diophantine sets, as shown in the following eample Eample 153 Consider the function rem(b, c), the remainder on dividing b by c Then notice that a = rem(b, c) a < c and c (b a) We have seen that the sets S 1 = { a, b, c a < c} and S 2 = { a, b, c c (b a)} are Diophantine sets Then since S = { a, b, c a = rem(b, c)} = S 1 S 2, S is Diophantine We will also mention that while the intersection and union of Diophantine sets remains Diophantine, the complement of a Diophantine set may not be Diophantine In fact, we will see that the unsolvability of Hilbert s Tenth Problem results in part from the fact that there are Diophantine sets whose complements are not Diophantine For eamples of such sets, see [1, pgs 57-66]

19 Chapter 2 More Eamples We have now eamined a number of Diophantine sets Throughout our proof of Hilbert s tenth problem, we will use the fact that these and many other sets are Diophantine Thus, we devote this chapter to a study of more eamples that will be useful later 21 More Diophantine Relations The relation equal to is Diophantine Eample 211 Let S = { a, b a = b}, and let D(a, b, ) = ( + 1)(a b) Then [D(a, b, ) = ( + 1)(a b) = 0] a = b a, b S, so S is a Diophantine set Likewise, the relation not equal to is Diophantine Eample 212 Let S = { a, b a b}, and let D(a, b, ) = ( + 1) + (a b) 2 Then [D(a, b, ) = (+1)+(a b) 2 = 0] [(a b) 2 = +1] a, b S, so S is a Diophantine set 12

20 13 Also, the relation does not divide is Diophantine Eample 213 Let S = { a, b a b} Note that a b rem(b, a) > 0 We have already seen that the set S 1 = { a, b, c c = rem(b, a)} is a Diophantine set Further, we see that the set S 2 = { a, b, c c > 0} is Diophantine by considering the equation D(a, b, c, ) = (c ( + 1))(a + 1)(b + 1), as [D(a, b, c, ) = (c ( + 1))(a + 1)(b + 1) = 0] [c ( + 1) = 0] [c = + 1] c > 0 Then S = S 1 S 2, where S 1 and S 2 are Diophantine sets, so S is a Diophantine set The congruence relation is also Diophantine Eample 214 Let S = { a, b, c a b(modc)} Note that a b(modc) rem(a, c) = rem(b, c) Then since we have seen that the rem function and equal to relation are Diophantine, we know that S is Diophantine as well 22 More Diophantine Functions Define the function arem(b, c) to be the least absolute value X among all numbers X congruent to b mod c In other words, arem(b, c) ±b mod c & 0 arem(b, c) c 2 For eample, while rem(8, 5) = 3, arem(8, 5) = 2 The function arem(b, c) is Diophantine Eample 221 Let S = { a, b, c a = arem(b, c)} Note that a = arem(b, c) 2a c & [c (b a) or c (b + a)] Since we have seen that the less than or equal to and divides relations are Diophantine, we know that S is Diophantine as well

21 14 The choose function is also Diophantine Eample 222 Let S = { a, b, c ( a b) = c} For a proof that this set is Diophantine, see [1,pgs 44-45] The function b div c, defined to be the integer part of b, is also Diophantine c Eample 223 Let S = { a, b, c a = b div c} Note that a = b div c ac + rem(b, c) = b Then since we have seen that multiplication, addition, and the rem function are Diophantine, we know that S is Diophantine as well 23 Another Diophantine Function:Eponentiation In contrast to the eamples that we have seen so far, it is actually quite difficult and technical to prove that eponentiation is a Diophantine function, ie that the set {< a, b, c > a = b c } is a Diophantine set At first glance, we might attempt to prove such a fact by using an equation such as D(a, b, c, ) = ( + 1)(b c a) The problem, however, is that this equation is not a polynomial, because of the term b c, and therefore is not Diophantine Thus, we will instead attempt to find a set of Diophantine conditions on a, b, and c that hold if and only if a = b c As it turns out, this is hard to do and requires a number of different steps In fact, the proof that eponentiation is Diophantine is perhaps one of the most difficult steps in proving the unsolvability of Hilbert s Tenth Problem We will now turn our attention to this proof 231 Overview of the Proof In order to prove that eponentiation is Diophantine, we must prove that the set { a, b, c a = b c } is a Diophantine set We note that the powers of any given b can be epressed as a recurrent sequence β b, where β b (0) = 1 and β b (n + 1) = bβ b (n)

22 15 Thus, another way to show that eponentiation is Diophantine would be to show that the set {< a, b, c > a = β b (c)} is a Diophantine set However, it turns out that it is actually easier to work with another recurrent sequence α b, where b 2 and α b (0) = 0 α b (1) = 1 and α b (n + 2) = bα b (n + 1) α b (n) First, we will eamine some important properties of α b Then, we will prove that the sets and S = {< a, b > b 2 & n[a = α b (n)]} S = {< a, b, c > b 4 & a = α b (c)} are Diophantine sets Lastly, we will attempt to epress β b in terms of α b in order to show that eponentiation is in fact Diophantine 232 Eamining the Sequence α b First, we will give a formal definition of α b Definition 231 Let the second-order recurrent sequence α b be defined for b 2 as α b (0) = 0 α b (1) = 1 α b (n + 2) = bα b (n + 1) α b (n) While this is the definition of α b we will use most often, in some cases it will be easier to use the equivalent form given in Proposition 232 Proposition 232 It is equivalent to define α b with α b (n 2) = bα b (n 1) α b (n) Proof From our initial definition of α b we have that α b (n) = bα b (n 1) α b (n 2) Rearranging the terms in this equality yields α b (n 2) = bα b (n 1) α b (n) With these definitions of α b in hand, we can start to investigate some of its properties One especially important property of α b is its increasing nature Proposition 233 The sequence α b is strictly increasing, ie 0 = α b (0) < α b (1) < < α b (n) < α b (n + 1) <

23 16 Proof For the base case, let n = 1 Since α b (1) = 1 and α b (0)=0, we have that α b (0) < α b (1) Then for the induction case, assume for all n k that We need to show that α b (k) < α b (k + 1) By our definition of α b, we know that α b (n 1) < α b (n) α b (k + 1) = bα b (k) α b (k 1), and by our induction hypothesis, we know that Therefore, we get that α b (k 1) < α b (k) α b (k + 1) = bα b (k) α b (k 1) > bα b (k) α b (k) = (b 1)α b (k) α b (k) Thus, by induction, for all n 1, α b (n 1) < α b (n) The increasing nature of α b can be seen in the following eample

24 17 Eample 234 For b=2, the first ten terms of α b are α 2 (0) = 0 α 2 (1) = 1 α 2 (2) = 2α 2 (1) α 2 (0) = 2(1) 0 = 2 α 2 (3) = 2α 2 (2) α 2 (1) = 2(2) 1 = 3 α 2 (4) = 2α 2 (3) α 2 (2) = 2(3) 2 = 4 α 2 (5) = 2α 2 (4) α 2 (3) = 2(4) 3 = 5 α 2 (6) = 6 α 2 (7) = 7 α 2 (8) = 8 α 2 (9) = 9 This eample also demonstrates a convenient pattern of α b that occurs when b = 2 Proposition 235 When b = 2, α b (n) = n for all n Proof For the base case, let n = 0 Then α 2 (0) = 0 Let n = 1 Then α 2 (1) = 1 For the induction case, assume that α 2 (n) = n for all n k We need to show that α 2 (k + 1) = k + 1 By our induction hypothesis, α 2 (k + 1) = 2α 2 (k) α 2 (k 1) = 2k (k 1) = k + 1 Thus, by induction, α 2 (n) = n for all n Unfortunately, Proposition 235 does not necessarily hold for larger values of b, as shown by the following eample Eample 236 For b = 3, the first ten terms of α 3 are α 3 (0) = 0 α 3 (1) = 1 α 3 (2) = 3α 3 (1) α 3 (0) = 3(1) 0 = 3 α 3 (3) = 3α 3 (2) α 3 (1) = 3(3) 1 = 8 α 3 (4) = 3α 3 (3) α 3 (2) = 3(8) 3 = 21 α 3 (5) = 3α 3 (4) α 3 (3) = 3(21) 8 = 55 α 3 (6) = 144 α 3 (7) = 377 α 3 (8) = 987 α 3 (9) = 2584

25 18 However, there is a weaker form of Proposition 235 that holds for general b Proposition 237 For all n, α b (n) n Proof We have seen that this proposition is true for b = 2, so assume that b 3 Our proof will proceed by induction For base cases, we have α b (0) = 0 and α b (1) = 1 For the induction case, assume that for all n k, k 0, that n α b (n) We need to show that k + 1 α b (k + 1) We find that α b (k + 1) = bα b (k) α b (k 1) by definition bα b (k) α b (k) since α b is increasing by Proposition233 = (b 1)α b (k) (b 1)k by our induction hypothesis 2k since we assumed b 3 k + 1 since k 0 Therefore, by induction, α b (n) n for all n We can also compare the values of α b for different values of b Proposition 238 If b 1 b 2 mod q, then α b1 (n) α b2 (n) mod q for all n Proof For the base case, let b 1 b 2 mod q Then α b1 (0) = 0 and α b2 (0) = 0, so α b1 (0) α b2 (0) mod q Also, α b1 (1) = 1 and α b2 (1) = 1, so α b1 (1) α b2 (1) mod q Further, α b1 (2) = b 1 α b1 (1) α b1 (0) = b 1 0 = b 1 and α b2 (2) = b 2 α b2 (1) α b2 (0) = b 2, so by our assumption α b1 (2) α b2 (2) mod q For the induction case, assume that α b1 (n) α b2 (n) mod q for all n k We need to show that α b1 (k + 1) α b2 (k + 1) mod q By definition, So α b1 (k + 1) = b 1 α b1 (k) α b1 (k 1) α b2 (k + 1) = b 2 α b2 (k) α b1 (k 1) α b1 (k + 1) α b2 (k + 1) = b 1 α b1 (k) α b1 (k 1) (b 2 α b2 (k) α b2 (k 1)) = b 1 α b1 (k) b 2 α b2 (k) (α b1 (k 1) α b2 (k 1)) Also, since b 1 b 2 mod q, there eist a, i, j Z such that b 1 = iq + a and b 2 = jq + a,

26 19 so we get that α b1 (k + 1) α b2 (k + 1) = (iq + a)α b1 (k) (jq + a)α b2 (k) (α b1 (k 1) α b2 (k 1)) = iqα b1 (k) jqα b2 (k) + aα b1 (k) aα b2 (k) (α b1 (k 1) α b2 (k 1) = q(iα b1 (k) jα b2 (k)) + a(α b1 (k) α b2 (k)) (α b1 (k 1) α b2 (k 1)) By our induction hypothesis, we know that α b1 (k) α b2 (k) mod q, so q (α b1 (k) α b2 (k)), and α b1 (k 1) α b2 (k 1) mod q, so q (α b1 (k 1) α b2 (k 1)) Thus, there eist r, s Z such that α b1 (k + 1) α b2 (k + 1) = q(iα b1 (k) jα b2 (k)) + a(qr) (qs) = q(iα b1 (k) jα b2 (k) + ar s) Thus, q (α b1 (k + 1) α b2 (k + 1)), so α b1 (k + 1) α b2 (k + 1) mod q So, for all n, if b 1 b 2 mod q, then α b1 (n) α b2 (n) mod q Corollary 239 For b > 2, α b (n) n mod b 2 for all n Proof Since (b 2) (b 2), it follows that b 2 mod b 2 Then, by Proposition 238, α b (n) α 2 (n) mod b 2, and further α b (n) n mod b 2 by Proposition 235 The nature of α b as described by Proposition 238 and Corollary 239 will be of great importance in the following sections We will also make use of the periodic nature of α b, described by the following proposition Proposition 2310 For any positive integer v, α b is periodic modulo v In Section 313, the following proposition will be useful as well Proposition 2311 For all b 2 and all n, (α b (n)) 2 α b (n + 1)α b (n 1) = 1 Proof First, we use the elements of α b to define a matri A b as follows: ( ) αb (n + 1) α A b (n) = b (n) α b (n) α b (n 1) with the convention that α b ( 1) = 0 We start our proof by showing that ( ) n b 1 A b (n) = 1 0

27 20 Notice that A b (n) = ( ) αb (n + 1) α b (n) α b (n) α b (n 1) = = ( ) bαb (n) α b (n 1) α b (n) bα b (n 1) α b (n 2) α b (n 1) ( ) ( ) αb (n) α b (n 1) b 1 α b (n 1) α b (n 2) 1 0 = A b (n 1) ( ) b For base cases, note that when n = 0, and when n = 1, A b (0) = A b (1) = A b (0) ( ) 1 0 = 0 1 ( ) b 1 = 1 0 ( ) 0 b For the induction case, assume that for all n k ( ) n b 1 A b (n) = 1 0 We must show that By our induction hypothesis, A b (k + 1) = A b (k) Thus, by induction, for all n, A b (k + 1) = ( ) b 1 = 1 0 A b (n) = ( ) k+1 b ( b ( ) 1 b ) k ( b ( ) n b ) = ( ) k+1 b 1 1 0

28 21 Now, notice that det ( ) b 1 = 1, 1 0 and since determinants are multiplicative, (( ) n ) b 1 det (A b (n)) = det = 1 n = Also, using our original definition of A b (n), we get that det (A b (n)) = (α b (n + 1)) ( α b (n 1)) (α b (n)) ( α b (n)) Thus we can conclude that = (α b (n)) 2 α b (n + 1)α b (n 1) (α b (n)) 2 α b (n + 1)α b (n 1) = 1 Lastly, in Section 234 we will use another characteristic of α b, described below Corollary 2312 At least one of any two consecutive terms of α b is odd Proof Suppose that both α b (n) and α b (n + 1) are even Then (α b (n)) 2 must be even, and α b (n + 1)α b (n 1) must be even as well Then (α b (n)) 2 α b (n + 1)α b (n 1) 2 But this is a contradiction to Proposition 2311 Thus, at least one of α b (n) and α b (n + 1) must be odd We have now finished eamining a number of characteristics of α b that we will require in the following sections With these characteristics in mind, we can move on to proving that the sets S and S are Diophantine 233 S is a Diophantine Set In this section, we will prove that the set S = {< a, b > b 2 and n[a = α b (n)]} is a Diophantine set

29 22 Lemma 2313 If < a, b > S, then the Diophantine equation D(a, b, ) = 2 ab + a 2 1 = 0 has a solution for Proof Let < a, b > S Then b 2 and there eists an n such that a = α b (n) Then let = α b (n 1) By Proposition 2311, we know that (α b (n)) 2 α b (n+1)α b (n 1) = 1 Then we have that D(a, b, ) = (α b (n 1)) 2 bα b (n)α b (n 1) + (α b (n)) 2 1 = (α b (n)) 2 bα b (n)α b (n 1) + (α b (n 1)) 2 1 = (α b (n)) 2 (bα b (n) α b (n 1))(α b (n 1)) 1 = (α b (n)) 2 α b (n + 1)α b (n 1) 1 = 1 1 = 0 Therefore, when < a, b > S, the Diophantine equation D(a, b, ) = 2 ab + a 2 1 = 0 has a solution for, namely = α b (n 1) Lemma 2314 For b 2, if 2 by + y 2 = 1 and y <, then there eists an m such that = α b (m + 1) and y = α b (m) Proof Our proof will follow by induction on y For the base case, let y = 0 Since 2 by + y 2 = 1, = 1 For m = 0, α b (0 + 1) = 1 =, and α b (0) = 0 = y, and thus our proposition holds For the induction case, fi y > 0, and assume that for any ŷ < y and ˆ > ŷ, if ˆ 2 bˆŷ + ŷ 2 = 1, then there eists an ˆm such that ˆ = α b ( ˆm + 1) and ŷ = α b ( ˆm) Suppose that there is an > y such that 2 by + y 2 = 1 We need to find an m such that y = α b (m) and = α b (m + 1) Notice that ( ) ( ) 1 y 2 y 2 1 = by + = by by Further, since > y, we know that y > y 2 > y 2 1, and thus y > y2 1 Then we get that ( ) ( ) 1 y 2 y 2 1 = by + = by > by y Now, define 1 = y and y 1 = by Then since > by y, we get that y 1 = by < by (by y) = y, and further that 1 = y > y 1

30 23 Also, notice that 2 1 b 1 y 1 + y 2 1 = y 2 by(by ) + (by ) 2 = y 2 b 2 y 2 + by + b 2 y 2 2by + 2 = y 2 by + 2 = 1 Thus 1 and y 1 are such that y 1 < y, 1 > y 1, and 2 1 b 1 y 1 + y 2 1 = 1 Therefore by our induction hypothesis, there must eist some m 1 such that Let m = m Then 1 = α b (m 1 + 1) and y 1 = α b (m 1 ) and = by y 1 = b 1 y 1 = bα b (m 1 + 1) α b (m 1 ) = α b (m 1 + 2) = α b (m + 1) y = 1 = α b (m 1 + 1) = α b (m) Thus, by induction, for any y 0 and > y, if 2 by + y 2 = 1, then there must eist some m such that = α b (m + 1) and y = α b (m) Theorem 2315 The set S = {< a, b > b 2 and n[a = α b (n)]} is a Diophantine set Proof It follows directly from Lemmas 2313 and 2314 that there eists a Diophantine equation, namely D(a, b, ) = 2 ba + a 2, such that < a, b > S D(a, b, ) = 0 has solutions for Thus, by definition, S must be a Diophantine set 234 S is a Diophantine Set In order to show that S = {< a, b, c > a = α b (c)} is a Diophantine set, we will find a system of Diophantine conditions that represents it This system will be slightly harder to determine than the equation we used to represent S Therefore, we will first work through the development of this system, before giving a formal proof that the system is solvable if and only if < a, b, c > S

31 24 Determining the System of Diophantine Conditions Step One: First, we can consider S to be the union of the terms of the following sequence for all b 4: < α b (0), b, 0 >,, < α b (n), b, n >, (231) Step Two: Net, we will rewrite Sequence 231 in such a way that n does not appear on its own, but rather only as an argument of α b Proposition 2316 The first b 2 terms of 231 correspond to the first b 2 terms of the following sequence, for all b 4: < α b (0), b, rem(α b (0), b 2) >,, < α b (n), b, rem(α b (n), b 2) >, (232) Proof First, note that n < b 2 for the first b 2 terms of 231 By Corollary 239, we know that α b (n) n mod b 2 Therefore, when n < b 2, rem(α b (n), b 2) = n Adjusting our sequence so that n only appears as an argument of α b is particularly useful Our only way to describe that a triple < a, b, c > belongs to the set of elements of sequence 231 is with the conditions b 4 and a = α b (c), and we do not know that the latter condition is diophantine However, we can say that a triple < a, b, c > belongs to the set of elements of sequence 232 if b 4 and n[a = α b (n)] and c = rem(a, b 2), all of which we have seen are Diophantine conditions The problem remains, however, that only the first b 2 members of 231 match 232 Step Three: Here again, we will adjust our initial sequence 231 by using three new variables, namely u, v, and w, and setting two conditions to be as follows: Condition 1 : w b mod v Conditon 2 : w 2 mod u Proposition 2317 If w b mod v and w 2 mod u and v > 2α b (k) and u > 2k, then the first k entries of 231 correspond to the first k members of the following sequence for all b 4: < arem(α b (0), v), b, arem(α b (0), u) >,, < arem(α b (n), v), b, arem(α b (n), u) >, (233)

32 25 Proof First, note that n < k for the first k entries of 231 Let w b mod v Then, from Proposition 238, we know that α w (n) α b (n) mod v Also, let v 2α b (k) Then when n < k, arem(α w (n), v) = α b (n) Further, let w 2 mod u Then, from Proposition 238, we know that α w (n) α 2 (n) mod u, and thus by Proposition 235, α w (n) n mod u Also, let u > 2k Then, when n < k, arem(α w (n), u) = n By requiring that v > 2α b (k) and u > 2k, our new sequence again only matches sequence 231 for a finite number of terms Therefore, rather than setting a particular k, we will require only that w b mod v and w 2 mod u, and then take the union of all terms of any sequence of the form 233 such that u, v, and w satisfy these conditons For each of these individual sequences in our union, there will be some k 0 such that v > 2α b (k) and u > 2k, and therefore each individual sequence will match sequence 231 for only k terms However, since we are taking the union of all sequences of the form sequence 233 with all u, v and w satisfying w b mod v and w 2 mod u, we can always find some v 1 > v and some u 1 > u such that that k 1 > k Therefore, we can be certain that all triples from our original sequence 231 will be included in this union However, our new set will contain etra triples as well, since for any individual sequence in our union, all of its terms will be included, not just the first k terms that match sequence 231 Therefore, our net goal will be to set additional conditions to eliminate these etra triples from our set Step Four: First, we will narrow down the location of our etra triples by using Proposition 2310, which states that for any positive v, the sequence α b (0),, α b (n), is periodic mod v Thus, we might predict that sequence 233 will be periodic as well By choosing a specific v, we can control the length of this period, and thus narrow down the location of all unique triples to a finite initial segment of sequence 233 We will choose this v to be as follows: Condition 3 : v = α b (m + 1) α b (m 1) Let s start by finding the period of the sequence arem(α w (0), v),, arem(α w (n), v), under this new condition on v Proposition 2318 If v = α b (m+1) α b (m 1), then the sequence α b (0),, α b (n),, mod v has a period length of 4m In particular, the terms of the sequence will be as follows:

33 26 n α b (n) mod v m 1 α b (m 1) m α b (m) m + 1 α b (m 1) 2m 1 1 2m 0 2m m 1 α b (m 1) 3m α b (m) 3m + 1 α b (m 1) 4m 1 1 Proof It is obvious that α b (0) α b (0) mod v α b (1) α b (1) mod v α b (m 1) α b (m 1) mod v α b (m) α b (m) mod v Since v = α b (m + 1) α b (m 1), we have that α b (m + 1) α b (m 1) mod v From this, along with our original definition of α b, and the equivalent definition of α b given in Proposition 232,we get that α b (m + 2) = bα b (m + 1) α b (m) (bα b (m 1) α b (m)) mod v α b (m 2) mod v

34 27 Continuing this process yields α b (m + 3) = bα b (m + 2) α b (m + 1) (bα b (m 2) α b (m 1)) mod v α b (m 3) mod v α b (2m 1) = α b (m + (m 1)) = bα b (m + (m 2)) α b (m + (m 3)) (bα b (m (m 2)) α b (m (m 3))) mod v (bα b (2) α b (3)) mod v α b (1) mod v 1 mod v α b (2m) = α b (m + m) = bα b (m + (m 1)) α b (m + (m 2)) (bα b (m (m 1)) α b (m (m 2))) mod v (bα b (1) α b (2)) mod v α b (0) mod v 0 mod v α b (2m + 1) = α b (m + (m + 1)) = bα b (m + m) α b (m + (m 1)) (bα b (m m) α b (m (m 1))) mod v (bα b (0) α b (1)) mod v α b (1) mod v 1 mod v To continue past α b (2m + 1), we will show by induction that for all n 2, α b (2m + n) α b (n) mod v For a base case, let n = 2 We have seen that α b (2m) 0 α b (2m) mod v

35 28 and that α b (2m + 1) 1 α b (2m 1) mod v Thus, α b (2m + 2) = bα b (2m + 1) α b (2m) by the definition of α b bα b (2m 1) ( α b (2m)) mod v (bα b (2m 1) α b (2m)) mod v α b (2m 2) mod v by Proposition 232 α b (m + (m 2)) mod v α b (m (m 2)) mod v by the initial part of this proof α b (2) mod v Now for the induction case, assume that for all n k that We must show that We find that α b (2m + n) α b (2m n) α b (n) mod v α b (2m + (k + 1)) α b (2m (k + 1)) α b (k + 1) mod v α b (2m + (k + 1)) = bα b (2m + k) α b (2m + (k 1)) by the definition of α b bα b (2m k) ( α b (2m (k 1))) mod v by our induction hypothesis (bα b (2m k) α b (2m (k 1))) mod v α b (2m (k + 1)) mod v by Proposition 232 α b (m + (m k 1)) mod v α b (m (m k 1)) mod v by the initial part of this proof α b (k + 1) mod v as desired Using the fact that α b (2m + n) α b (n) mod v for any n, we can now calculate

36 29 the rest of the terms of α b mod v to be as follows: α b (2m + 1) 1 mod v α b (3m 1) α b (2m + (m 1)) α b (m 1) mod v α b (3m) α b (2m + (m)) α b (m) mod v α b (3m + 1) α b (2m + (m + 1)) α b (m + 1) α b (m 1) mod v α b (4m 1) α b (2m + (2m 1)) α b (2m 1) 1 mod v α b (4m) α b (2m + 2m) α b (2m) 0 mod v α b (4m + 1) α b (2m + (2m + 1)) α b (2m + 1) ( 1) 1 mod v At α b (4m) we can see that the terms of α b mod v begin to repeat, resulting in a period length of 4m Corollary 2319 If w b mod v and v = α b (m + 1) α b (m 1), then the sequence α w mod v has period 4m, with the same terms as the sequence α b mod v Proof Since w b mod v, by Proposition 238, we know that α b (n) α w (n) mod v Corollary 2320 If w b mod v and v = α b (m + 1) α b (m 1), then the sequence arem(α w (0), v),, arem(α w (n), v), has period 2m In particular, the sequence will be as follows: n arem(α w (n), v) m 1 α b (m 1) m α b (m) m + 1 α b (m 1) 2m 1 1

37 30 Proof First, notice that v = α b (m + 1) α b (m 1) = (bα b (m) α b (m 1)) α b (m 1) = bα b (m) 2α b (m 1) Then since b 4 and α b (m) > α b (m 1) by Proposition 233, we get that v 2α b (m) Then Corollary 2320 follows from Corollary 2319 and the definition of the arem function We have now found the period of the sequence arem(α w (0), v),, arem(α w (n), v),, whose elements are the first terms in the triples of sequence 233 Net, let s find the period of the sequence arem(α w (0), u),, arem(α w (n), u),, whose elements are the third terms in the triples of sequence 233 We will begin with the following proposition Proposition 2321 If w 2 mod u, then the sequence α w (0),, α w (n),, mod u has a period of length u In particular, the terms of the sequence will be as follows: n α w (n) mod u u 1 u 1 Proof Since w 2 mod u, we know that α w (n) α w (2) mod u by Proposition 238 Then by Proposition 235, we can conclude that α w (n) n mod u Corollary 2322 If w 2 mod u, then the sequence arem(α w (0), u),, arem(α w (n), u), has a period length of u In particular, the terms of the sequence will be

38 31 n arem(α w (n), u) u 1 u u 2 u 2 u + 1 u u 1 1 u 0 for even u, and n arem(α w (n), u) u 1 u u 1 2 u 1 2 u 1 u u 1 1 u 0 for odd u Proof For n = 0 to n = u, n u, so arem(α 2 2 w(n), u) = n For n = u + 1 to n = u, 2

39 32 n > u 2, so arem(α w(n)) = u n Thus we see that ( u ) ) ( u ) arem (α w 2 + 1, u = u = u 2 1 arem(α w (u 1), u) = u (u 1) = 1 arem(α w (u), u) = u u = 0 Now that we have eamined the periods of the terms of the triples in sequence 233, we can eamine the period of the entire sequence First, we will add a new condition on u, namely Condition 4 : u m With the addition of this condition, we arrive at the following theorem Proposition 2323 If w b mod v, w 2 mod u, v = α b (m + 1) α b (m 1) and u m, then sequence 233 has a period length of 2m In particular, the terms of sequence 233 will be as follows:

40 33 n < arem(α w (n), v), b, arem(α w (n), u) > 0 < 0, b, 0 > 1 < 1, b, 1 > u 1 < α 2 b( u 1), b, u 1 > 2 2 u < α 2 b ( u), b, u > 2 2 u 2 < α b( u + 1), b, u 1 > 2 2 u 1 < α b (u 1), b, 1 > u < α b (u), b, 0 > u + 1 < α b (u + 1), b, 1 > m 1 < α b (m 1), b, 1 > m < α b (m), b, 0 > m + 1 < α b (m 1), b, 1 > 2m 1 < 1, b, 1 > Proof From Corollary 2320, we know that arem(α w (0), v),, arem(α w (n), v), has a period length of 2m, and from Corollary 2322, we know that arem(α w (0), u),, arem(α w (n), u), has a period length of u Then since u m, we know that u 2m,and thus the period of arem(α w (0), u),, arem(α w (n), u), fits into the period of arem(α w (0), v),, arem(α w (n), v), Putting these terms together in the triples of 233 therefore yields a sequence with a period of 2m The specific terms of 233 follow directly from Corrolaries 2320 and 2322 as well Corollary 2324 If w b mod v, w 2 mod u, v = α b (m + 1) α b (m 1) and u m, then the etra triples of sequence 233 can be found in its first m + 1 entries Proof From the specifc terms of arem(α w (0), v),, arem(α w (n), v), given in Corollary 2320, and the specific terms of arem(α w (0), u),, arem(α w (n), u), given in Corollary 2322, we can see that both sequences are symmetric Thus, the unique terms of sequence arem(α w (0), v),, arem(α w (n), v), can be found in its first m+1 terms, and the unique terms of arem(α w (0), u),, arem(α w (n), u), can be found

41 34 in its first u/2 + 1 terms Since u m, 233 will be symmetric as well, with its unique terms located within the first m + 1 elements Now that we have located the etra triples, we just need to determine condition(s) to eliminate these triples from our set Step Five: To eliminate the etra triples, we will require that Condition 5 : 2arem(α w (n), v) < u Proposition 2325 The conditions that v = α b (m + 1) α b (m 1), u m, and 2arem(α w (n), v) < u eliminate the etra triples from our set Proof Recall that v = α b (m + 1) α b (m 1) = ((bα b (m) α b (m 1)) α b (m 1) = bα b (m) 2α b (m 1) 2α b (m) Thus when n < m+1, arem(α w (n), v) = α b (n) Since all the etra triples are located within the first m + 1 terms of 233, the condition 2arem(α w (n), v) < u can be rewritten as 2α b (n) < u Then by Proposition 233, we know that n < α b (n), and therefore 2n < u This implies by Corollary 2322 that arem(α w (n), u) = n Thus, the conditions that we implemented do in fact ensure that arem(α w (n), u) = n and arem(α w (n), v) = α b (n) By eliminating these etra triples, we have now constructed a set that is equivalent to the set of elements of sequence 231 Step Si: The previous steps have given us a set of conditions that can be used represent the set of elements of sequence 231 The only thing left to do is check that these conditions are Diophantine In Eample 214, we saw that the equivalence relation is Diophantine, and thus we know that Condition 1 and Condition 2 are Diophantine conditions Also, in Eample 221, we saw that the arem function is Diophantine, and thus we know that Condition 5 is Diophantine as well A slight problem occurs, however, with Condtions 3 and 4, as we do not know that the condition u m together with v = α b (m + 1) = α b (m 1) is Diophantine From the previous section, we know

42 35 that the conditions s 2 bsr + r 2 = 1 together with r < s are Diophantine conditions that imply that m[s = α b (m) and r = α b (m 1)] Thus we can epress the condition v = α b (m + 1) α b (m 1) = bα b (m) 2α b (m 1) in a diophantine way with the condtions s 2 bsr + r 2 = 1, r < s, and v = bs 2r However, we can only ensure the eistance of m; we cannot set any conditions on m itself in a Diophantine way Thus we cannot guarantee that the condition u m is Diophantine In order to fi this problem, we will use the following proposition Proposition 2326 If (α b (k)) 2 α b (m), then α b (k) m Proof Suppose that α b (k) 2 α b (m) If α b (m) = 0, then m = 0, so α b (k) mthen assume that α b (m) 0 Then since α b (k) 2 α b (m), α b (k) < α b (m), so k < m Then by the division algorithm, we know that we can find some l and n such that m = kl + n 0 n < k Recall from the proof of Proposition 2311 that we ( can define ) the matri A b whose b 1 elements are defined by α b and for any, A b () = Then we find that 1 0 ( ) m b 1 A b (m) = 1 0 ( b 1 = 1 0 ( ) n b 1 = 1 0 ) n+kl = A b (n) A b (k) l Plugging in for each matri, we find that ( ) kl b ( ) ( ) ( ) l αb (m + 1) α b (m) αb (n + 1) α = b (n) αb (k + 1) α b (k) α b (m) α b (m 1) α b (n) α b (n 1) α b (k) α b (k 1) Then it follows that ( ) ( ) ( ) l αb (m + 1) α b (m) αb (n + 1) α b (n) αb (k + 1) 0 mod (α α b (m) α b (m 1) α b (n) α b (n 1) 0 α b (k 1) b (k)) Then considering this congruence element-wise, we find that α b (m) α b (n) α b (k + 1) l α b (n 1) 0 α b (n) α b (k + 1) l mod (α b (k))

43 36 Therefore, α b (k) (α b (m) α b (n)α b (k + 1) 2 ) Now, suppose that there is some d such that d α b (k) and d α b (k + 1) Recall that by Proposition 2311, α b (k) 2 α b (k + 1)α b (k 1) = 1 Then d 1, so d = 1 Therefore, we know that α b (k) and α b (k+1) are coprime Putting this together with the fact that α b (k) (α b (m) α b (n)α b (k + 1) 2 ), we find that α b (k) α b (n) However, we also know that n < k, and therefore by Proposition 233, α b (n) < α b (k) So, α b (n) = 0, meaning n = 0, and m = kl With this, we find that ( ) 1 0 A b (n) = 0 1 Then it follows that A b (m) = [A b (k)] l = = = [( )] l αb (k + 1) α b (k) α b (k) α b (k 1) [( ) ( )] l αb (k)b α b (k) αb (k 1) 0 α b (k) 0 0 α b (k 1) [ α b (k) ( ) b 1 α 1 0 b (k 1) ( )] l = l ( ) ( l b 1 ( 1) l i α b (k) i α b (k 1) l i i 1 0 i=0 ) i

44 37 Then if we consider this equivalence mod α b (k) 2, all terms in the summation become 0 ecept for those with i = 0 and i = 1 Thus ( ) αb (m + 1) α A b (m) = b (m) α b (m) α b (m 1) ( ) ( ) 1 0 b 1 ( 1) l α b (k 1) l + ( 1) l 1 lα 0 1 b (k)α b (k 1) l 1 mod (α 1 0 b (k) 2 ) Then if we consider this congruence element-wise, we find that and thus that α b (m) ( 1) l 1 lα b (k)α b (k 1) l 1 mod (α b (k) 2 ) α b (k) 2 [ α b (m) [ ( 1) l 1 lα b (k)α b (k 1) l 1]] Then since we know also that α b (k) 2 α b (m), we find that meaning α b (k) 2 lα b (k)α b (k 1) l 1, α b (k) lα b (k 1) l 1 Now, again by Proposition 2311, we know that α b (k) 2 α b (k + 1)α b (k 1) = 1 and thus α b (k) and α b (k 1) are coprime by the same reasoning used to show that α b (k) and α b (k + 1) are coprime Thus, it follows that α b (k) l, and since m = lk, as desired α b (k) α b (m) Therefore, we can use the diophantine condition that u 2 but + t 2 = 1 to guarantee that k[u = α b (k)] and then set the Diophantine condition that u 2 s, where s = α b (m) for some m We are now prepared to give a formal proof that S is a Diophantine set

45 38 Formal Proof Lemma 2327 Given a,b,c, suppose there eist s, r, u, t, v, w such that the following conditions hold: Then a = α b (c) b 4, u 2 but + t 2 = 1 s 2 bsr + r 2 = 1 r < s u 2 s v =bs 2r v w b u w 2 w > 2 2 wy + y 2 = 1 2a < u a =arem(, v) c =arem(, u) Proof First, by Section 233, we know that } b 4 u 2 but + t 2 k[u = α b (k)] = 1 Likewise, by Section 233, we know that b 4 r < s s 2 bsr + r 2 = 1 m[s = α b(m) and r = α b (m 1)] Also, by Proposition 2326, we have that u 2 s u = α b (k) u m s = α b (m) Net, by Definition 231, we get that

46 39 v = bs 2r s = α b (m) v = bα b(m) 2α b (m 1) = α b (m + 1) α b (m 1) r = α b (m 1) Further, by Section 233, we know that } w > 2 2 wy + y 2 = 1 n[ = α w (n)] Net, note that the condition v (w b) implies that w b mod v and the condition u (w 2) implies that w 2 mod u Therefore, by Propositions 235 and 238, we get that = α w (n) v (w b) α b(n) mod v and n mod u u (w 2) Now, note that for any n and m, we can let n = 2lm ± j with j m for some l and jin particular, this is always possible because we allow that l = 0 Given this, we can now show that α b (n) ±α b (j) mod v First, recall that we can define the matri A b (n) using α b as follows: ( ) αb (n + 1) α A b (n) = b (n) α b (n) α b (n 1) Further, recall that in the proof of Proposition 2311, we found that ( ) n b 1 A b (n) = 1 0 Then given that n = 2lm ± j, we find that

47 40 ( ) n b 1 A b (n) = 1 0 ( b 1 = 1 0 ( ) m2l b 1 = 1 0 ) 2lm±j ( ) ±j b = A b (m) 2l A b (j) ±1 Further, because we know that v = α b (m + 1) α b (m 1), we know that v (α b (m + 1) α b (m 1)), meaning α b (m + 1) α b (m 1) mod v Therefore, A b (m) = ( ) αb (m + 1) α b (m) α b (m) α b (m 1) ( ) αb (m 1) α b (m) mod v α b (m) α b (m + 1) ( ) αb (m 1) α b (m) mod v α b (m) α b (m + 1) ( ) d f Now, suppose we attempted to find the inverse of A b (m), ie the matri such g h that ( ) ( ) ( ) αb (m + 1) α b (m) d f 1 0 = α b (m) α b (m 1) g h 0 1 Then we end up with the system of equations α b (m + 1)d α b (m)g = 1 α b (m)d α b (m 1)g = 0 α b (m + 1)f α b (m)h = 0 α b (m)f α b (m 1)h = 1

48 41 which we can solve to find that meaning A b (m) 1 = d = α b (m 1) f = α b (m) g = α b (m) h = α b (m + 1) ( ) αb (m 1) α b (m) α b (m) α b (m + 1) Putting this together with our previous result, we now have that and thus A b (m) A b (m) 1 mod v A b (m) 2 A b (m) 1 A b (m) Putting this together with our result that we find that A b (n) = A b (m) 2l A b (j) ±1, A b (n) ±A b (j) ±1 mod v ( ) 1 0 mod v 0 1 where all four combinations of the + and signs are possible Then eamining this congruence element-wise, we find that α b (n) ±α b (j) mod v, as desired Net, we can eamine the following sequence of inequalities to find that 2α b (j) v: 2α b (j) 2α b (m) since j m (b 2)α b (m) since b 4 = bα b (m) 2α b (m) < bα b (m) 2α b (m 1) since α b is increasing = v

49 42 Therefore, we now have that 2α b (j) < v a = arem(, v) α b (n) mod v a = arem(, v) = arem(α b(n), v) = α b (j) Then, with this result and the fact that α b, we have that } a = α b (j) 2j 2α b (j) = 2a < u 2a < u With this, we have that c = arem(, u) n mod u n = 2lm ± j u m 2j < u c = arem(, u) = arem(n, u) = j And finally, we arrive at our desired result, as } a = α b (j) a = α b (c) c = j Lemma 2328 Let a = α b (c) and b 4 Then there eist s, r, u, t, v, w such that: u 2 but + t 2 = 1 s 2 bsr + r 2 = 1 r < s u 2 s v =bs 2r v w b u w 2 w > 2 2 wy + y 2 = 1 2a < u a =arem(, v) c =arem(, u)