Hilbert’s Tenth Problem

Nicole Bowen

B.S., 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

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Copyright by

Nicole Bowen

May 2014

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APPROVAL PAGE

Bachelor of Science Honors Thesis

Hilbert’s Tenth Problem

Presented by

Nicole Bowen, B.S. Math

Honors Major AdvisorWilliam Abikoff

Honors Thesis AdvisorDavid Reed Solomon

University of Connecticut

May 2014

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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

expected.

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Hilbert’s Tenth Problem

Nicole Bowen, B.S.

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

exist, 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 explanation of the unsolvability of Hilbert’s Tenth

Problem.

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Contents

I Introduction 1

II Background: Diophantine Equations and Sets 3

Ch. 1. Key Definitions and Concepts 4

1.1 Diophantine Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Simplifying Hilbert’s Problem . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Diophantine Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4 Diophantine Functions, Relations, and Properties . . . . . . . . . . . 7

1.5 Unions and Intersections of Diophantine Sets . . . . . . . . . . . . . . 9

Ch. 2. More Examples 12

2.1 More Diophantine Relations . . . . . . . . . . . . . . . . . . . . . . . 12

2.2 More Diophantine Functions . . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Another Diophantine Function:Exponentiation . . . . . . . . . . . . 142.3.1 Overview of the Proof . . . . . . . . . . . . . . . . . . . . . . 142.3.2 Examining the Sequence αb . . . . . . . . . . . . . . . . . . . 152.3.3 S is a Diophantine Set. . . . . . . . . . . . . . . . . . . . . . . 212.3.4 S∗ is a Diophantine Set . . . . . . . . . . . . . . . . . . . . . . 232.3.5 The set {< a, b, c > |a = bc} is Diophantine . . . . . . . . . . . 44

III Background: Computability Theory 56

Ch. 3. Key Concepts and Definitions 57

3.1 Register Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3.2 Computable and Computably Enumberable Sets . . . . . . . . . . . . 59

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IV The Proof 61

Ch. 4. Comparison of Diophantine and C.E. Sets 62

4.1 Further Examination of Register Machines . . . . . . . . . . . . . . . 62

4.2 Preparation for Determining Diophantine Conditions . . . . . . . . . 64

4.3 Determining the Diophantine Conditions . . . . . . . . . . . . . . . . 65

Ch. 5. Hilbert’s Tenth Problem is Unsolvable 83

Bibliography 84

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Part I

Introduction

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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, onemay attempt to find solutions that are more restricted, perhaps to certain subsets ofthe reals. For example, say we are dealing with an equation whose x and y variablesrepresent the number of horses and sheep on a farm. In this example, a solution of sayx =√

2 and y = −16 is not so meaningful, so we may want to restrict our solutionsto the natural numbers.

Such examples were the focus for a Greek mathematician named Diophantus (200’sA.D.), who was concerned with finding only natural or positive rational number solu-tions to equations, as he considered other values to be nonsensical. Thus, equationswhose solutions are restricted to the natural numbers, integers, or rationals are oftencalled 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 thedegree of an equation. As either of these increase, it becomes more and more difficultto solve the equation, to the point where we still do not have complete methods todo so. One particular difficulty when attempting to solve Diophantine equations isthat we may not be sure that a solution even exists, and thus any attempt to findone may be in vain.

In 1900, David Hilbert asked for a method to help solve this dilemma in whatcame to be known as Hilbert’s tenth problem. In particular, the problem was givenas follows:

10. DETERMINATION OF THE SOLVABILITY OF A DIOPHANTINEEQUATION

Given a diophantine equation with any number of unknown quantities and withrational integral numerical coefficients: To devise a process according to which it

can be determined by a finite number of operations whether the equation is solvablein rational integers.

In 1970, a Russian mathematician named Yuri Matiyasevich found that such aprocess does not exist, with the help of his colleagues Martin Davis, Julia Robin-son, and Hilary Putnam. His proof, which has had many applications in moderncomputability theory, is described in detail in this thesis.

Part II

Background: DiophantineEquations and Sets

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Chapter 1

Key Definitions and Concepts

1.1 Diophantine Equations

In order to understand Hilbert’s tenth problem, we must first know what aDiophantine equation is. In general, a Diophantine equation is classified not onlyby its form, but also by the range of its unknowns, which are often restricted to therationals, integers, or natural numbers. For our purposes, we will define a Diophantineequation 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 candefine a Diophantine equation as follows:

Definition 1.1.1. A Diophantine equation is an equation of the form

D(x1, ..., xm) = 0

where D is a polynomial with integer coefficients, and where solutions for x1, ..., xmare restricted to the integers.

1.2 Simplifying Hilbert’s Problem

As previously stated, when Hilbert posed his tenth problem, he wanted a methodfor determining whether or not integer solutions exist for any Diophantine equation.

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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 particularDiophantine equation, the problem of deciding whether or not it has integer solutionsis a different problem from deciding whether or not it has natural number solutions.However, in terms of deciding whether or not a general process exists for checkingthe solvability of an equation, it is equivalent to work with the natural numbers. Inother words, we will see in this section that there is a general process for determiningwhether or not Diophantine equations have natural number solutions if and only ifthere is a general process for determining whether or not Diophantine equations haveinteger solutions.

First, we will show that if there is no general process for checking the existance ofnatural number solutions, there there is no general process for checking the existanceof integer solutions. Note that we can take a system of Diophantine equations andcompress it into a single Diophantine equation. In particular, the system

D1(x1, ..., xm) = 0

...

Dk(x1, ..., xm) = 0

has an integer solution x1, ..., xm if and only if the Diophantine equation

D21(x1, ..., xm) + ...+D2

k(x1, ..., xm) = 0

has an integer solution x1, ..., xm.Now, letD(x1, ..., xm) = 0 be any arbitrary Diophantine equation, and let E(x1, ..., xm,

y1,1, ..., ym,4) = 0 be the Diophantine equation formed by compressing the system

D(x1, ...,xm) = 0

x1 = y21,1 + y21,2 + y21,3 + y21,4...

xm = y2m,1 + y2m,2 + y2m,3 + y2m,4.

IfD(x1, ..., xm) = 0 has a solution in the natural numbers, then E(x1, ..., xm, y1,1, ..., ym,4)= 0 must have a solution in the integers, since any natural number can be writtenas the sum of four squares. Likewise, if E(x1, ..., xm, y1,1, ..., ym,4) = 0 has a so-lution in the integers, then D(x1, ..., xm) = 0 must have a solution in the naturalnumbers. Thus, any arbitrary Diophantine equation D(x1, ..., xm) = 0 has naturalnumber solutions if and only if E(x1, ..., xm, y1,1, ..., ym,4) = 0 has an integer solu-

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tion. So, if we were to find that there is no method of checking whether or notD(x1, ..., xm) = 0 has natural number solutions, then there is no way of checkingwhether or not E(x1, ..., xm, y1,1, ..., ym,4) = 0 has integer solutions. This means thatwe cannot check the existance of integer solutions for any arbitrary Diophantine equa-tion,as E is one such equation.

Next, we will show that if there is a general process for checking natural numbersolutions, then there is a general process for checking integer solutions. Suppose thatan arbitrary Diophantine equatio D(x1, ..., xm) = 0 has a solution in the integers.Note that any integer xk can be written as the difference of two natural numbersak and bk. Thus, the Diophantine equation D(a1 − b1, ..., am − bm) = 0 must have asolution for a1, ..., am, b1, ..., bm in the natural numbers. Likewise, if D(a1−b1, ..., am−bm) = 0 has a solution for a1, ..., am, b1, ..., bm in the natural numbers, then x1 =a1 − b1, ..., xm = am − bm is a solution to D(x1, ..., xm) = 0 in the integers. Thus,any arbitrary Diophantine equation D(x1, ..., xm) = 0 is solvable in the integers ifand only if the Diophantine equation D(a1 − b1, ..., am − bm) = 0 is solvable in thenatural numbers. So, if we were to find that there is a method of checking whetheror not any Diophantine equation if solvable in the natural numbers, then we wouldknow that there is a method of finding whether or not D(a1 − b1, ..., am − bm) = 0 issolvable in the natural numbers. And if we can check the existance of solutions forD(a1 − b1, ..., am − bm) = 0 in the natural numbers, then we can check the existanceof integer solutions for any arbitrary D(x1, ..., xm) = 0.

With this, the question of the solvability of Hilbert’s problem in the integers isreducible to the question of its solvability in the natural numbers. In general, thiswill make our work in proving that Hilbert’s tenth problem is unsolvable easier, as itallows 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.

1.3 Diophantine Sets

With the definition of a Diophantine equation in hand, we can define anotherimportant object of study, called a Diophantine set.

Definition 1.3.1. Let S be a set of n-tuples of natural numbers. Then S is calleda Diophantine set if there exists some Diophantine equation D(a1, ..., an, x1, ..., xm),with parameters a1, ..., an and unknowns x1, ..., xm, such that

〈a1, ..., an〉 ∈ S ⇐⇒ ∃x1, ..., xm[D(a1, ..., an, x1, ..., xm) = 0].

In order to understand this definition, it is perhaps best to look at some examples.

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Example 1.3.2. Let S = {3}, and consider the equation

D(a, x) = (x+ 1)(a− 3).

First, suppose a ∈ S. Then a = 3, so D(a, x) = (x+ 1)(a− 3) = 0 for all x. Instead,suppose a /∈ S. Then a 6= 3, so a−3 6= 0. Then, since x ∈ N, D(a, x) = (x+1)(a−3) 6=0 for all x. Therefore, we can conclude that S is a Diophantine set, since we havefound a Diophantine equation D(a, x) such that a ∈ S ⇐⇒ ∃x[D(a, x) = 0].

Example 1.3.3. Let S = {3, 5}. Then consider the Diophantine equation

D(a, x) = (x+ 1)(a− 3)(a− 5).

By the same reasoning described in the previous example, a ∈ S ⇐⇒ ∃x[D(a, x) =0]. Therefore, S is a Diophantine set.

1.4 Diophantine Functions, Relations, and Prop-

erties

In the previous section, we found that the sets S = {3} and S = {3, 5} wereDiophantine. In some cases, rather than showing that a set of particular n-tuples isDiophantine, we may want to show that a set of n-tuples with certain properties isDiophantine, as in the next example.

Example 1.4.1. Let S = {〈a, b, c〉|a+ b = c}, and let

D(a, b, c, x) = (x+ 1)(a+ b− c).

Then 〈a, b, c〉 ∈ S ⇐⇒ ∃x[D(a, b, c, x) = 0], so S is a Diophantine set.

In this example, rather than concluding that the set S = {〈a, b, c〉|a+ b = c} is aDiophantine set, we might instead state that addition is a Diophantine function.

Definition 1.4.2. A function of natural numbers is a Diophantine function whenits set of solutions is a Diophantine set.

In the next example, we show that multiplication is a Diophantine function.

Example 1.4.3. Let S = {〈a, b, c〉|ab = c}, and let

D(a, b, c, x) = (x+ 1)(ab− c).

Then 〈a, b, c〉 ∈ S ⇐⇒ ∃x[D(a, b, c, x) = 0], so S is a Diophantine set.

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Similarly, we can show that many relations are Diophantine.

Definition 1.4.4. A relation between n natural numbers is a Diophantine relationwhen the set of all n-tuples for which the relation holds is a Diophantine set.

In the following example, we show that the less than (and likewise greater than)relation is Diophantine.

Example 1.4.5. Let S = {〈a, b〉|a < b}, and let

D(a, b, x) = (b− a)− (x+ 1).

Then ∃x[D(a, b, x) = (b − a) − (x + 1) = 0] ⇐⇒ ∃x[b − a = x + 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.

Example 1.4.6. Let S = {〈a, b〉|a ≤ b}. Then consider the equation

D(a, b, x) = (b− a)− x.

Then ∃x[D(a, b, x) = (b− a)− x = 0] ⇐⇒ ∃x[b− a = x] ⇐⇒ a ≤ b. Therefore, Sis a Diophantine set.

The relation of divisibility is also Diophantine.

Example 1.4.7. Let S = {〈a, b〉 | a|b}, and let

D(a, b, x) = ax− b.

Then ∃x[D(a, b, x) = ax − b = 0] ⇐⇒ ∃x[ax = b] ⇐⇒ a|b. Therefore, S is aDiophantine set.

Further, we can show that certain properties are Diophantine.

Definition 1.4.8. A property of natural numbers is a Diophantine property whenthe set of numbers for which this property holds is Diophantine.

In the following example, we see that the property is an even number is Diophan-tine.

Example 1.4.9. . Let S be the set of all even numbers, and let

D(a, x) = 2x− a.

Then a ∈ S ⇐⇒ ∃x[D(a, x) = 0], so S is a Diophantine set.

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We will also note that for any Diophantine function, relation, or propertyX(a1, ..., an),the condition

∃xk+1, ..., xn[X(a1, ..., ak, xk+1, ..., xn)]

is also Diophantine. In other words, if the set

{〈a1, ..., an〉 | X(a1, ..., an)}

is a Diophantine set, then the set

{〈a1, ..., ak〉 | ∃xk+1, ..., xn[X(a1, ..., ak, xk+1, ..., xn)]}

is also a Diophantine set. For example, suppose there is some Diophantine relationR. Then there must exist a Diophantine equation D(a1, ..., an, x1, ..., xm) such that

R(a1, ..., an) ⇐⇒ ∃x1, ..., xm[D(a1, ..., an, x1, ..., xm) = 0].

Then it follows that

∃x1, ..., xm+k[D(a1, ..., an−k, x1, ..., xm+k) = 0] ⇐⇒ ∃x1, ..., xk[R(a1, ..., an−k, x1, ..., xk)].

Thus the set

{〈a1, ..., an−k〉 | ∃x1, ..., xk[R(a1, ..., an−k, x1, ..., xk)]}

is a Diophantine set as well.

1.5 Unions and Intersections of Diophantine Sets

In this section, we will see that the union and intersection of Diophantine sets isalso Diophantine.

Proposition 1.5.1. The union of two Diophantine sets of n-tuples is Diophantine.

Proof. Suppose that S1 and S2 are two Diophantine sets of n-tuples. Then there mustbe Diophantine equations D1 and D2 such that

〈a1, ..., an〉 ∈ S1 ⇐⇒ ∃x1, ..., xm[D1(a1, ..., an, x1, ..., xm) = 0]

and〈a1, ..., an〉 ∈ S2 ⇐⇒ ∃y1, ..., yl[D2(a1, ..., an, y1, ..., yl) = 0].

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Then consider the equation

D3(a1, ..., an, x1, ..., xm, y1, ..., yl) = D1(a1, ..., an, x1, ..., xm) ·D2(a1, ..., an, y1, ..., yl).

Then

∃x1, ..., xm, y1, ..., yl[D3(a1, ..., an, x1, ..., xm, y1, ..., yl) = 0]

⇐⇒ ∃x1, ...,xm[D1(a1, ..., an, x1, ..., xm) = 0] or ∃y1, ..., ym[D2(a1, ..., an, y1, ..., yl) = 0]

⇐⇒ 〈a1, ..., an〉 ∈ S1 or 〈a1, ..., an〉 ∈ S2

⇐⇒ 〈a1, ..., an〉 ∈ S1 ∪ S2.

Therefore, S1 ∪ S2 is a Diophantine set.

Further note that this proposition can be interpreted in terms of Diophantinefunctions, relations, and properties using the conjunction “or”.

Proposition 1.5.2. The intersection of two Diophantine sets of n-tuples is Diophan-tine.

Proof. Suppose that S1 and S2 are two Diophantine sets of n-tuples. Then there mustbe Diophantine equations D1 and D2 such that

〈a1, ..., an〉 ∈ S1 ⇐⇒ ∃x1, ..., xm[D1(a1, ..., an, x1, ..., xm) = 0]

and〈a1, ..., an〉 ∈ S2 ⇐⇒ ∃y1, ..., yl[D2(a1, ..., an, y1, ..., yl) = 0].

Then consider the equation

D3(a1, ..., an, x1, ..., xm, y1, ..., yl) = D21(a1, ..., an, x1, ..., xm) +D2

2(a1, ..., an, y1, ..., yl).

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Then

∃x1, ..., xm, y1, ..., yl[D3(a1, ..., an, x1, ..., xm, y1, ..., yl) = 0]

⇐⇒ ∃x1, ...,xm[D1(a1, ..., an, x1, ..., xm) = 0] and ∃y1, ..., ym[D2(a1, ..., an, y1, ..., yl) = 0]

⇐⇒ 〈a1, ..., an〉 ∈ S1 and 〈a1, ..., an〉 ∈ S2

⇐⇒ 〈a1, ..., an〉 ∈ S1 ∩ S2.

Therefore, S1 ∩ S2 is a Diophantine set.

Further note that this proposition can be interpreted in terms of Diophantinefunctions, relations, and properties using the conjunction “and”.

Up until now, in order to show that a set is Diophantine, we have had to find aparticular Diophantine equation that fits Defintion 1.3.1. However, with the resultsof this section, we can show that a set is Diophantine by representing it as the unionor intersection of other Diophantine sets, as shown in the following example.

Example 1.5.3. 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 S1 = {〈a, b, c〉 | a < c} and S2 = {〈a, b, c〉 | c|(b − a)}are Diophantine sets. Then since S = {〈a, b, c〉 | a = rem(b, c)} = S1 ∩ S2, S isDiophantine.

We will also mention that while the intersection and union of Diophantine setsremains 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 partfrom the fact that there are Diophantine sets whose complements are not Diophantine.For examples of such sets, see [1, pgs 57-66].

Chapter 2

More Examples

We have now examined a number of Diophantine sets. Throughout our proofof Hilbert’s tenth problem, we will use the fact that these and many other sets areDiophantine. Thus, we devote this chapter to a study of more examples that will beuseful later.

2.1 More Diophantine Relations

The relation equal to is Diophantine.

Example 2.1.1. Let S = {〈a, b〉 | a = b}, and let

D(a, b, x) = (x+ 1)(a− b).

Then ∃x[D(a, b, x) = (x + 1)(a − b) = 0] ⇐⇒ a = b ⇐⇒ 〈a, b〉 ∈ S, so S is aDiophantine set.

Likewise, the relation not equal to is Diophantine.

Example 2.1.2. Let S = {〈a, b〉 | a 6= b}, and let

D(a, b, x) = (x+ 1) + (a− b)2.

Then ∃x[D(a, b, x) = (x+1)+(a−b)2 = 0] ⇐⇒ ∃x[(a−b)2 = x+1] ⇐⇒ 〈a, b〉 ∈ S,so S is a Diophantine set.

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Also, the relation does not divide is Diophantine.

Example 2.1.3. Let S = {〈a, b〉 | a 6 | b}. Note that

a 6 | b ⇐⇒ rem(b, a) > 0.

We have already seen that the set S1 = {〈a, b, c〉 | c = rem(b, a)} is a Diophantineset. Further, we see that the set S2 = {〈a, b, c〉 | c > 0} is Diophantine by consideringthe equation D(a, b, c, x) = (c− (x+ 1))(a+ 1)(b+ 1), as

∃x[D(a, b, c, x) = (c− (x+ 1))(a+ 1)(b+ 1) = 0] ⇐⇒ ∃x[c− (x+ 1) = 0]

⇐⇒ ∃x[c = x+ 1]

⇐⇒ c > 0.

Then S = S1 ∩ S2, where S1 and S2 are Diophantine sets, so S is a Diophantineset.

The congruence relation is also Diophantine.

Example 2.1.4. 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.

2.2 More Diophantine Functions

Define the function arem(b, c) to be the least absolute value |X| among all numbersX congruent to b mod c. In other words,

arem(b, c) ≡ ±b mod c & 0 ≤ arem(b, c) ≤ c

2.

For example, while rem(8, 5) = 3, arem(8, 5) = 2. The function arem(b, c) is Dio-phantine.

Example 2.2.1. 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 Diophan-tine, we know that S is Diophantine as well.

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The choose function is also Diophantine.

Example 2.2.2. Let S = {〈a, b, c〉 |(ab

)= 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 bc, is also Diophantine.

Example 2.2.3. 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 areDiophantine, we know that S is Diophantine as well.

2.3 Another Diophantine Function:Exponentiation

In contrast to the examples that we have seen so far, it is actually quite difficultand technical to prove that exponentiation is a Diophantine function, i.e that theset {< a, b, c > |a = bc} is a Diophantine set. At first glance, we might attempt toprove such a fact by using an equation such as D(a, b, c, x) = (x + 1)(bc − a). Theproblem, however, is that this equation is not a polynomial, because of the term bc,and therefore is not Diophantine. Thus, we will instead attempt to find a set ofDiophantine conditions on a, b, and c that hold if and only if a = bc. As it turns out,this is hard to do and requires a number of different steps. In fact, the proof thatexponentiation is Diophantine is perhaps one of the most difficult steps in provingthe unsolvability of Hilbert’s Tenth Problem. We will now turn our attention to thisproof.

2.3.1 Overview of the Proof

In order to prove that exponentiation is Diophantine, we must prove that the set

{〈a, b, c〉|a = bc}

is a Diophantine set. We note that the powers of any given b can be expressed as arecurrent sequence βb, where

βb(0) = 1 and βb(n+ 1) = bβb(n).

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Thus, another way to show that exponentiation is Diophantine would be to show thatthe set

{< a, b, c > |a = βb(c)}

is a Diophantine set. However, it turns out that it is actually easier to work withanother recurrent sequence αb, where b ≥ 2 and

αb(0) = 0 αb(1) = 1 and αb(n+ 2) = bαb(n+ 1)− αb(n).

First, we will examine some important properties of αb. Then, we will prove thatthe sets

S = {< a, b > |b ≥ 2 & ∃n[a = αb(n)]}

andS∗ = {< a, b, c > |b ≥ 4 & a = αb(c)}

are Diophantine sets. Lastly, we will attempt to express βb in terms of αb in order toshow that exponentiation is in fact Diophantine.

2.3.2 Examining the Sequence αb

First, we will give a formal definition of αb.

Definition 2.3.1. Let the second-order recurrent sequence αb be defined for b ≥ 2as

αb(0) = 0 αb(1) = 1 αb(n+ 2) = bαb(n+ 1)− αb(n).

While this is the definition of αb we will use most often, in some cases it will beeasier to use the equivalent form given in Proposition 2.3.2.

Proposition 2.3.2. It is equivalent to define αb with

αb(n− 2) = bαb(n− 1)− αb(n).

Proof. From our initial definition of αb we have that αb(n) = bαb(n− 1)− αb(n− 2).Rearranging the terms in this equality yields αb(n− 2) = bαb(n− 1)− αb(n).

With these definitions of αb in hand, we can start to investigate some of its prop-erties. One especially important property of αb is its increasing nature.

Proposition 2.3.3. The sequence αb is strictly increasing, i.e.

0 = αb(0) < αb(1) < ... < αb(n) < αb(n+ 1) < ...

16

Proof. For the base case, let n = 1.Since αb(1) = 1 and αb(0)=0, we have that

αb(0) < αb(1).

Then for the induction case, assume for all n ≤ k that

αb(n− 1) < αb(n).

We need to show that αb(k) < αb(k + 1).By our definition of αb, we know that

αb(k + 1) = bαb(k)− αb(k − 1),

and by our induction hypothesis, we know that

αb(k − 1) < αb(k).

Therefore, we get that

αb(k + 1) = bαb(k)− αb(k − 1) > bαb(k)− αb(k) = (b− 1)αb(k) ≥ αb(k).

Thus, by induction, for all n ≥ 1,

αb(n− 1) < αb(n).

The increasing nature of αb can be seen in the following example.

17

Example 2.3.4. For b=2, the first ten terms of αb are

α2(0) = 0

α2(1) = 1

α2(2) = 2α2(1)− α2(0) = 2(1)− 0 = 2

α2(3) = 2α2(2)− α2(1) = 2(2)− 1 = 3

α2(4) = 2α2(3)− α2(2) = 2(3)− 2 = 4

α2(5) = 2α2(4)− α2(3) = 2(4)− 3 = 5

α2(6) = 6

α2(7) = 7

α2(8) = 8

α2(9) = 9

This example also demonstrates a convenient pattern of αb that occurs when b = 2.

Proposition 2.3.5. When b = 2, αb(n) = n for all n.

Proof. For the base case, let n = 0. Then α2(0) = 0. Let n = 1. Then α2(1) = 1.For the induction case, assume that α2(n) = n for all n ≤ k. We need to show thatα2(k + 1) = k + 1. By our induction hypothesis, α2(k + 1) = 2α2(k) − α2(k − 1) =2k − (k − 1) = k + 1. Thus, by induction, α2(n) = n for all n.

Unfortunately, Proposition 2.3.5 does not necessarily hold for larger values of b, asshown by the following example.

Example 2.3.6. For b = 3, the first ten terms of α3 are

α3(0) = 0

α3(1) = 1

α3(2) = 3α3(1)− α3(0) = 3(1)− 0 = 3

α3(3) = 3α3(2)− α3(1) = 3(3)− 1 = 8

α3(4) = 3α3(3)− α3(2) = 3(8)− 3 = 21

α3(5) = 3α3(4)− α3(3) = 3(21)− 8 = 55

α3(6) = 144

α3(7) = 377

α3(8) = 987

α3(9) = 2584

18

However, there is a weaker form of Proposition 2.3.5 that holds for general b.

Proposition 2.3.7. For all n, αb(n) ≥ n.

Proof. We have seen that this proposition is true for b = 2, so assume that b ≥ 3. Ourproof will proceed by induction. For base cases, we have αb(0) = 0 and αb(1) = 1.For the induction case, assume that for all n ≤ k, k 6= 0, that n ≤ αb(n). We need toshow that k + 1 ≤ αb(k + 1). We find that

αb(k + 1) = bαb(k)− αb(k − 1) by definition

≥ bαb(k)− αb(k) since αb is increasing by Proposition2.3.3

= (b− 1)αb(k)

≥ (b− 1)k by our induction hypothesis

≥ 2k since we assumed b ≥ 3

≥ k + 1 since k 6= 0.

Therefore, by induction, αb(n) ≥ n for all n.

We can also compare the values of αb for different values of b.

Proposition 2.3.8. If b1 ≡ b2 mod q, then αb1(n) ≡ αb2(n) mod q for all n.

Proof. For the base case, let b1 ≡ b2 mod q. Then αb1(0) = 0 and αb2(0) = 0, soαb1(0) ≡ αb2(0) mod q. Also, αb1(1) = 1 and αb2(1) = 1, so αb1(1) ≡ αb2(1) mod q.Further, αb1(2) = b1αb1(1)−αb1(0) = b1− 0 = b1 and αb2(2) = b2αb2(1)−αb2(0) = b2,so by our assumption αb1(2) ≡ αb2(2) mod q .

For the induction case, assume that αb1(n) ≡ αb2(n) mod q for all n ≤ k. We needto show that αb1(k + 1) ≡ αb2(k + 1) mod q.By definition,

αb1(k + 1) = b1αb1(k)− αb1(k − 1)

αb2(k + 1) = b2αb2(k)− αb1(k − 1).

So

αb1(k + 1)− αb2(k + 1) = b1αb1(k)− αb1(k − 1)− (b2αb2(k)− αb2(k − 1))

= b1αb1(k)− b2αb2(k)− (αb1(k − 1)− αb2(k − 1)).

Also, since b1 ≡ b2 mod q, there exist a, i, j ∈ Z such that b1 = iq+a and b2 = jq+a,

19

so we get that

αb1(k + 1)− αb2(k + 1) = (iq + a)αb1(k)− (jq + a)αb2(k)− (αb1(k − 1)− αb2(k − 1))

= iqαb1(k)− jqαb2(k) + aαb1(k)− aαb2(k)− (αb1(k − 1)− αb2(k − 1)

= q(iαb1(k)− jαb2(k)) + a(αb1(k)− αb2(k))− (αb1(k − 1)− αb2(k − 1)).

By our induction hypothesis, we know that αb1(k) ≡ αb2(k) mod q, so q|(αb1(k) −αb2(k)), and αb1(k−1) ≡ αb2(k−1) mod q, so q|(αb1(k−1)−αb2(k−1)). Thus, thereexist r, s ∈ Z such that

αb1(k + 1)− αb2(k + 1) = q(iαb1(k)− jαb2(k)) + a(qr)− (qs)

= q(iαb1(k)− jαb2(k) + ar − s).

Thus, q|(αb1(k + 1) − αb2(k + 1)), so αb1(k + 1) ≡ αb2(k + 1) mod q. So, for all n, ifb1 ≡ b2 mod q, then αb1(n) ≡ αb2(n) mod q.

Corollary 2.3.9. For b > 2, αb(n) ≡ n mod b− 2 for all n.

Proof. Since (b − 2)|(b − 2), it follows that b ≡ 2 mod b − 2. Then, by Proposition2.3.8, αb(n) ≡ α2(n) mod b − 2, and further αb(n) ≡ n mod b − 2 by Proposition2.3.5.

The nature of αb as described by Proposition 2.3.8 and Corollary 2.3.9 will be of greatimportance in the following sections. We will also make use of the periodic nature ofαb, described by the following proposition.

Proposition 2.3.10. For any positive integer v, αb is periodic modulo v.

In Section 3.1.3, the following proposition will be useful as well.

Proposition 2.3.11. For all b ≥ 2 and all n,

(αb(n))2 − αb(n+ 1)αb(n− 1) = 1.

Proof. First, we use the elements of αb to define a matrix Ab as follows:

Ab(n) =

(αb(n+ 1) −αb(n)αb(n) −αb(n− 1)

)with the convention that αb(−1) = 0. We start our proof by showing that

Ab(n) =

(b −11 0

)n.

20

Notice that

Ab(n) =

(αb(n+ 1) −αb(n)αb(n) −αb(n− 1)

)

=

(bαb(n)− αb(n− 1) −αb(n)

bαb(n− 1)− αb(n− 2) −αb(n− 1)

)

=

(αb(n) −αb(n− 1)

αb(n− 1) −αb(n− 2)

)(b −11 0

)

= Ab(n− 1)

(b −11 0

).

For base cases, note that when n = 0,

Ab(0) =

(1 00 1

)=

(b −11 0

)0

and when n = 1,

Ab(1) = Ab(0)

(b −11 0

)=

(b −11 0

)1

.

For the induction case, assume that for all n ≤ k

Ab(n) =

(b −11 0

)n.

We must show that

Ab(k + 1) =

(b −11 0

)k+1

.

By our induction hypothesis,

Ab(k + 1) = Ab(k)

(b −11 0

)=

(b −11 0

)k (b −11 0

)=

(b −11 0

)k+1

.

Thus, by induction, for all n,

Ab(n) =

(b −11 0

)n.

21

Now, notice that

det

(b −11 0

)= 1,

and since determinants are multiplicative,

det (Ab(n)) = det

((b −11 0

)n)= 1n = 1.

Also, using our original definition of Ab(n), we get that

det (Ab(n)) = (αb(n+ 1)) (−αb(n− 1))− (αb(n)) (−αb(n))

= (αb(n))2 − αb(n+ 1)αb(n− 1)

Thus we can conclude that

(αb(n))2 − αb(n+ 1)αb(n− 1) = 1.

Lastly, in Section 2.3.4 we will use another characteristic of αb, described below.

Corollary 2.3.12. At least one of any two consecutive terms of αb is odd.

Proof. Suppose that both αb(n) and αb(n+ 1) are even. Then (αb(n))2 must be even,and αb(n+ 1)αb(n−1) must be even as well. Then (αb(n))2−αb(n+ 1)αb(n−1) ≥ 2.But this is a contradiction to Proposition 2.3.11. Thus, at least one of αb(n) andαb(n+ 1) must be odd.

We have now finished examining a number of characteristics of αb that we will requirein the following sections. With these characteristics in mind, we can move on toproving that the sets S and S∗ are Diophantine.

2.3.3 S is a Diophantine Set.

In this section, we will prove that the set

S = {< a, b > |b ≥ 2 and ∃n[a = αb(n)]}

is a Diophantine set.

22

Lemma 2.3.13. If < a, b > ∈ S, then the Diophantine equation D(a, b, x) = x2 −abx+ a2 − 1 = 0 has a solution for x.

Proof. Let < a, b > ∈ S. Then b ≥ 2 and there exists an n such that a = αb(n). Thenlet x = αb(n−1). By Proposition 2.3.11, we know that (αb(n))2−αb(n+1)αb(n−1) =1. Then we have that

D(a, b, x) = (αb(n− 1))2 − bαb(n)αb(n− 1) + (αb(n))2 − 1

= (αb(n))2 − bαb(n)αb(n− 1) + (αb(n− 1))2 − 1

= (αb(n))2 − (bαb(n)− αb(n− 1))(αb(n− 1))− 1

= (αb(n))2 − αb(n+ 1)αb(n− 1)− 1

= 1− 1

= 0.

Therefore, when < a, b > ∈ S, the Diophantine equation D(a, b, x) = x2− abx+ a2−1 = 0 has a solution for x, namely x = αb(n− 1).

Lemma 2.3.14. For b ≥ 2, if x2 − bxy + y2 = 1 and y < x, then there exists an msuch that x = αb(m+ 1) and y = αb(m).

Proof. Our proof will follow by induction on y.For the base case, let y = 0.Since x2 − bxy + y2 = 1, x = 1. For m = 0, αb(0 + 1) = 1 = x, and αb(0) = 0 = y,and thus our proposition holds.For the induction case, fix y > 0, and assume that for any y < y and x > y, ifx2 − bxy + y2 = 1, then there exists an m such that x = αb(m+ 1) and y = αb(m).Suppose that there is an x > y such that x2 − bxy + y2 = 1. We need to find an msuch that y = αb(m) and x = αb(m+ 1). Notice that

x = by +

(1− y2

x

)= by −

(y2 − 1

x

)≤ by.

Further, since x > y, we know that xy > y2 > y2 − 1, and thus y > y2−1x

. Then weget that

x = by +

(1− y2

x

)= by −

(y2 − 1

x

)> by − y.

Now, define x1 = y and y1 = by − x. Then since x > by − y, we get that

y1 = by − x < by − (by − y) = y,

and further thatx1 = y > y1.

23

Also, notice that

x21 − bx1y1 + y21 = y2 − by(by − x) + (by − x)2

= y2 − b2y2 + bxy + b2y2 − 2bxy + x2

= y2 − bxy + x2

= 1.

Thus x1 and y1 are such that y1 < y, x1 > y1, and x21 − bx1y1 + y21 = 1. Therefore byour induction hypothesis, there must exist some m1 such that

x1 = αb(m1 + 1) and y1 = αb(m1).

Let m = m1 + 1. Then

x = by − y1= bx1 − y1= bαb(m1 + 1)− αb(m1)

= αb(m1 + 2)

= αb(m+ 1)

andy = x1 = αb(m1 + 1) = αb(m).

Thus, by induction, for any y ≥ 0 and x > y, if x2 − bxy + y2 = 1, then there mustexist some m such that x = αb(m+ 1) and y = αb(m).

Theorem 2.3.15. The set S = {< a, b > |b ≥ 2 and ∃n[a = αb(n)]} is a Diophantineset.

Proof. It follows directly from Lemmas 2.3.13 and 2.3.14 that there exists a Dio-phantine equation, namely D(a, b, x) = x2 − bax + a2, such that < a, b > ∈ S ⇐⇒D(a, b, x) = 0 has solutions for x. Thus, by definition, S must be a Diophantineset.

2.3.4 S∗ is a Diophantine Set

In order to show that S∗ = {< a, b, c > |a = αb(c)} is a Diophantine set, wewill find a system of Diophantine conditions that represents it. This system will beslightly harder to determine than the equation we used to represent S. Therefore, wewill first work through the development of this system, before giving a formal proofthat the system is solvable if and only if < a, b, c >∈ S∗.

24

Determining the System of Diophantine Conditions

Step One:

First, we can consider S∗ to be the union of the terms of the following sequencefor all b ≥ 4:

< αb(0), b, 0 >, ..., < αb(n), b, n >, .... (2.3.1)

Step Two:

Next, we will rewrite Sequence 2.3.1 in such a way that n does not appear on itsown, but rather only as an argument of αb.

Proposition 2.3.16. The first b−2 terms of 2.3.1 correspond to the first b−2 termsof the following sequence, for all b ≥ 4:

< αb(0), b, rem(αb(0), b− 2) >, ..., < αb(n), b, rem(αb(n), b− 2) >, ... (2.3.2)

Proof. First, note that n < b−2 for the first b−2 terms of 2.3.1. By Corollary 2.3.9, weknow that αb(n) ≡ n mod b−2. Therefore, when n < b−2, rem(αb(n), b−2) = n.

Adjusting our sequence so that n only appears as an argument of αb is particu-larly useful. Our only way to describe that a triple < a, b, c > belongs to the set ofelements of sequence 2.3.1 is with the conditions b ≥ 4 and a = αb(c), and we donot 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 2.3.2 if b ≥ 4 and ∃n[a = αb(n)]and c = rem(a, b − 2), all of which we have seen are Diophantine conditions. Theproblem remains, however, that only the first b− 2 members of 2.3.1 match 2.3.2.

Step Three:

Here again, we will adjust our initial sequence 2.3.1 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 2.3.17. If w ≡ b mod v and w ≡ 2 mod u and v > 2αb(k) and u > 2k,then the first k entries of 2.3.1 correspond to the first k members of the followingsequence for all b ≥ 4:

< arem(αb(0), v), b, arem(αb(0), u) >, ..., < arem(αb(n), v), b, arem(αb(n), u) >, ...(2.3.3)

25

Proof. First, note that n < k for the first k entries of 2.3.1. Let w ≡ b mod v. Then,from Proposition 2.3.8, we know that αw(n) ≡ αb(n) mod v. Also, let v ≥ 2αb(k).Then when n < k, arem(αw(n), v) = αb(n). Further, let w ≡ 2 mod u. Then, fromProposition 2.3.8, we know that αw(n) ≡ α2(n) mod u, and thus by Proposition 2.3.5,αw(n) ≡ n mod u. Also, let u > 2k. Then, when n < k, arem(αw(n), u) = n.

By requiring that v > 2αb(k) and u > 2k, our new sequence again only matchessequence 2.3.1 for a finite number of terms. Therefore, rather than setting a par-ticular k, we will require only that w ≡ b mod v and w ≡ 2 mod u, and then takethe union of all terms of any sequence of the form 2.3.3 such that u, v, and w sat-isfy these conditons. For each of these individual sequences in our union, there willbe some k ≥ 0 such that v > 2αb(k) and u > 2k, and therefore each individualsequence will match sequence 2.3.1 for only k terms. However, since we are takingthe union of all sequences of the form sequence 2.3.3 with all u, v and w satisfyingw ≡ b mod v and w ≡ 2 mod u, we can always find some v1 > v and some u1 > usuch that that k1 > k. Therefore, we can be certain that all triples from our originalsequence 2.3.1 will be included in this union. However, our new set will contain extratriples as well, since for any individual sequence in our union, all of its terms will beincluded, not just the first k terms that match sequence 2.3.1. Therefore, our nextgoal will be to set additional conditions to eliminate these “extra” triples from our set.

Step Four:

First, we will narrow down the location of our “extra” triples by using Proposi-tion 2.3.10, which states that for any positive v, the sequence αb(0), ..., αb(n), ... isperiodic mod v. Thus, we might predict that sequence 2.3.3 will be periodic as well.By choosing a specific v, we can control the length of this period, and thus narrowdown the location of all unique triples to a finite initial segment of sequence 2.3.3.We will choose this v to be as follows:

Condition 3 : v = αb(m+ 1)− αb(m− 1).

Let’s start by finding the period of the sequence arem(αw(0), v), ..., arem(αw(n), v), ...under this new condition on v.

Proposition 2.3.18. If v = αb(m+1)−αb(m−1), then the sequence αb(0), ..., αb(n), ..., modv has a period length of 4m. In particular, the terms of the sequence will be as follows:

26

n αb(n) mod v0 01 1...

...m− 1 αb(m− 1)m αb(m)

m+ 1 αb(m− 1)...

...2m− 1 1

2m 02m+ 1 −1

......

3m− 1 −αb(m− 1)3m −αb(m)

3m+ 1 −αb(m− 1)...

...4m− 1 −1

Proof. It is obvious that

αb(0) ≡ αb(0) mod v

αb(1) ≡ αb(1) mod v

...

αb(m− 1) ≡ αb(m− 1) mod v

αb(m) ≡ αb(m) mod v

Since v = αb(m+ 1)− αb(m− 1), we have that

αb(m+ 1) ≡ αb(m− 1) mod v.

From this, along with our original definition of αb, and the equivalent definition of αbgiven in Proposition 2.3.2,we get that

αb(m+ 2) = bαb(m+ 1)− αb(m)

≡ (bαb(m− 1)− αb(m)) mod v

≡ αb(m− 2) mod v

27

Continuing this process yields

αb(m+ 3) = bαb(m+ 2)− αb(m+ 1)

≡ (bαb(m− 2)− αb(m− 1)) mod v

≡ αb(m− 3) mod v

...

αb(2m− 1) = αb(m+ (m− 1))

= bαb(m+ (m− 2))− αb(m+ (m− 3))

≡ (bαb(m− (m− 2))− αb(m− (m− 3))) mod v

≡ (bαb(2)− αb(3)) mod v

≡ αb(1) mod v

≡ 1 mod v

αb(2m) = αb(m+m)

= bαb(m+ (m− 1))− αb(m+ (m− 2))

≡ (bαb(m− (m− 1))− αb(m− (m− 2))) mod v

≡ (bαb(1)− αb(2)) mod v

≡ αb(0) mod v

≡ 0 mod v

αb(2m+ 1) = αb(m+ (m+ 1))

= bαb(m+m)− αb(m+ (m− 1))

≡ (bαb(m−m)− αb(m− (m− 1))) mod v

≡ (bαb(0)− αb(1)) mod v

≡ −αb(1) mod v

≡ −1 mod v

To continue past αb(2m+ 1), we will show by induction that for all n ≥ 2,

αb(2m+ n) ≡ −αb(n) mod v.

For a base case, let n = 2. We have seen that

αb(2m) ≡ 0 ≡ −αb(2m) mod v

28

and thatαb(2m+ 1) ≡ −1 ≡ −αb(2m− 1) mod v.

Thus,

αb(2m+ 2) = bαb(2m+ 1)− αb(2m) by the definition of αb

≡ −bαb(2m− 1)− (−αb(2m)) mod v

≡ −(bαb(2m− 1)− αb(2m)) mod v

≡ −αb(2m− 2) mod v by Proposition 2.3.2

≡ −αb(m+ (m− 2)) mod v

≡ −αb(m− (m− 2)) mod v by the initial part of this proof

≡ −αb(2) mod v

Now for the induction case, assume that for all n ≤ k that

αb(2m+ n) ≡ −αb(2m− n) ≡ −αb(n) mod v.

We must show that

αb(2m+ (k + 1)) ≡ −αb(2m− (k + 1)) ≡ −αb(k + 1) mod v.

We find that

αb(2m+ (k + 1)) = bαb(2m+ k)− αb(2m+ (k − 1)) by the definition of αb

≡ −bαb(2m− k)− (−αb(2m− (k − 1))) mod v by our induction hypothesis

≡ −(bαb(2m− k)− αb(2m− (k − 1))) mod v

≡ −αb(2m− (k + 1)) mod v by Proposition 2.3.2

≡ −αb(m+ (m− k − 1)) mod v

≡ −αb(m− (m− k − 1)) mod v by the initial part of this proof

≡ −αb(k + 1) mod v

as desired.Using the fact that αb(2m+ n) ≡ −αb(n) mod v for any n, we can now calculate

29

the rest of the terms of αb mod v to be as follows:

αb(2m+ 1) ≡ −1 mod v

...

αb(3m− 1) ≡ αb(2m+ (m− 1)) ≡ −αb(m− 1) mod v

αb(3m) ≡ αb(2m+ (m)) ≡ −αb(m) mod v

αb(3m+ 1) ≡ αb(2m+ (m+ 1)) ≡ −αb(m+ 1) ≡ −αb(m− 1) mod v

...

αb(4m− 1) ≡ αb(2m+ (2m− 1)) ≡ −αb(2m− 1) ≡ −1 mod v

αb(4m) ≡ αb(2m+ 2m) ≡ −αb(2m) ≡ 0 mod v

αb(4m+ 1) ≡ αb(2m+ (2m+ 1)) ≡ −αb(2m+ 1) ≡ −(−1) ≡ 1 mod v

...

At αb(4m) we can see that the terms of αb mod v begin to repeat, resulting in a periodlength of 4m.

Corollary 2.3.19. If w ≡ b mod v and v = αb(m+1)−αb(m−1), then the sequenceαw mod v has period 4m, with the same terms as the sequence αb mod v.

Proof. Since w ≡ b mod v, by Proposition 2.3.8, we know that αb(n) ≡ αw(n) modv.

Corollary 2.3.20. If w ≡ b mod v and v = αb(m+1)−αb(m−1), then the sequencearem(αw(0), v), ..., arem(αw(n), v), ... has period 2m. In particular, the sequence willbe as follows:

n arem(αw(n), v)0 01 1...

...m− 1 αb(m− 1)m αb(m)

m+ 1 αb(m− 1)...

...2m− 1 1

30

Proof. First, notice that

v = αb(m+ 1)− αb(m− 1)

= (bαb(m)− αb(m− 1))− αb(m− 1)

= bαb(m)− 2αb(m− 1).

Then since b ≥ 4 and αb(m) > αb(m− 1) by Proposition 2.3.3, we get that

v ≥ 2αb(m).

Then Corollary 2.3.20 follows from Corollary 2.3.19 and the definition of the aremfunction.

We have now found the period of the sequence arem(αw(0), v), ..., arem(αw(n), v), ...,whose elements are the first terms in the triples of sequence 2.3.3. Next, let’s findthe period of the sequence arem(αw(0), u), ..., arem(αw(n), u), ..., whose elements arethe third terms in the triples of sequence 2.3.3. We will begin with the followingproposition.

Proposition 2.3.21. If w ≡ 2 mod u, then the sequence αw(0), ..., αw(n), ...,mod uhas a period of length u. In particular, the terms of the sequence will be as follows:

n αw(n) mod u0 01 1...

...u− 1 u− 1

Proof. Since w ≡ 2 mod u, we know that αw(n) ≡ αw(2) mod u by Proposition 2.3.8.Then by Proposition 2.3.5, we can conclude that αw(n) ≡ n mod u.

Corollary 2.3.22. If w ≡ 2 mod u, then the sequence arem(αw(0), u), ..., arem(αw(n), u), ...has a period length of u. In particular, the terms of the sequence will be

31

n arem(αw(n), u)0 01 1...

...u2− 1 u

2− 1

u2

u2

u2

+ 1 u2− 1

......

u− 1 1u 0

for even u, and

n arem(αw(n), u)0 01 1...

...u−12− 1 u−1

2− 1

u−12

u−12

u−12

+ 1 u−12− 1

......

u− 1 1u 0

for odd u.

Proof. For n = 0 to n = u2, n ≤ u

2, so arem(αw(n), u) = n. For n = u

2+ 1 to n = u,

32

n > u2, so arem(αw(n)) = u− n. Thus we see that

arem(αw

(u2

+ 1), u)

= u−(u

2+ 1)

=u

2− 1

...

arem(αw(u− 1), u) = u− (u− 1) = 1

arem(αw(u), u) = u− u = 0

Now that we have examined the periods of the terms of the triples in sequence 2.3.3,we can examine the period of the entire sequence. First, we will add a new conditionon u, namely

Condition 4 : u|m.

With the addition of this condition, we arrive at the following theorem.

Proposition 2.3.23. If w ≡ b mod v, w ≡ 2 mod u, v = αb(m + 1) − αb(m − 1)and u|m, then sequence 2.3.3 has a period length of 2m. In particular, the terms ofsequence 2.3.3 will be as follows:

33

n < arem(αw(n), v), b, arem(αw(n), u) >0 < 0, b, 0 >1 < 1, b, 1 >...

...u2− 1 < αb(

u2− 1), b, u

2− 1 >

u2

< αb(u2), b, u

2>

u2

+ 1 < αb(u2

+ 1), b, u2− 1 >

......

u− 1 < αb(u− 1), b, 1 >u < αb(u), b, 0 >

u+ 1 < αb(u+ 1), b, 1 >...

...m− 1 < αb(m− 1), b, 1 >m < αb(m), b, 0 >

m+ 1 < αb(m− 1), b, 1 >...

......

......

...2m− 1 < 1, b, 1 >

Proof. From Corollary 2.3.20, we know that arem(αw(0), v), ..., arem(αw(n), v), ... hasa period length of 2m, and from Corollary 2.3.22, we know that arem(αw(0), u), ...,arem(αw(n), u), ... has a period length of u. Then since u|m, we know that u|2m,andthus the period of arem(αw(0), u), ..., arem(αw(n), u), ... fits into the period ofarem(αw(0), v), ..., arem(αw(n), v), .... Putting these terms together in the triples of2.3.3 therefore yields a sequence with a period of 2m. The specific terms of 2.3.3follow directly from Corrolaries 2.3.20 and 2.3.22 as well.

Corollary 2.3.24. If w ≡ b mod v, w ≡ 2 mod u, v = αb(m + 1) − αb(m − 1) andu|m, then the“extra” triples of sequence 2.3.3 can be found in its first m+ 1 entries.

Proof. From the specifc terms of arem(αw(0), v), ..., arem(αw(n), v), ... given in Corol-lary 2.3.20, and the specific terms of arem(αw(0), u), ..., arem(αw(n), u), ... given inCorollary 2.3.22, we can see that both sequences are symmetric. Thus, the uniqueterms of sequence arem(αw(0), v), ..., arem(αw(n), v), ... can be found in its first m+1terms, and the unique terms of arem(αw(0), u), ..., arem(αw(n), u), ... can be found

34

in its first u/2 + 1 terms. Since u|m, 2.3.3 will be symmetric as well, with its uniqueterms located within the first m+ 1 elements.

Now that we have located the “extra” triples, we just need to determine condi-tion(s) to eliminate these triples from our set.

Step Five:

To eliminate the extra triples, we will require that

Condition 5 : 2arem(αw(n), v) < u.

Proposition 2.3.25. The conditions that v = αb(m + 1) − αb(m − 1), u|m, and2arem(αw(n), v) < u eliminate the “extra” triples from our set.

Proof. Recall that

v = αb(m+ 1)− αb(m− 1)

= ((bαb(m)− αb(m− 1))− αb(m− 1)

= bαb(m)− 2αb(m− 1)

≥ 2αb(m)

Thus when n < m+1, arem(αw(n), v) = αb(n). Since all the extra triples are locatedwithin the first m + 1 terms of 2.3.3, the condition 2arem(αw(n), v) < u can berewritten as 2αb(n) < u. Then by Proposition 2.3.3, we know that n < αb(n), andtherefore 2n < u. This implies by Corollary 2.3.22 that arem(αw(n), u) = n. Thus,the conditions that we implemented do in fact ensure that arem(αw(n), u) = n andarem(αw(n), v) = αb(n).

By eliminating these extra triples, we have now constructed a set that is equiva-lent to the set of elements of sequence 2.3.1.

Step Six:

The previous steps have given us a set of conditions that can be used representthe set of elements of sequence 2.3.1. The only thing left to do is check that theseconditions are Diophantine. In Example 2.1.4, we saw that the equivalence relationis Diophantine, and thus we know that Condition 1 and Condition 2 are Diophantineconditions. Also, in Example 2.2.1, 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 togetherwith v = αb(m+ 1) = αb(m− 1) is Diophantine. From the previous section, we know

35

that the conditions s2 − bsr+ r2 = 1 together with r < s are Diophantine conditionsthat imply that ∃m[s = αb(m) and r = αb(m−1)]. Thus we can express the conditionv = αb(m + 1) − αb(m − 1) = bαb(m) − 2αb(m − 1) in a diophantine way with thecondtions s2 − bsr + r2 = 1, r < s, and v = bs − 2r. However, we can only ensurethe existance 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 fix thisproblem, we will use the following proposition.

Proposition 2.3.26. If (αb(k))2|αb(m), then αb(k)|m.

Proof. Suppose that αb(k)2 | αb(m). If αb(m) = 0, then m = 0, so αb(k)|m.Thenassume that αb(m) 6= 0. Then since αb(k)2 | αb(m), αb(k) < αb(m), so k < m. Thenby 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 2.3.11 that we can define the matrix Ab whose

elements are defined by αb and for any x, Ab(x) =

(b −11 0

)x. Then we find that

Ab(m) =

(b −11 0

)m=

(b −11 0

)n+kl=

(b −11 0

)n·(b −11 0

)kl= Ab(n) · Ab(k)l

Plugging in for each matrix, we find that(αb(m+ 1) −αb(m)αb(m) −αb(m− 1)

)=

(αb(n+ 1) −αb(n)αb(n) −αb(n− 1)

)·(αb(k + 1) −αb(k)αb(k) −αb(k − 1)

)l.

Then it follows that(αb(m+ 1) −αb(m)αb(m) −αb(m− 1)

)≡(αb(n+ 1) −αb(n)αb(n) −αb(n− 1)

)·(αb(k + 1) 0

0 −αb(k − 1)

)lmod (αb(k)).

Then considering this congruence element-wise, we find that

αb(m) ≡ αb(n) · αb(k + 1)l − αb(n− 1) · 0 ≡ αb(n) · αb(k + 1)l mod (αb(k)).

36

Therefore,αb(k) | (αb(m)− αb(n)αb(k + 1)2).

Now, suppose that there is some d such that d|αb(k) and d|αb(k+ 1). Recall thatby Proposition 2.3.11,

αb(k)2 − αb(k + 1)αb(k − 1) = 1.

Then d|1, so d = 1. Therefore, we know that αb(k) and αb(k+1) are coprime. Puttingthis together with the fact that αb(k) | (αb(m)− αb(n)αb(k + 1)2), we find that

αb(k) | αb(n).

However, we also know that n < k, and therefore by Proposition 2.3.3,

αb(n) < αb(k).

So, αb(n) = 0, meaning n = 0, and m = kl. With this, we find that

Ab(n) =

(1 00 1

). Then it follows that

Ab(m) = [Ab(k)]l

=

[(αb(k + 1) −αb(k)αb(k) −αb(k − 1)

)]l

=

[(αb(k)b −αb(k)αb(k) 0

)−(αb(k − 1) 0

0 αb(k − 1)

)]l

=

[αb(k)

(b −11 0

)− αb(k − 1)

(1 00 1

)]l

=l∑

i=0

(−1)l−i(l

i

)αb(k)iαb(k − 1)l−i

(b −11 0

)i

37

Then if we consider this equivalence modαb(k)2 , all terms in the summation become0 except for those with i = 0 and i = 1. Thus

Ab(m) =

(αb(m+ 1) −αb(m)αb(m) −αb(m− 1)

)

≡ (−1)lαb(k − 1)l(

1 00 1

)+ (−1)l−1lαb(k)αb(k − 1)l−1

(b −11 0

)mod (αb(k)2).

Then if we consider this congruence element-wise, we find that

αb(m) ≡ (−1)l−1lαb(k)αb(k − 1)l−1 mod (αb(k)2)

and thus that

αb(k)2 |[αb(m)−

[(−1)l−1lαb(k)αb(k − 1)l−1

]].

Then since we know also that αb(k)2 | αb(m), we find that

αb(k)2 | lαb(k)αb(k − 1)l−1,

meaningαb(k) | lαb(k − 1)l−1.

Now, again by Proposition 2.3.11, we know that

αb(k)2 − αb(k + 1)αb(k − 1) = 1

and thus αb(k) and αb(k − 1) are coprime by the same reasoning used to show thatαb(k) and αb(k + 1) are coprime. Thus, it follows that

αb(k) | l,

and since m = lk,αb(k) | αb(m)

as desired.

Therefore, we can use the diophantine condition that u2 − but+ t2 = 1 to guaranteethat ∃k[u = αb(k)] and then set the Diophantine condition that u2|s, where s = αb(m)for some m.

We are now prepared to give a formal proof that S∗ is a Diophantine set.

38

Formal Proof

Lemma 2.3.27. Given a,b,c, suppose there exist s, r, u, t, v, w such that the followingconditions hold:

b ≥ 4,

u2−but+ t2 = 1

s2−bsr + r2 = 1

r < s

u2|sv =bs− 2r

v|w − bu|w − 2

w > 2

x2−wxy + y2 = 1

2a < u

a =arem(x, v)

c =arem(x, u).

Then a = αb(c).

Proof. First, by Section 2.3.3, we know that

b ≥ 4

u2−but+ t2 = 1

}∃k[u = αb(k)]

Likewise, by Section 2.3.3, we know that

b ≥ 4

r < s

s2−bsr + r2 = 1

∃m[s = αb(m) and r = αb(m− 1)]

Also, by Proposition 2.3.26, we have that

u2|su = αb(k)

s = αb(m)

u|m

Next, by Definition 2.3.1, we get that

39

v = bs− 2r

s = αb(m)

r = αb(m− 1)

v = bαb(m)− 2αb(m− 1) = αb(m+ 1)− αb(m− 1)

Further, by Section 2.3.3, we know that

w > 2

x2−wxy + y2 = 1

}∃n[x = αw(n)]

Next, note that the condition v|(w − b) implies that w ≡ b mod v and the conditionu|(w − 2) implies that w ≡ 2 mod u. Therefore, by Propositions 2.3.5 and 2.3.8, weget that

x = αw(n)

v|(w − b)u|(w − 2)

x ≡ αb(n) mod v and x ≡ n mod u

Now, note that for any n and m, we can let

n = 2lm± j with j ≤ m

for some l and j.In particular, this is always possible because we allow that l = 0.Given this, we can now show that

x ≡ αb(n) ≡ ±αb(j) mod v.

First, recall that we can define the matrix Ab(n) using αb as follows:

Ab(n) =

(αb(n+ 1) −αb(n)αb(n) −αb(n− 1)

)Further, recall that in the proof of Proposition 2.3.11, we found that

Ab(n) =

(b −11 0

)nThen given that n = 2lm± j, we find that

40

Ab(n) =

(b −11 0

)n=

(b −11 0

)2lm±j

=

(b −11 0

)m2l

·(b −11 0

)±j

= Ab(m)2l · Ab(j)±1

Further, because we know that v = αb(m+ 1)− αb(m− 1), we know that v|(αb(m+1)− αb(m− 1)), meaning αb(m+ 1) ≡ αb(m− 1) mod v. Therefore,

Ab(m) =

(αb(m+ 1) −αb(m)αb(m) −αb(m− 1)

)

≡(αb(m− 1) −αb(m)αb(m) −αb(m+ 1)

)mod v

≡ −(−αb(m− 1) αb(m)−αb(m) αb(m+ 1)

)mod v

Now, suppose we attempted to find the inverse of Ab(m), i.e the matrix

(d fg h

)such

that (αb(m+ 1) −αb(m)αb(m) −αb(m− 1)

)·(d fg h

)=

(1 00 1

).

Then we end up with the system of equations

αb(m+ 1)d− αb(m)g = 1

αb(m)d− αb(m− 1)g = 0

αb(m+ 1)f − αb(m)h = 0

αb(m)f − αb(m− 1)h = 1

41

which we can solve to find that

d = −αb(m− 1)

f = αb(m)

g = −αb(m)

h = αb(m+ 1)

meaning

Ab(m)−1 =

(−αb(m− 1) αb(m)−αb(m) αb(m+ 1)

).

Putting this together with our previous result, we now have that

Ab(m) ≡ −Ab(m)−1 mod v

and thus

Ab(m)2 ≡ −Ab(m)−1Ab(m) ≡ −(

1 00 1

)mod v.

Putting this together with our result that

Ab(n) = Ab(m)2l · Ab(j)±1,

we find that

Ab(n) ≡ ±Ab(j)±1 mod v

where all four combinations of the + and − signs are possible. Then examining thiscongruence element-wise, we find that

x ≡ αb(n) ≡ ±αb(j) mod v,

as desired.Next, we can examine the following sequence of inequalities to find that 2αb(j) ≤ v:

2αb(j) ≤ 2αb(m) since j ≤ m

≤ (b− 2)αb(m) since b ≥ 4

= bαb(m)− 2αb(m)

< bαb(m)− 2αb(m− 1) since αb is increasing

= v

42

Therefore, we now have that

2αb(j) < v

a = arem(x, v)

x ≡ αb(n) mod v

a = arem(x, v) = arem(αb(n), v) = αb(j)

Then, with this result and the fact that αb , we have that

a = αb(j)

2a < u

}2j ≤ 2αb(j) = 2a < u

With this, we have that

c = arem(x, u)

x ≡ n mod u

n = 2lm± ju|m2j < u

c = arem(x, u) = arem(n, u) = j

And finally, we arrive at our desired result, as

a = αb(j)

c = j

}a = αb(c)

Lemma 2.3.28. Let a = αb(c) and b ≥ 4. Then there exist s, r, u, t, v, w such that:

u2−but+ t2 = 1

s2−bsr + r2 = 1

r < s

u2|sv =bs− 2r

v|w − bu|w − 2

w > 2

x2−wxy + y2 = 1

2a < u

a =arem(x, v)

c =arem(x, u).

43

Proof. First, we choose u such that u = αb(k) for some k, u is odd, and 2a < u. Wecan guarantee that we can find such a u, because by Proposition 2.3.3, we know thatαb is an increasing sequence, and by Corollary 2.3.12, we know that at least one ofany two consecutive terms of αb is odd.

Next, we let t = αb(k + 1). Then by Section 2.3.3, we know that the condition

u2 − but+ t2 = 1

holds.Further, we letm = uk, and choose r and s such that s = αb(m) and r = αb(m−1).

Then by Proposition 2.3.3 and Section 2.3.3, we know that the conditions

s2 − bsr + r2 = 1 and r < s

both hold.

Then by the proof of Proposition 2.3.26, we find that

s = αb(uk) ≡ (−1)u−1uαb(k)αb(k − 1)u−1 mod (αb(k)2).

Since u is odd, u − 1 is even, meaning (−1)u−1 = 1. Further, since u = αb(k), wehave that

s ≡ u2αb(k − 1)u−1 mod u2,

and thus the condition u2 | s must hold.

Next, since we know that s = αb(m) and r = αb(m− 1), we can find that

bs− 2r = bαb(m)− 2αb(m− 1)

≥ 4αb(m)− 2αb(m− 1)

> 4αb(m)− 2αb(m)

= 2αb(m) > 0

so we can choose v = bs− 2r.Next, we’ll guarantee that there is some w that satisfies the conditions w > 2,

v|w − b, and u|w − 2. In particular, this requires that the system w ≡ b mod v andw ≡ u mod u has a solution for w. By the Chinese Remainder Theorem, we knowthat such a solution exists as long as u and v are coprime. To check that this is in factthe case, suppose that there is some d such that d|u and d|v. Since u2|s, it followsthat d|s. Since v = bs−2r, d|2r, and since u is odd, d is odd, so d|r. Therefore, sinces2 − bsr + r2 = 1, d|1, so d = 1. Therefore, u and v are in fact coprime.

44

Further, we can choose x and y such that x = αw(c) and y = αw(c+ 1). Then bySection 2.3.3, we know that the condition

x2 − wxy + y2 = 1

holds.Next, by the condition v|w − b, we know that w ≡ b mod v, and therefore by

Proposition 2.3.8, we know that αw(c) ≡ αb(c) mod v. Then since a = αb(c) andx = αw(c), we find that x ≡ a mod v. Also, above we saw that v > 2αb(m). Then2a < u and m = uk, we know that a < m < αb(m), meaning v > 2a. Thus, we canconclude that a = arem(x, v).

Lastly, since w > 2, we know by Corollary 2.3.9 that x = αw(c) ≡ c mod w − 2.Then since u|w − 2, x ≡ c mod u. Further, we know that 2c ≤ 2αb(c) = 2a ≤ u.Therefore, we can conclude that c = arem(x, u).

Theorem 2.3.29. S∗ is a Diophantine set.

Proof. Since all conditions stated in Proposition 2.3.27 and 2.3.28 are Diophantineconditions, these two propositions together prove that S∗ is a Diophantine set.

2.3.5 The set {< a, b, c > |a = bc} is Diophantine

We have now completed our analysis of of αb, with the important result thatS∗ = {< a, b, c > |a = αb(c)} is a Diophantine set. In this section, we will compareS∗ to the set {< a, b, c > |a = bc}. In particular, we will attempt to rewrite bc interms of αb, and thus prove that exponentiation is Diophantine. For convenience, wewill define 00 = 1 throughout this section.

We will begin our comparison with the following proposition.

Proposition 2.3.30. (b− 1)n ≤ αb(n+ 1) ≤ bn

Proof. Our proof will proceed by induction.Let n = 0. Then (b − 1)n = 1, αb(n + 1) = 1, and bn = 1, and since 1 ≤ 1 ≤ 1, theproposition holds. Let n = 1. Then (b− 1)n = b− 1, αb(n+ 1) = b, and bn = b, andsince b− 1 ≤ b ≤ b, the proposition holds. For the induction case, assume (b− 1)n ≤αb(n + 1) ≤ bn for all n ≤ k. We must show that (b− 1)k+1 ≤ αb(k + 1 + 1) ≤ bk+1.By our induction hypothesis, we know that

αb(k + 1) ≤ bk

45

and sob · αb(k + 1) ≤ b · bk,

meaningb · αb(k + 1)− αb(k) ≤ b · bk.

Thus,αb(k + 2) ≤ bk+1.

Also, by our induction hypothesis we know that

(b− 1)k ≤ αb(k + 1)

and therefore

(b− 1) · (b− 1)k ≤ (b− 1) · αb(k + 1) = bαb(k + 1)− αb(k + 1).

Then since αb is increasing, we know that

(b− 1) · (b− 1)k ≤ bαb(k + 1)− αb(k),

meaning(b− 1)k+1 ≤ αb(k + 2).

Thus, we have found that

(b− 1)k+1 ≤ αb(k + 2) ≤ bk+1.

This implies, by induction, that for all n ≥ 0,

(b− 1)n ≤ αb(n+ 1) ≤ bn.

Next, we will prove a similar but slightly more complex inequality. Although itmay seem strange at first, it will actually be quite useful in developing a Diophantineway to express bc in terms of αb.

Proposition 2.3.31. For all b ≥ 0 and all c ≥ 0, if x > 16(c+ 1)(b+ 1)c,then

bc ≤ αbx+4(c+ 1)

αx(c+ 1)< bc + 1.

Proof. First, we note that since x > 16(c+ 1)(b+ 1)c, x ≥ 16. Our proof will be splitinto two cases, one for b = 0, and one for b > 0.

46

Case 1: Let b = 0.Say c = 0. Then for any x,

αbx+4(c+ 1)

αx(c+ 1)=αbx+4(1)

αx(1)=

1

1= 1.

Then since we have defined 00 = 1, we obtain that

1 ≤ αbx+4(c+ 1)

αx(c+ 1)< 2,

and thus our proposition holds.Then instead say c > 0. By Proposition 2.3.30, we know that

αbx+4(c+ 1) ≤ (bx+ 4)c = 4c

andαx(c+ 1) ≥ (x− 1)c,

and thereforeαbx+4(c+ 1)

αx(c+ 1)≤ 4c

(x− 1)c.

Then since x > 16(c+ 1)(b+ 1)c, it must be true that x > 5, and thus

4c

(x− 1)c< 1.

Then

0 ≤ αbx+4(c+ 1)

αx(c+ 1)< 1,

so again our proposition holds.

Case 2: Let b > 0.First, we will show that

αbx+4(c+ 1)

αx(c+ 1)< bc + 1.

Will will proceed by proving a series of inequalities.

(a.)

αbx+4(c+ 1)

αx(c+ 1)≤ (bx+ 4)c

(x− 1)c

47

This inequality follows from Proposition 2.3.30, which tells us that

αbx+4(c+ 1) ≤ (bx+ 4)c

andαx(c+ 1) ≥ (x− 1)n.

.............................................................................................................

(b.)

(bx+ 4)c

(x− 1)c≤(1 + 4

x

)cbc(

1− 1x

)cNotice that

(bx+ 4)c

(x− 1)c=

(bx(1 + 4

bx

))c(x(1− 1

x

))c =bcxc

(1 + 4

bx

)cxc(1− 1

x

)c =bc(1 + 4

bx

)c(1− 1

x

)cThen since 4

bx≤ 4

x, it follows that

(bx+ 4)c

(x− 1)c≤(1 + 4

x

)cbc(

1− 1x

)cas desired.

.............................................................................................................

(c.) (1 + 4

x

)cbc(

1− 1x

)c ≤ bc(1− 1

x

)c (1− 4

x

)c

48

Notice that (1 + 4

x

)cbc(

1− 1x

)c ≤ bc(1− 1

x

)c (1− 4

x

)c⇐⇒

(1 +

4

x

)c≤ 1(

1− 4x

)c⇐⇒

(1 +

4

x

)c(1− 4

x

)c≤ 1 since

(1− 4

x

)c> 0

⇐⇒((

1 +4

x

)(1− 4

x

))c≤ 1

⇐⇒(

1− 4

x+

4

x− 16

x2

)c≤ 1

⇐⇒(

1− 16

x2

)c≤ 1.

Since x ≥ 16, we know that16

x2< 1,

meaning

0 <

(1− 16

x2

)< 1

and so (1− 16

x2

)c≤ 1.

Thus (1 + 4

x

)cbc(

1− 1x

)c ≤ bc(1− 1

x

)c (1− 4

x

)cas desired.

.............................................................................................................

49

(d.)

bc(1− 1

x

)c (1− 4

x

)c ≤ bc(1− 4

x

)2cWe know that

1

x<

4

x,

meaning (1− 1

x

)>

(1− 4

x

)and therefore (

1− 1

x

)c≥(

1− 4

x

)c.

It then follows that

bc(1− 1

x

)c (1− 4

x

)c ≤ bc(1− 4

x

)c (1− 4

x

)c =bc(

1− 4x

)2cas desired.

.............................................................................................................

(e.)

bc(1− 4

x

)2c ≤ bc

1− 8cx

Here, we must show that (1− 4

x

)2c

≥ 1− 8c

x

for all c. This inequality will follow by induction.Let c=0. Then (

1− 4

x

)2c

= 1 ≥ 1 = 1− 8c

x.

50

Then assume that for c ≤ k, (1− 4

x

)2c

≥ 1− 8c

x.

Then let c = k + 1. Then(1− 4

x

)2(k+1)

=

(1− 4

x

)2k+2

=

(1− 4

x

)2k (1− 4

x

)2

≥(

1− 8k

x

)(1− 4

x

)2

by our induction hypothesis.

= 1 −(

8 + 8k

x

)+

(16 + 64k

x2

)−(

128k

x3

).

Notice that (16 + 64k

x2

)−(

128k

x3

)≥ 0

⇐⇒ 16x + 64xk − 128k ≥ 0

⇐⇒ x + 4xk − 8k ≥ 0

⇐⇒ x (1 + 4k) − 8k ≥ 0.

Then by our initial assumption, we know that when c = k+1, x ≥ 16, so in particular,x ≥ 8k, and further x ≥ 8k

1+4k. Therefore

x (1 + 4k) − 8k ≥ 0,

meaning (16 + 64k

x2

)−(

128k

x3

)≥ 0.

51

Then we get that

1 −(

8 + 8k

x

)+

(16 + 64k

x2

)−(

128k

x3

)

≥ 1 − 8 + 8k

x

= 1 − 8 (k + 1)

x.

Then when c = k + 1, it holds that(1− 4

x

)2c

≥ 1− 8c

x.

Therefore, by induction, for all c ≥ 0,(1− 4

x

)2c

≥ 1− 8c

x.

We can then conclude that

bc(1− 4

x

)2c ≤ bc

1− 8cx

as desired.

.............................................................................................................

(f.)

bc

1− 8cx

≤ bc(

1 +16c

x

)

52

Notice that

bc

1− 8cx

≤ bc(

1 +16c

x

)

⇐⇒ 1

1− 8cx

≤ 1 +16c

x

⇐⇒ x

x− 8c≤ 1 +

16c

x

⇐⇒ x2

x− 8c≤ x+ 16c

⇐⇒ x2 ≤ (x+ 16c) (x− 8c) since x > 8c

⇐⇒ x2 ≤ x2 + 8cx− 128c2

⇐⇒ 0 ≤ 8cx− 128c2

⇐⇒ 0 ≤ 8c (x− 16c)

⇐⇒ c = 0 or 0 ≤ x− 16c.

Then since x > 16(c+ 1)(b+ 1)c, it must always be true that

0 < x− 16c

and thus

bc

1− 8cx

≤ bc(

1 +16c

x

)as desired.

.............................................................................................................

53

(g.)

bc(

1 +16c

x

)< bc + 1

Notice that

bc(

1 +16c

x

)< bc + 1

⇐⇒ bc +

(bc16c

x

)< bc + 1

⇐⇒(

16cbc

x

)< 1

⇐⇒ 16cbc < x.

Then since x > 16(c+ 1)(b+ 1)c, it must be true that x > 16cbc, and thus

bc(

1 +16c

x

)< bc + 1

as desired.

.............................................................................................................

Putting the seven previous inequalities together, we get that

αbx+4(c+ 1)

αx(c+ 1)< bc + 1.

To prove our proposition, it remains to show that

bc <αbx+4(c+ 1)

αx(c+ 1).

This is easily done since, by Proposition 2.3.30, we know that

αbx+4(c+ 1) ≥ (bx+ 3)c

54

andαx(c+ 1) ≤ xc,

meaning

αbx+4(c+ 1)

αx(c+ 1)≥ (bx+ 3)c

xc=

(bx+ 3

x

)c= (b+ 3x)c ≥ bc.

Thus, combining Case 1 and Case 2, we can finally conclude that for all b ≥ 0 andall c ≥ 0, if x > 16(c+ 1)(b+ 1)c, then

bc ≤ αbx+4(c+ 1)

αx(c+ 1)< bc + 1.

Corollary 2.3.32. If x > 16(c+ 1)(b+ 1)c, then

bc = αbx+4(c+ 1)divαx(c+ 1).

Proof. By Proposition 2.3.31, we know that

bc ≤ αbx+4(c+ 1)

αx(c+ 1)< bc + 1.

for x > 16(c+ 1)(b+ 1)c. Then the integer part of αbx+4(c+1)

αx(c+1)must be bc.

Recall that in Example 2.3.32, we found that the div function is diophantine.Therefore, Corollary 2.3.32 provides us with the desired diophantine way of expressingbc in terms of αb. The only slight problem left to deal with is expressing the conditionx > 16(c + 1)(b + 1)c in a Diophantine way. We solve this problem easily with thefollowing theorem.

Proposition 2.3.33. If x = 16(c+ 1)αb+4(c+ 1), then x > 16(c+ 1)(b+ 1)c.

Proof. Using Proposition 2.3.30, we get that

αb+4(c+ 1) ≥ (b+ 3)c ≥ (b+ 1)c .

Note that there are obviously an infinite number of choices of x that would fulfillthe requirement that x > 16(c+ 1)(b+ 1)c. The reason for using αb+4 is simply that

55

is automatically fulfills the condition that the base of α be greater than or equal to4.

We have now seen that we can represent the set

{〈a, b, c〉 | a = bc}

by the set

{〈a, b, c〉 | x = 16(c+ 1)αb+4(c+ 1) and a = αbx+4(c+ 1)divαx(c+ 1)}.

In order to make the conditions

x = 16(c+ 1)αb+4(c+ 1) and a = αbx+4(c+ 1)divαx(c+ 1)

Diophantine, we can replace αb+4, αbx+4, and αx(c+1) each with the set of conditionsgiven in Proposition 2.3.27. Thus, the set {〈a, b, c〉 | a = bc} is Diophantine.

Part III

Background: ComputabilityTheory

56

Chapter 3

Key Concepts and Definitions

3.1 Register Machines

In order to understand Hilbert’s tenth problem, we first needed to understandDiophantine equations and Diophantine sets. In addition, we need to specify whatis meant by “devis[ing] a process” that determines if Diophantine equations havesolutions. In modern times, this “process” can be thought of as an algorithm, so thequestion of whether or not a Diophantine equation has solutions can be thought ofas a question of computability. Thus, we now turn to giving a background on thecomputability concepts necessary in understanding Hilbert’s tenth problem and itsnegative solution. In this section, we will examine a model of computation known asa register machine.

Definition 3.1.1. A register machine M consists of a finite number of registersR1, ..., Rn, each of which can hold a natural number, and a finite length programof instructions L0, ..., Lm, where each instruction Li must have one of the followingforms:

• Li: Rk → Rk + 1 (and GO TO Li+1)

• Li: If Rk 6= 0, then Rk → Rk − 1 and GO TO LjIf Rk = 0, then GO TO Li+1

57

58

• Li: HALT

In order to understand this definition better, let’s look at some examples. Ourfirst example is a machine that can be used to add two natural numbers together.

Example 3.1.2. Let M consist of two registers R1 and R2 and the following program:

• L0: R1 → R1 + 1 (and GO TO L1)

• L1: If R2 6= 0, then R2 → R2 − 1 and GO TO L0

(else GO TO L2)

• L2: If R1 6= 0, then R1 → R1 − 1 and GO TO L3

(else GO TO L3)

• L3: HALT

Then suppose the initial input in R1 is 2 and the initial input in R2 is 3. Thenthe following diagram represents the progression of M :

R1 : 2 3 3 4 4 5 5 6 6 5R2 : 3 3 2 2 1 1 0 0 0 0

Next Instruction: L0 L1 L0 L1 L0 L1 L0 L1 L2 L3

Instead, suppose the initial input in R1 is 13 and the initial input in R2 is 4. Thenthe following diagram represents the progression of M :

R1 : 13 14 14 15 15 16 16 17 17 18 18 17R2 : 4 4 3 3 2 2 1 1 0 0 0 0

Next Instruction: L0 L1 L0 L1 L0 L1 L0 L1 L0 L1 L2 L3

For a third case, suppose the initial input in R1 is 5 and the initial input in R2 is0. Then the following diagram represents the progression of M :

R1 : 5 6 6 5R2 : 0 0 0 0

Next Instruction: L0 L1 L2 L3

59

Note that the progression of the register machine M is different for different initialinputs. In particular, the order of the insturctions used, the contents of R1 and R2

at each step, and the number of steps used before reaching the halt instruction, allvary depending on the initial input.

It is also possible that, on some or all inputs, a register machines will never reachthe halt instruction, as shown in the following example.

Example 3.1.3. Let M consist of one register R1 and the following program:

• L0: R1 → R1 + 1

• L1: If R1 6= 0 then R1 → R1 − 1 and GO TO L0

• L2: HALT

Then let the initial input of R1 be some natural number n. Then the followingdiagram represents the progression of M :

R1 : n n+ 1 n n+ 1 n ...Next Instruction: L0 L1 L0 L1 L0 ...

Thus, on any input, this register machine enters an infinite loop and never reachesHALT.

3.2 Computable and Computably Enumberable Sets

Now that we know the definition of a register machine, we can define a computablyenumerable set.

Definition 3.2.1. A set S of n-tuples is called computably enumerable when thereexists a register machine M such that M reaches a halt on input a1, ..., an if and onlyif 〈a1, ..., an〉 ∈ S.

Note that this definition requires that M runs forever on any input that is not inS. However, as a register machine progresses on some input, we may not be able totell if the machine will run forever, or if it simply has not reached a halt yet. Thus,a stronger classification of sets is the computable set, defined as follows:

60

Definition 3.2.2. A set S of n-tuples is called RM − computable when there existsa register machine M such that on input a1, ..., an, M halts with output 1 if and onlyif 〈a1, ..., an〉 ∈ S and with output 0 if and only if 〈a1, ..., an〉 /∈ S.

Thus, if a set is computable, we have a definitive way of checking whether anygiven n-tuple is in the set. To show that Hilbert’s problem is unsolvable, we willshow that Diophantine sets are not computable, which implies that we have no wayof checking whether or not the Diophantine equation that represents the set hasnatural number solutions (if there was a way, then the set would be computable). Inorder to do this, we will make use of the following well-known theorem.

Theorem 3.2.3. A set S is computable if and only if S and Sc are computablyenumerable.

Now that we have a background on Diophantine sets and the appropriate com-putability concepts, we are prepared to prove the unsolvability of Hilbert’s TenthProblem. We will begin with a comparison of Diophantine and computably enumer-able sets.

Part IV

The Proof

61

Chapter 4

Comparison of Diophantine andC.E. Sets

In this chapter, we will see that a set is Diophantine if and only if it is computablyenumerable. In general, it is a trivial fact that any Diophantine set is computablyenumerable. If a set is Diphantine, then we know that there is some Diophantineequation that represents the set. Therefore, we can build a register machine, basedon this equation, which will halt if and only if an input is a solution, making the setcomputably enumerable. On the other hand, it is not so obvious that a computablyenumerable set is Diophantine. Our goal in the following sections will be to provethat it is.

4.1 Further Examination of Register Machines

By definition, we know that if a set S is computably enumerable, then there isa register machine M such that on input x, M reaches a halt if and only if x ∈ S.We also know that M must have some finite number of registers, call them R1, R2,...,Rn, and some finite number of instructions, call them L0, L1, ..., Lm. To pictureMx, the progression of M on some input x ∈ S, we can use the following diagram:

62

63

t=0 t=1 t=2 ... t=sR1 r1,0 = x r1,1 r1,2 ... r1,sR2 r2,0 = 0 r2,1 r2,2 ... r2,s...

......

......

...Rn rn,0 = 0 rn,1 rn,2 ... rn,s

L0 l0,0 = 1 l0,1 l0,2 ... l0,sL1 l1,0 = 0 l1,1 l1,2 ... l1,s...

......

......

...Lm lm,0 = 0 lm,1 lm,2 ... lm,s

where

• s is the number of steps that it takes for M to reach a halt on input x ∈ S

• each ri,t is the entry in register Ri at time t

• and each

li,t =

{1, if Lj is the next instruction at time t

0, else.

In order for Mx to be a valid progression of M , the values of each ri,t and each lj,tmust fit a number of characteristics, including the following:

• The first instruction must be L0,i.e. l0,0 = 1.

• At any particular time t, there must be exactly one instruction that comes next,i.e. for any t, there exists some j ∈ [0,m] such that lj,t = 1 and for all k ∈ [0,m],k 6= i, lk,t = 0.

• The only halt instruction is the last instruction,i.e. lm,s = 1, where Lm is the halt instruction, and for all t 6= s, lm,t = 0.

• At any time t, the values of r1,t, ..., rn,t and l0,t, ..., lm,t result from applying oneof the possible RM instructions, as described in Definition 3.1.1, to the valuesof r1,t−1,...,rn,t−1 and l0,t−1,...,lm,t−1.

Note that for any x ∈ S, the values of each ri,t and each lj,t will have theseproperties, and if there are values of each ri,t and each lj,t that have these propertiesfor some x, then x ∈ S. Therefore, to show that S is Diophantine, we will rewritethe above conditons as Diophantine conditions that hold if and only if x ∈ S.

64

4.2 Preparation for Determining Diophantine Con-

ditions

In order to determine the system of Diophantine conditions that are met if andonly if x ∈ S, we must first determine what constants, parameters, and variables wewill need. Since we have a particular register machine M associated with our set S,the values of m and n are constants. Our parameter will be x, representing any inputwe might put into M . Since the number of steps in Mx varies with each input x, swill be a variable. We might also attempt to make each value of ri,t and lj,t variablesas well, but since s changes with each step, each input x would require a differentnumber of variables. Instead, note that we could take our register machine and definethe values R∗

1, ..., R∗n, L∗

0, ..., L∗m, where

• for any i ∈ [1, n], R∗i =

s∑t=0

ri,t ·Qt for some Q ∈ N

• for any j ∈ [0,m], L∗j =

s∑t=0

lj,t ·Qt for some Q ∈ N.

Note that as long as Q is a large enough base, each ri,t and each lj,t could be uniquelyrecovered from each R∗

i and each L∗j . With this value of Q, we could also define the

value I, where

• I =s∑t=0

1 ·Qt.

So, the variables in our Diophantine conditions will include s, Q, I, R∗1, ..., R∗

n, andL∗0, ..., L∗

m.

In addition, we will also make use of a new binary relation, �.

Definition 4.2.1. Let r and s be two numbers written in base 2 notation, i.e.

r =

y∑i=0

ri2i (0 ≤ ri ≤ 1) and s =

y∑i=0

si2i (0 ≤ si ≤ 1)

for some y ∈ N. Then r � s if and only if ri ≤ si for all i ∈ [0, y].

In order to show that � is a Diophantine relation, we will use the following lemma,without proof.

65

Lemma 4.2.2. r � s if and only if(sr

)= 1 mod 2.

Theorem 4.2.3. � is a Diophantine relation.

Proof. In Example 2.2.2, we saw that the choose function is Diophantine, and inExample 2.1.4, we saw that the congruence relation is Diophantine. Therefore, byLemma 4.2.2, � is a Diophantine relation.

With these preparations in place, we can now determine the Diophantine condi-tions that represent S.

4.3 Determining the Diophantine Conditions

We will proceed by examining each desired characteristic individually and the appro-priate Diophantine condition(s) that can be used to represent them.

.............................................................................................................

Property: Q must be a sufficiently large base to uniquely determine each ri,t and eachlj,t.

Diophantine Conditions: x+ s < Q2

and m+ 1 < Q.

Reasoning: At any time t, ri,t cannot be any larger than x+ t. Thus, any Q > x+ swould suffice to uniquely determine each ri,t. We instead require that Q > 2(x + s)for later reasons. Further, since each lj,t is no larger than 1, Q > 1 is sufficientlylarge. We instead require that Q > m+ 1 for later reasons.

.............................................................................................................

Note that our intention is to set condtions on the values of each ri,t and lj,t, andthe above property allows us to do so by setting conditions on the values of R∗

1, ..., R∗n,

L∗0, ..., L∗

m when examined base Q. However, the Diophantine relation that we haveat our disposal is �, which compares values base 2 rather than base Q. Therefore,we will set the following condition on Q to make it easier for us to convert back andforth between values written in base 2 and base Q.

.............................................................................................................

Property: Q is some power of 2.

66

Diophantine Condition: Q � 2Q− 1

Reasoning: Take any value Q ∈ N. Then the values Q, 2Q, and 2Q − 1 can bewritten in base 2 as

... ×2k+1 ×2k ... ×21 ×20

Q = ... ak+1 ak ... a1 a0

2Q = ... ak ak−1 ... a0 0

1 = 0 0 0 0 0 1

where each 0 ≤ ai ≤ 1. Also, note that Q is a power of 2 if and only if there is aunique i such that ai = 1.

Since we know that Q > 1, we know that there is at least one i such that ai = 1.Then let k be the lowest value of i such that ai = 1. Then in subtracting 1 from 2Q,we obtain the following:

... ×2k+1 ×2k ... ×21 ×20

Q = ... ak+1 1 0 0 0

2Q = ... 1 0 0 0 0−

1 = 0 0 0 0 0 1

2Q− 1 = ... 0 1 1 1 1

Note that for all values of i that are visible in the above chart, the conditionQ � 2Q − 1 holds. Now suppose that there is some h > k such that ah = 1 as well.If h = k + 1, i.e. if ak+1 = 1, we can see from the above diagram that our conditionQ � 2Q−1 is no longer satisfied, as the coefficient on 2k+1 in Q would be 1, while thecoefficient on 2k+1 in 2Q−1 would be 0. Thus, we can assume that ah−1 = 0, meaningthat after subtraction, the coefficient on 2h in 2Q − 1 is 0. Again, this contradictsour condition, as we assumed that the coefficient on 2h in Q is 1. This contradictioncan be illustrated as follows:

67

... ×2h ×2h−1 ... ×2k+1 ×2k ... ×21 ×20

Q = ... 1 0 ... 0 1 0 0 0

2Q = ... 0 ... ... 1 0 0 0 0−

1 = 0 0 0 0 0 0 0 0 1

2Q− 1 = ... 0 ... ... 0 1 1 1 1

Therefore, the condition Q � 2Q− 1 guarantees that there is only one value of i suchthat ai = 1, and thus that Q is a power of 2.

.............................................................................................................

Property: An arbitraty I ∈ N has the form I =s∑t=0

Qt

Diophantine Condition: 1 + (Q− 1)I = Qs+1

Reasoning: Using the formula for a geometric series, we get that

I =Qs+1 − 1

Q− 1.

.............................................................................................................

For the remaining properties, it will be useful to examine the conversion betweena number written in base Q and that number written in base 2. Let N ∈ N. First,let’s examine N in base Q. We know that we can write any N as

N =∑t

qt ·Qt where 0 ≤ qt < Q.

Then since Q is some k-th power of 2, we can write N out in base Q as follows:

... ×Q2 ×Q1 ×Q0 ... ×(2k)2 ×(2k)1 ×(2k)0

N = ... q2 q1 q0 = ... q2 q1 q0

68

Next, let’s examine N in base 2. We know that we can write N as

N =∑v

uv · 2v where 0 ≤ uv ≤ 1.

Then N written in base 2 will be as follows:

... ×22k ×22k−1 ... ×2k+1 ×2k ×2k−1 ... ×21 ×20

N = ... u2k u2k−1 ... uk+1 uk uk−1 ... u1 u0

Now, suppose we want to convert between N written in base Q and N written inbase 2. For a particular t, we know that the value of qt can range from 0 to 2k−1 + 1.Then let’s consider the value of qt ·Qt for the various possibilities of qt.

qt qt ·Qt value qt ·Qt written in base 2

... ×2kt+(k−1) ... ×2kt+2 ×2kt+1 ×2kt ...0 (0) · 2kt 0 0 0 0 0 0 01 (20) · 2kt 0 0 0 0 0 1 02 (21) · 2kt 0 0 0 0 1 0 03 (21 + 20) · 2kt 0 0 0 0 1 1 04 (22) · 2kt 0 0 0 1 0 0 05 (22 + 20) · 2kt 0 0 0 1 0 1 06 (22 + 21) · 2kt 0 0 0 1 1 0 07 (22 + 21 + 20) · 2kt 0 0 0 1 1 1 0...

......

......

......

......

2k−1 (2k−1) · 2kt 0 1 0 0 0 0 02k−1 + 1 (2k−1 + 20) · 2kt 0 1 0 0 0 1 0

The important thing to notice here is that for any t, the value qt ·Qt written in base2 only ever has a 1 as the coefficients on 2kt through 2kt+(k−1). Therefore, each qtdetermines exactly the values of ukt through ukt+(k−1). For example, q0 determines u0through uk−1, q1 determines uk through u2k−1, q2 determines u2k through u3k−1, andso on. In particular, for any t,

ukt+(k−1)...ukt+1ukt

69

is the value of qt written out in base 2. Now that we understand this relationship, wecan continue examining the remaining properties.

.............................................................................................................

Property: An arbitrary R∗i ∈ N has the form R∗

i =s∑t=0

ri,t ·Qt (0 ≤ ri,t ≤ Q2

)

Diophantine Condition: R∗i � (Q

2− 1)I for all i = 1, 2, ..., n

Reasoning: We know that we can write any arbitrary R∗i ∈ N as R∗

i =∑t

ri,t · Qt

with 0 ≤ ri,t < Q for all t. Therefore, we just need to show that the conditionR∗i � (Q

2− 1)I guarantees that s is a large enough bound for this summation, i.e.

that for all t > s, ri,t = 0, and that for all t ∈ [0, s], ri,t <Q2

. Since we know that Q

is some k-th power of 2 and I =s∑t=0

Qt , we find that

(Q

2− 1

)I =

s∑t=0

(2k−1 − 1

)·Qt.

Then since 2k−1 − 1 < Q, so the coefficients on Q0 through Qs when (Q2− 1)I is

written in base Q will be 2k−1− 1. Also, note that the value 2k−1− 1 written in base2 can be determined as follows:

×2k−1 ... ×21 ×20

2k−1 = 1 0 0 0−

1 = 0 0 0 1

2k−1 − 1 = 0 1 1 1

Therefore, by our previous examination of converting between base Q and base 2,we know that we can write

(Q2− 1)I in base 2 as

70×2

(s+2)k−

1

... ×2(s+1)k+

1

×2(s+1)k

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

= 0 0 0 0 0 1 1 1 ... 0 1 1 1 0 1 1 1

Then by the condition R∗i � (Q

2− 1)I, R∗

i written in base 2 and Q notation is

×2(s+2)k−

1

... ×2(s+1)k+

1

×2(s+1)k

×2(s+1)k−

1... ×2

sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

= 0 0 0 0 0 ? ? ? ... 0 ? ? ? 0 ? ? ?

×Qs+1 ×Qs ×Q1 ×Q0

= 0 ri,s ... ri,1 ri,0

Therefore, for all t > s, ri,t = 0, and for all t ∈ [0, s], ri,t <Q2

(if it weren’t, thenthe coefficient on each 2tk−1 in R∗

i would not be a 0). Thus, our condition guaranteesthat R∗

i ∈ N has the desired form.

.............................................................................................................

Property: An arbitrary L∗j ∈ N has the form L∗

j =s∑t=0

lj,t ·Qt (0 ≤ lj,t ≤ 1).

Diophantine Condition: L∗j � I for all j = 0, 1, ...,m

Reasoning: Again, any L∗j,t ∈ N can be written as

∑t=0

lj,t ·Qt (0 ≤ lj,t < Q). Therefore,

we just need to show that our condition guarantees that s is a large enough bound forthis summation, i.e. that for all t > s, lj,t = 0, and that for all t ∈ [0, s], 0 ≤ lj,t ≤ 1.First, note that I in base 2 and base Q is

71

×Qs+1 ×Qs ×Q1 ×Q0

= 0 1 ... 1 1

×2(s+2)k−

1

... ×2(s+1)k+

1

×2(s+1)k

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

= 0 0 0 0 0 0 0 1 ... 0 0 0 1 0 0 0 1

Then by the condition L∗j � I, we find that L∗

j in base 2 and base Q is

×2(s+2)k−

1

... ×2(s+1)k+

1

×2(s+1)k

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

= 0 0 0 0 0 0 0 ? ... 0 0 0 ? 0 0 0 ?

×Qs+1 ×Qs ×Q1 ×Q0

= 0 1 or 0 ... 1 or 0 1 or 0

Therefore, the condition L∗j � I guarantees that L∗

j has the desired form.

.............................................................................................................

Property: For any t ∈ [0, s], there is exactly one j ∈ [0,m] such that lj,t = 1.

Diophantine Condition: I =m∑j=0

L∗j (Note that this is a Diophantine condition be-

cause m is a constant.)

Reasoning: As we just examined, the condition L∗j � I for j = 0, 1, ...,m guaran-

tees that for j = 0, 1, ...,m, L∗j in base 2 is

72

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

L∗j = 0 0 0 ? ... 0 0 0 ? 0 0 0 ?

Then the condition I =m∑j=0

L∗j guarantees that

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

L∗1 = 0 0 0 ? ... 0 0 0 ? 0 0 0 ?

+

L∗2 = 0 0 0 ? ... 0 0 0 ? 0 0 0 ?

+...

......

...+

L∗m = 0 0 0 ? ... 0 0 0 ? 0 0 0 ?

I = 0 0 0 1 ... 0 0 0 1 0 0 0 1

Note that each of the columns of the coefficients on 20, 2k, 22k, ..., 2sk (we willrefer to these as the columns of interest) has k − 1 columns of all zeroes to its left.Also, note that at most, any column of interest could have m + 1 1’s. If this werethe case, then during addition we would carry a value of m+1

2or m

2(whichever is an

integer) over to the next column to the left. If this value were large enough, then wewould end up carrying a value of m+1

22or m

22over to the next column to the left, and

so on. Our concern would be that this value eventually gets carried all the way tothe next column of interest. In order for this to happen, either m+1

2k> 1, or m

2k> 1.

However, recall that in our first condition, we set Q = 2k > m + 1. Therefore, thepossibility of any value getting carried into a column of interest during addition iseliminated. Thus, for any column of interest,

• if there are no 1’s, the value of I in this column will be a 0, a contradiction toour condition.

73

• if there is a single 1, the value of I in this column is a 1, as required by ourcondition.

• if the number of 1’s is greater than 1 and even, the value of I in this columnwill be a 0, a contradiction to our condition.

• if the number of 1’s is greater than 1 and odd, then the value of I in some columnto the left (but before the next column of interest) will be a 1, a contradictionto our condition.

Thus, the condition I =m∑j=0

L∗j guarantees that for any t ∈ [0, s], there is exactly one

j ∈ [0,m] such that lj,t = 1.

.............................................................................................................

Property: l0,0 = 1.

Diophantine Condition: 1 � L∗0

Reasoning: We know that 1 in base 2 is

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

1 = 0 0 0 0 ... 0 0 0 0 0 0 0 1

Then the condition 1 � L∗0 guarantees that L∗

0 in base 2 must be

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

L∗0 = ? ? ? ? ... ? ? ? ? ? ? ? 1

Thus, l0,0 = 1, as desired.

74

.............................................................................................................

Property: lm,t = 1 if and only if t = s, where Lm is the halt instruction.

Diophantine Condition: L∗m = Qs

Reasoning: The condition L∗m = Qs implies that L∗

m in base Q is

×Qs ×Qs−1 ×Q1 ×Q0

L∗m = 1 0 ... 0 0

Thus lm,s = 1 and for all t 6= s, lm,t = 0, as desired.

.............................................................................................................

The conditions that follow can be used to ensure the property that at any timet, the values of r1,t, ..., rn,t, l0,t, ..., lm,t are determined by applying the appropriateinstruction of M to the values r1,t−1, ..., rn,t−1, l0,t−1, ..., lm,t−1. Thus, these conditionswill depend on the particular instructions L0, ..., Lm that make up M .

.............................................................................................................

For any j ∈ [0,m] such that the instruction Lj is GO TO Lp, we include a copyof the following condition.

Diophantine Condition: QL∗j � L∗

p

Reasoning: In base Q, we know that L∗j is

×Qs ×Qs−1 ×Q1 ×Q0

L∗j = 0 lj,s ... lj,1 lj,0

Then QL∗j in base Q and base 2 is

75

×Qs+1 ×Qs ×Q1 ×Q0

= lj,s lj,s−1 ... lj,0 0

×2(s+2)k−

1

... ×2(s+1)k+

1

×2(s+1)k

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

= 0 0 0 lj,s 0 0 0 lj,s−1 ... 0 0 0 lj,0 0 0 0 0

Further, L∗p in base Q and base 2 is

×Qs+1 ×Qs ×Q1 ×Q0

= 0 lp,s ... lp,1 lp,0

×2(s+2)k−

1

... ×2(s+1)k+

1

×2(s+1)k

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

= 0 0 0 0 0 0 0 lp,s ... 0 0 0 lp,1 0 0 0 lp,0

Thus, for any t, the value lj,t is compared with the value lp,t+1. Therefore, thecondition QL∗

j � L∗p guarantees that for any t, if lj,t = 1, then lp,t+1 = 1.

.............................................................................................................

For any j ∈ [0,m] such that the instruction Lj is IF Ri = 0, GO TO Lp (IFRi 6= 0, GO TO Lj+1), we include a copy of the following conditions.

Diophantine Conditions: QL∗j � L∗

p + L∗j+1 and QL∗

j � L∗j+1 +QI − 2R∗

j

Reasoning:First, we will show that the condition QL∗

j � L∗p + L∗

j+1 guarantees that for anyt such that lj,t = 1, either lj,t+1 = 1 or lp,t+1 = 1. We know that QL∗

j in base Q andbase 2 is

76

×Qs+1 ×Qs ×Q1 ×Q0

= lj,s lj,s−1 ... lj,0 0

×2(s+2)k−

1

... ×2(s+1)k+

1

×2(s+1)k

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

x20

= 0 0 0 lj,s 0 0 0 lj,s−1 ... 0 0 0 lj,0 0 0 0 0

Further, L∗p in base Q and base 2 is

×Qs+1 ×Qs ×Q1 ×Q0

= 0 lp,s ... lp,1 lp,0

×2(s+2)k−

1

... ×2(s+1)k+

1

×2(s+1)k

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

= 0 0 0 0 0 0 0 lp,s ... 0 0 0 lp,1 0 0 0 lp,0

And lastly, L∗j+1 in base Q and base 2 is

×Qs+1 ×Qs ×Q1 ×Q0

= 0 lj+1,s ... lj+1,1 lj+1,0

×2(s+2)k−

1

... ×2(s+1)k+

1

×2(s+1)k

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

= 0 0 0 0 0 0 0 lj+1,s ... 0 0 0 lj+1,1 0 0 0 lj+1,0

Then we can see that when QL∗j , L

∗p, and L∗

j+1 are written in base 2, any lj,tlines up with lp,t+1 and lj+1,t+1. Therefore, for any t such that lj,t = 1, the conditionQL∗

j � L∗p + L∗

j+1 guarantees that either lp,t+1 = 1 or lj+1,t+1 = 1 (we know from aprevious condition that they cannot both be 1).

77

Next, we will see that the condition QL∗j � L∗

j+1 + QI − 2R∗j specifies that if

ri,t = 0, then it is lp,t+1 = 1, whereas if ri,t 6= 0, then it is lj+1,t+1 = 1. First, we knowthat QI in base Q and base 2 is

×Qs+1 ×Qs ×Q1 ×Q0

= 1 1 ... 1 0

×2(s+2)k−

1

... ×2(s+1)k+

1

×2(s+1)k

×2(s+1)k−

1... ×2

sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

= 0 0 0 1 0 0 0 1 ... 0 0 0 1 0 0 0 0

Then recall that in our first condition, we set ri,t <Q2

, so any 2ri,t < Q. Further,note that any value 2ri,0, 2ri,1, ..., 2ri,s is divisible by 2, so any of these values writtenout in base 2 will have a coefficient of 0 on the term 20. Thus, the value 2R∗

j in base2 has a coefficient of 0 on all 20, 2k, 22k, ..., 2sk. In particular, 2R∗

i in base Q andbase 2 is

×Qs+1 ×Qs ×Q1 ×Q0

= 0 2ri,s ... 2ri,1 2ri,0

×2(s+2)k−

1

... ×2(s+1)k+

1

×2(s+1)k

×2(s+1)k−

1

... ×2sk+1

×2sk

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

= 0 0 0 0 ? ? ? 0 ... ? ? ? 0 ? ? ? 0

Now, suppose that for some t such that lj,t = 1, ri,t = 0. Then our set up to findthe value L∗

j+1 + QI − 2R∗i in base 2 will look like one of the following two cases,

depending on the specific value of t.

78

Case A: t 6= 0

×2(t+2)k−

1

... ×2(t+1)k+

1

×2(t+1)k

×2(t+1)k−

1

... ×2tk+1

×2tk

L∗j+1 = ... 0 0 0 lj+1,t+1 0 0 0 0 ...

+

QI = ... 0 0 0 1 0 0 0 1 ...

L∗j+1 +QI = ... ? ? ? a 0 0 0 1 ...

−2R∗

j = ... ? ? ? 0 0 0 0 0 ...

L∗j+1 +QI − 2R∗

j = ... ? ? ? b ? ? ? ? ...

Case B: t = 0

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

L∗j+1 = ... 0 0 0 lj+1,t+1 0 0 0 0

+

QI = ... 0 0 0 1 0 0 0 0

L∗j+1 +QI = ... ? ? ? a 0 0 0 0

−2R∗

j = ... ? ? ? 0 0 0 0 0

L∗j+1 +QI − 2R∗

j = ... ? ? ? b ? ? ? ?

79

In case A, we don’t know for certain what the coefficients on any of the terms tothe right of 2tk are. So, it is possible that borrowing would have to occur during thesubtraction. However, this borrowing will stop at the 2tk column, because the valueof L∗

j+1 + QI here is a 1. In case B, we know no borrowing occurs to the right of

2(t+1)k, because we can see that all of these coefficients are 0’s. Thus, in either case,we know that the coefficients on 2(t+1)k remain undisturbed. Further, note that thecondition QL∗

j � L∗j+1 + QI − 2R∗

j requires that b = 1. If lj+1,t+1 = 1, then a = 0,and so b = 0, a contradiction to our condition. If lj+t,t+1 = 0, then a = 1, so b = 1,as required. Therefore, we can conclude that if ri,t = 0, then lj+1,t+1 = 0, and thusby the condition QL∗

j � L∗p + L∗

j+1, lp,t+1 = 1.On the other hand, suppose that for t such that lj,t = 1, ri,t 6= 0. Then again we

must consider the following two cases, depending on the specific value of t.

Case A: t 6= 0

×2(t+2)k−

1

... ×2(t+1)k+

1

×2(t+1)k

×2(t+1)k−

1... ×2

tk+1

×2tk

L∗j+1 = ... 0 0 0 lj+1,t+1 0 0 0 0 ...

+

QI = ... 0 0 0 1 0 0 0 1 ...

L∗j+1 +QI = ... ? ? ? a 0 0 0 1 ...

−2R∗

j = ... ? ? ? 0 at least one 1 0 ...

L∗j+1 +QI − 2R∗

j = ... ? ? ? b ? ? ? ? ...

80

Case B: t = 0

×22k−1

... ×2k+

1

×2k

×2k−

1

... ×21

×20

L∗j+1 = ... 0 0 0 lj+1,t+1 0 0 0 0

+

QI = ... 0 0 0 1 0 0 0 0

L∗j+1 +QI = ... ? ? ? a 0 0 0 0

−2R∗

j = ... ? ? ? 0 at least one 1 0

L∗j+1 +QI − 2R∗

j = ... ? ? ? b ? ? ? ?

Just as before, in either case, we know that any borrowing that occurs to theright of the 2tk column is stopped at the 2tk. But since ri,t 6= 0, we know that atleast one of the coefficients on 2tk+1, 2tk+2, ..., 2(t+1)k−1 in 2R∗

j will be a 1. Thus,the borrowing begins again wherever this first 1 is placed, and will continue throughto the coefficient of 2(t+1)k in L∗

j+1 + QI. If lj+1,t+1 = 1, then a = 0. Then, duringborrowing, a will become a 1, making b = 1, as required. If lj+1,t+1 = 0, then a = 1.Then, during borrowing, a will become a 0, making b = 0, a contradiction to ourcondition. Thus, if ri,t 6= 0, lj+1,t+1 = 1.

.............................................................................................................

Lastly, we have to make sure that any restrictions Lj of the form Ri → Ri + 1or Ri → Ri − 1 are followed correctly. In order to do this, we can use the followingconditions.

Diophantine Conditions:

For i = 1,

R∗i = QR∗

i +∑f

QL∗f −

∑g

QL∗g + x

81

and for i = 2, 3, ..., n,

R∗i = QR∗

i +∑f

QL∗f −

∑g

QL∗g

where∑f

QL∗f is the sum over all f such that Lf : Ri → Ri + 1 and

∑g

QL∗g is the

sum over all g such that Lg : Ri → Ri − 1.

Reasoning: For i = 2, 3, ..., n, we can picture the condition R∗i = QR∗

i +∑f

QL∗f −∑

g

QL∗g as follows:

×Qs+1 ×Qs ×Q1 ×Q0

QR∗i = ri,s ri,s−1 ... ri,0 0

+

QLf1 = lf1,s lf1,s−1 ... lf1,0 0...

...+

QLfα = lfα,s lfα,s−1 ... lfα,0 0−

QLg1 = lg1,s lg1,s−1 ... lg1,0 0...

...−

QLgβ = lgβ ,s lgβ ,s−1 ... lgβ ,0 0

R∗i = ri,s+1 ri,s ... ri,1 ri,0

Note that this picture would be the same for the condition R∗1 = QR∗

1 +∑f

QL∗f −∑

g

QL∗g + x, with the addition of x in the Q0 column.

Then ri,0 = 0 for all i = 1, 2, ...n, and r1,0 = x, as desired. For any t 6= 0,ri,t = ri,t−1 + lf1,t−1 + ... + lfα,t−1 − lg1,t−1 − ... − lgβ ,t−1. At time t, if the instructionat time t − 1 was to add 1 to Ri, then one of the lf1,t−1, ..., lfα,t−1 will be a 1,and ri,t = ri,t−1 + 1. If the instruction was to subtract 1 from Ri, then one of thelg1,t−1 , ..., lgβ ,t−1 will be a 1, and ri,t = ri,t−1 − 1. For any other type of instruction,ri,t = ri,t−1. Thus, our conditions guarantee that each for any time t, each register

82

has the appropriate entry.

.............................................................................................................

To conclude, given a computably enumerable set S, there must be some registermachine M such that M halts on input x if and only if x ∈ S. We have now seen agroup of Diophantine conditions that will be satisfied if and only if there is a progres-sion of M that halts on input x. Therefore, these Diophantine conditions are satisfiedif and only if x ∈ S, so S is a Diophantine set.

We have now seen that a set is Diophantine if and only if it is computably enumer-able. With this result, we can proceed to prove the unsolvability of Hilbert’s tenthproblem.

Chapter 5

Hilbert’s Tenth Problem isUnsolvable

From our background in computability theory, we know that there are many setswhich are computably enumerable but not computable. By the remarkable resultproved in Chapter 4, we now know that these sets are Diophantine sets that are notcomputable. Then if we consider the Diophantine equation that represents any suchset, we know that there is no process to determine whether or not the eqauation issolvable in the natural numbers. Thus, the process desired by Hilbert does not exist,and his tenth problem is proven unsolvable.

83

Bibliography

[1] Matiyasevich, Y. Hilbert’s Tenth Problem. MIT Press, (1993), Massachusetts.

[2] Jones, J.P, and Matiyasevich, Y.V. Register Machine Proof of the Theorem onExponential Diophantine Representation of Enumerable Sets. The Journal ofSymbolic Logic, vol. 49, (1984), pp.818-828.

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