Hilberts Tenth Problem John Lindsay Orr Department of Mathematics - - PowerPoint PPT Presentation

hilbert s tenth problem
SMART_READER_LITE
LIVE PREVIEW

Hilberts Tenth Problem John Lindsay Orr Department of Mathematics - - PowerPoint PPT Presentation

Introduction Sketch of Proof Going Into the Details Hilberts Tenth Problem John Lindsay Orr Department of Mathematics Univesity of NebraskaLincoln September 15, 2005 John Lindsay Orr Hilberts Tenth Problem Introduction Sketch of


slide-1
SLIDE 1

Introduction Sketch of Proof Going Into the Details

Hilbert’s Tenth Problem

John Lindsay Orr

Department of Mathematics Univesity of Nebraska–Lincoln

September 15, 2005

John Lindsay Orr Hilbert’s Tenth Problem

slide-2
SLIDE 2

Introduction Sketch of Proof Going Into the Details

Outline

1

Introduction Disclaimer History and Statement of the Problem

2

Sketch of Proof Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

3

Going Into the Details Working with Diophantine Sets Coding n-tuples

John Lindsay Orr Hilbert’s Tenth Problem

slide-3
SLIDE 3

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Outline

1

Introduction Disclaimer History and Statement of the Problem

2

Sketch of Proof Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

3

Going Into the Details Working with Diophantine Sets Coding n-tuples

John Lindsay Orr Hilbert’s Tenth Problem

slide-4
SLIDE 4

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Disclaimer

I don’t know what I’m talking about! This guy does: Yuri Matiyasevich, “Hilbert’s Tenth Problem”

John Lindsay Orr Hilbert’s Tenth Problem

slide-5
SLIDE 5

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Outline

1

Introduction Disclaimer History and Statement of the Problem

2

Sketch of Proof Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

3

Going Into the Details Working with Diophantine Sets Coding n-tuples

John Lindsay Orr Hilbert’s Tenth Problem

slide-6
SLIDE 6

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Hilbert’s Problems

Hilbert’s twenty-three problems Second International Congress of Mathematicians held in Paris, 1900 Included Continuum Hypothesis and Riemann Hypothesis Included general projects such as “Can physics be axiomatized”?

John Lindsay Orr Hilbert’s Tenth Problem

slide-7
SLIDE 7

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Hilert’s Tenth Problem

  • 10. Determination of the Solvability of a Diophantine Equation

Given a diophantine equation with any number of unknown quantities and with rational integral numerical coefficients: To devise a process according to which it can be determined by a finite number of operations whether the equation is solvable in rational integers.

John Lindsay Orr Hilbert’s Tenth Problem

slide-8
SLIDE 8

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

A diophantine equation is a polynomial equation of the form D(x1, . . . , xm) = 0 where D is a polynomial with integer coefficients. Example. x2 + y2 − z2 = 0 Example. x3 + y3 − z3 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-9
SLIDE 9

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Can we find an algorithm which you can then present with any diophantine equation, D(x1, . . . , xm) = 0, and be sure that you will get a “Yes” or “No” answer as to whether the equation has solutions over Nm?

John Lindsay Orr Hilbert’s Tenth Problem

slide-10
SLIDE 10

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Can we find an algorithm which you can then present with any diophantine equation, D(x1, . . . , xm) = 0, and be sure that you will get a “Yes” or “No” answer as to whether the equation has solutions over Nm? The Answer: NO, WE CAN’T

John Lindsay Orr Hilbert’s Tenth Problem

slide-11
SLIDE 11

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Notes

Determining solvability isn’t the same as finding a solution This wouldn’t answer Fermat’s Last Theorem By N I mean {0, 1, 2, 3, . . .} By “solution” I almost always mean “solution in N,” not in Z.

John Lindsay Orr Hilbert’s Tenth Problem

slide-12
SLIDE 12

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Notes

Determining solvability isn’t the same as finding a solution This wouldn’t answer Fermat’s Last Theorem By N I mean {0, 1, 2, 3, . . .} By “solution” I almost always mean “solution in N,” not in Z.

John Lindsay Orr Hilbert’s Tenth Problem

slide-13
SLIDE 13

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Notes

Determining solvability isn’t the same as finding a solution This wouldn’t answer Fermat’s Last Theorem By N I mean {0, 1, 2, 3, . . .} By “solution” I almost always mean “solution in N,” not in Z.

John Lindsay Orr Hilbert’s Tenth Problem

slide-14
SLIDE 14

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Notes

Determining solvability isn’t the same as finding a solution This wouldn’t answer Fermat’s Last Theorem By N I mean {0, 1, 2, 3, . . .} By “solution” I almost always mean “solution in N,” not in Z.

John Lindsay Orr Hilbert’s Tenth Problem

slide-15
SLIDE 15

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Why Only Over N?

Over N: D(x1, x2, . . . , xn) = 0 Over Z: D(x1,x2, . . . , xn)2 +(y2

1,1 + y2 1,2 + y2 1,3 + y2 1,4 − x1)2

+(y2

2,1 + y2 2,2 + y2 2,3 + y2 2,4 − x2)2

. . . +(y2

n,1 + y2 n,2 + y2 n,3 + y2 n,4 − xn)2 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-16
SLIDE 16

Introduction Sketch of Proof Going Into the Details Disclaimer History and Statement of the Problem

Also study diophantine equation with parameters D(a1, . . . , an, x1, . . . , xm) = 0 and ask for which values of (a1, . . . , an) does the equation have a solution. Example. ax − by − 1 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-17
SLIDE 17

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Outline

1

Introduction Disclaimer History and Statement of the Problem

2

Sketch of Proof Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

3

Going Into the Details Working with Diophantine Sets Coding n-tuples

John Lindsay Orr Hilbert’s Tenth Problem

slide-18
SLIDE 18

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

What is a Turing Machine?

It’s a model for a computer Church-Turing Thesis says it models any computer What does it look like? The machine scans a (singly) infinite tape The machine takes states from X = {x1, . . . , xm}. The tape holds values from Y = {y1, . . . , yn}

John Lindsay Orr Hilbert’s Tenth Problem

slide-19
SLIDE 19

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

What is a Turing Machine?

It’s a model for a computer Church-Turing Thesis says it models any computer What does it look like? The machine scans a (singly) infinite tape The machine takes states from X = {x1, . . . , xm}. The tape holds values from Y = {y1, . . . , yn}

John Lindsay Orr Hilbert’s Tenth Problem

slide-20
SLIDE 20

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

What is a Turing Machine?

It’s a model for a computer Church-Turing Thesis says it models any computer What does it look like? The machine scans a (singly) infinite tape The machine takes states from X = {x1, . . . , xm}. The tape holds values from Y = {y1, . . . , yn}

1 3 2 4 5 6 7 8 Head Tape

y3 y2 y3 y1 y2 y1 xi ∗ y1

John Lindsay Orr Hilbert’s Tenth Problem

slide-21
SLIDE 21

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

What is a Turing Machine?

It’s a model for a computer Church-Turing Thesis says it models any computer What does it look like? The machine scans a (singly) infinite tape The machine takes states from X = {x1, . . . , xm}. The tape holds values from Y = {y1, . . . , yn}

1 3 2 4 5 6 7 8 Head Tape

y3 y2 y3 y1 y2 y1 xi ∗ y1

John Lindsay Orr Hilbert’s Tenth Problem

slide-22
SLIDE 22

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

What is a Turing Machine?

It’s a model for a computer Church-Turing Thesis says it models any computer What does it look like? The machine scans a (singly) infinite tape The machine takes states from X = {x1, . . . , xm}. The tape holds values from Y = {y1, . . . , yn}

1 3 2 4 5 6 7 8 Head Tape

y3 y2 y3 y1 y2 y1 xi ∗ y1

John Lindsay Orr Hilbert’s Tenth Problem

slide-23
SLIDE 23

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

What is a Turing Machine?

It’s a model for a computer Church-Turing Thesis says it models any computer What does it look like? The machine scans a (singly) infinite tape The machine takes states from X = {x1, . . . , xm}. The tape holds values from Y = {y1, . . . , yn}

1 3 2 4 5 6 7 8 Head Tape

y3 y2 y3 y1 y2 y1 xi ∗ y1

John Lindsay Orr Hilbert’s Tenth Problem

slide-24
SLIDE 24

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How a Turing Machine Works

At each step the machine:

1

scans the current cell while in state x

2

reads the value (y) from that cell

3

writes a value W(x, y) to the cell

4

moves in direction D(x, y)

5

enters state S(x, y) So the machine is determined by three finite functions: W : X×Y − → Y, D : X×Y − → {−1, 0, 1}, and S : X×Y − → X The machine also has a single initial state x1 and some final states.

John Lindsay Orr Hilbert’s Tenth Problem

slide-25
SLIDE 25

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How a Turing Machine Works

At each step the machine:

1

scans the current cell while in state x

2

reads the value (y) from that cell

3

writes a value W(x, y) to the cell

4

moves in direction D(x, y)

5

enters state S(x, y) So the machine is determined by three finite functions: W : X×Y − → Y, D : X×Y − → {−1, 0, 1}, and S : X×Y − → X The machine also has a single initial state x1 and some final states.

John Lindsay Orr Hilbert’s Tenth Problem

slide-26
SLIDE 26

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How a Turing Machine Works

At each step the machine:

1

scans the current cell while in state x

2

reads the value (y) from that cell

3

writes a value W(x, y) to the cell

4

moves in direction D(x, y)

5

enters state S(x, y) So the machine is determined by three finite functions: W : X×Y − → Y, D : X×Y − → {−1, 0, 1}, and S : X×Y − → X The machine also has a single initial state x1 and some final states.

John Lindsay Orr Hilbert’s Tenth Problem

slide-27
SLIDE 27

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How a Turing Machine Works

At each step the machine:

1

scans the current cell while in state x

2

reads the value (y) from that cell

3

writes a value W(x, y) to the cell

4

moves in direction D(x, y)

5

enters state S(x, y) So the machine is determined by three finite functions: W : X×Y − → Y, D : X×Y − → {−1, 0, 1}, and S : X×Y − → X The machine also has a single initial state x1 and some final states.

John Lindsay Orr Hilbert’s Tenth Problem

slide-28
SLIDE 28

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How a Turing Machine Works

At each step the machine:

1

scans the current cell while in state x

2

reads the value (y) from that cell

3

writes a value W(x, y) to the cell

4

moves in direction D(x, y)

5

enters state S(x, y) So the machine is determined by three finite functions: W : X×Y − → Y, D : X×Y − → {−1, 0, 1}, and S : X×Y − → X The machine also has a single initial state x1 and some final states.

John Lindsay Orr Hilbert’s Tenth Problem

slide-29
SLIDE 29

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How a Turing Machine Works

At each step the machine:

1

scans the current cell while in state x

2

reads the value (y) from that cell

3

writes a value W(x, y) to the cell

4

moves in direction D(x, y)

5

enters state S(x, y) So the machine is determined by three finite functions: W : X×Y − → Y, D : X×Y − → {−1, 0, 1}, and S : X×Y − → X The machine also has a single initial state x1 and some final states.

John Lindsay Orr Hilbert’s Tenth Problem

slide-30
SLIDE 30

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How a Turing Machine Works

At each step the machine:

1

scans the current cell while in state x

2

reads the value (y) from that cell

3

writes a value W(x, y) to the cell

4

moves in direction D(x, y)

5

enters state S(x, y) So the machine is determined by three finite functions: W : X×Y − → Y, D : X×Y − → {−1, 0, 1}, and S : X×Y − → X The machine also has a single initial state x1 and some final states.

John Lindsay Orr Hilbert’s Tenth Problem

slide-31
SLIDE 31

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How a Turing Machine Works

At each step the machine:

1

scans the current cell while in state x

2

reads the value (y) from that cell

3

writes a value W(x, y) to the cell

4

moves in direction D(x, y)

5

enters state S(x, y) So the machine is determined by three finite functions: W : X×Y − → Y, D : X×Y − → {−1, 0, 1}, and S : X×Y − → X The machine also has a single initial state x1 and some final states.

John Lindsay Orr Hilbert’s Tenth Problem

slide-32
SLIDE 32

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How to Program a Turing Machine

Build simple machines that do basic operations, like: LEFT or RIGHT WRITE(y) READ(y) STOP or NEVERSTOP Learn how to compose machines: if ( M1 ) { M2 }

  • r

while ( M1 ) { M2 }

John Lindsay Orr Hilbert’s Tenth Problem

slide-33
SLIDE 33

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How to Program a Turing Machine

Build simple machines that do basic operations, like: LEFT or RIGHT WRITE(y) READ(y) STOP or NEVERSTOP Learn how to compose machines: if ( M1 ) { M2 }

  • r

while ( M1 ) { M2 }

John Lindsay Orr Hilbert’s Tenth Problem

slide-34
SLIDE 34

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How to Program a Turing Machine

Build simple machines that do basic operations, like: LEFT or RIGHT WRITE(y) READ(y) STOP or NEVERSTOP Learn how to compose machines: if ( M1 ) { M2 }

  • r

while ( M1 ) { M2 }

John Lindsay Orr Hilbert’s Tenth Problem

slide-35
SLIDE 35

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How to Program a Turing Machine

Build simple machines that do basic operations, like: LEFT or RIGHT WRITE(y) READ(y) STOP or NEVERSTOP Learn how to compose machines: if ( M1 ) { M2 }

  • r

while ( M1 ) { M2 }

John Lindsay Orr Hilbert’s Tenth Problem

slide-36
SLIDE 36

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How to Program a Turing Machine

Build simple machines that do basic operations, like: LEFT or RIGHT WRITE(y) READ(y) STOP or NEVERSTOP Learn how to compose machines: if ( M1 ) { M2 }

  • r

while ( M1 ) { M2 }

John Lindsay Orr Hilbert’s Tenth Problem

slide-37
SLIDE 37

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How to Program a Turing Machine

Build simple machines that do basic operations, like: LEFT or RIGHT WRITE(y) READ(y) STOP or NEVERSTOP Learn how to compose machines: if ( M1 ) { M2 }

  • r

while ( M1 ) { M2 }

John Lindsay Orr Hilbert’s Tenth Problem

slide-38
SLIDE 38

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Say that a set S ⊆ N is Turing decidable if there is a Turing machine M such that, whenever M is started with initial data on the tape encoding a n ∈ N: M halts in state q2 if n ∈ S M halts in state q3 if n ∈ S

John Lindsay Orr Hilbert’s Tenth Problem

slide-39
SLIDE 39

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

How to Answer Hilbert’s Tenth Problem

Imagine indexing all possible diophantine equations in some

  • rder. E.g. D1, D2, D3, . . ..

Let S = {k : Dk has a solution}. Hilbert’s 10th problem becomes: Question Is S Turing decidable?

John Lindsay Orr Hilbert’s Tenth Problem

slide-40
SLIDE 40

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Say that a set S ⊆ N is Turing semidecidable if there is a Turing machine M such that, whenever M is started with initial data on the tape encoding a n ∈ N: if n ∈ S then M eventually halts if n ∈ S then M never halts

John Lindsay Orr Hilbert’s Tenth Problem

slide-41
SLIDE 41

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Lemma If S is Turing decidable then S and Sc are Turing semidecidable. Proof. Let M be a machine that decides S. To semidecide S use the machine: if ( M ) { STOP }; NEVERSTOP To semidecide Sc use the machine: if ( M ) { NEVERSTOP } STOP;

John Lindsay Orr Hilbert’s Tenth Problem

slide-42
SLIDE 42

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Lemma If S is Turing decidable then S and Sc are Turing semidecidable. Proof. Let M be a machine that decides S. To semidecide S use the machine: if ( M ) { STOP }; NEVERSTOP To semidecide Sc use the machine: if ( M ) { NEVERSTOP } STOP;

John Lindsay Orr Hilbert’s Tenth Problem

slide-43
SLIDE 43

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Lemma If S is Turing decidable then S and Sc are Turing semidecidable. Proof. Let M be a machine that decides S. To semidecide S use the machine: if ( M ) { STOP }; NEVERSTOP To semidecide Sc use the machine: if ( M ) { NEVERSTOP } STOP;

John Lindsay Orr Hilbert’s Tenth Problem

slide-44
SLIDE 44

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Theorem The set S is Turing decidable if and only if S and Sc are Turing semidecidable.

John Lindsay Orr Hilbert’s Tenth Problem

slide-45
SLIDE 45

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Outline

1

Introduction Disclaimer History and Statement of the Problem

2

Sketch of Proof Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

3

Going Into the Details Working with Diophantine Sets Coding n-tuples

John Lindsay Orr Hilbert’s Tenth Problem

slide-46
SLIDE 46

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Definition

Say that a set S ⊆ Nk is diophantine if there exists a diophantine equation D(a1, . . . , ak, x1, . . . , xn) = 0 such that (a1, . . . , ak) ∈ S if and only if D(a1, . . . , ak, x1, . . . , xn) = 0 has a solution in Nn.

John Lindsay Orr Hilbert’s Tenth Problem

slide-47
SLIDE 47

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Definition

Say that a set S ⊆ Nk is diophantine if there exists a diophantine equation D(a1, . . . , ak, x1, . . . , xn) = 0 such that (a1, . . . , ak) ∈ S if and only if D(a1, . . . , ak, x1, . . . , xn) = 0 has a solution in Nn.

  • Example. The set

{(a, b) : gcd(a, b) = 1} is diophantine.

John Lindsay Orr Hilbert’s Tenth Problem

slide-48
SLIDE 48

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Definition

Say that a set S ⊆ Nk is diophantine if there exists a diophantine equation D(a1, . . . , ak, x1, . . . , xn) = 0 such that (a1, . . . , ak) ∈ S if and only if D(a1, . . . , ak, x1, . . . , xn) = 0 has a solution in Nn.

  • Example. The set

{(a, b) : gcd(a, b) = 1} is diophantine. (Take D(a, b, x, y) = ax − by − 1.)

John Lindsay Orr Hilbert’s Tenth Problem

slide-49
SLIDE 49

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

  • Example. The set

{a : a is not a prime} is diophantine.

  • Proof. Let

D(a, x, y) = (x + 2)(y + 2) − a

John Lindsay Orr Hilbert’s Tenth Problem

slide-50
SLIDE 50

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

In fact, the set {a : a is a prime} is diophantine.

  • Factoid. A set S ⊆ N is diophantine if and only if S is the set of

non-negative values taken by some integer-coefficient polynomial as its variables range over N. Thus, incredibly, {prime numbers} = N ∩ {D(x1, . . . , xn : x1, . . . , xn ∈ N}

John Lindsay Orr Hilbert’s Tenth Problem

slide-51
SLIDE 51

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Lemma Every diophantine set is Turing semidecidable.

John Lindsay Orr Hilbert’s Tenth Problem

slide-52
SLIDE 52

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Proof. S has a diophantine representation D(a1, . . . , ak, x1, . . . , xn) = 0 Initialize the tape with (a1, . . . , ak) ∈ Nk, and run: foreach x = (x1, . . . , xn) ∈ Nn { if( D(a1, . . . , ak, x1, . . . , xn) = 0 ) {

STOP

} }

John Lindsay Orr Hilbert’s Tenth Problem

slide-53
SLIDE 53

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Theorem Every Turing semidecidable set is diophantine. Corollary A set is diophantine ⇐ ⇒ it is Turing semidecidable.

John Lindsay Orr Hilbert’s Tenth Problem

slide-54
SLIDE 54

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Theorem Every Turing semidecidable set is diophantine. Corollary A set is diophantine ⇐ ⇒ it is Turing semidecidable.

John Lindsay Orr Hilbert’s Tenth Problem

slide-55
SLIDE 55

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Summary

What have we learned? S is decidable ⇐ ⇒S, Sc are semidecidable ⇐ ⇒S, Sc are diophantine So one way to show a set is not decidable is to show that one

  • f S or Sc is not diophantine.

John Lindsay Orr Hilbert’s Tenth Problem

slide-56
SLIDE 56

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Outline

1

Introduction Disclaimer History and Statement of the Problem

2

Sketch of Proof Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

3

Going Into the Details Working with Diophantine Sets Coding n-tuples

John Lindsay Orr Hilbert’s Tenth Problem

slide-57
SLIDE 57

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Definition

The integer-coefficient polynomial U(a1, . . . , ak, c, y1, . . . , yw) is a universal diophantine polynomial if, for any diophantine equation D(a1, . . . , ak, x1, . . . , xn) = 0 we can find a code c ∈ N such that ∃x1, . . . , xn with D(a1, . . . , ak, x1, . . . , xn) = 0 ⇐ ⇒ ∃y1, . . . , yw with U(a1, . . . , ak, c, y1, . . . , yw) = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-58
SLIDE 58

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Theorem For each k, there exists a universal diophantine equation Uk(a1, . . . , ak, c, y1, . . . , yw) Let H0 = {c : U0(c, y1, . . . , yv) = 0 has a solution} This is our “enumeration of the solvable diophantine equations”. We shall show that H0 is diophantine and Hc

0 is not!

John Lindsay Orr Hilbert’s Tenth Problem

slide-59
SLIDE 59

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Theorem For each k, there exists a universal diophantine equation Uk(a1, . . . , ak, c, y1, . . . , yw) Let H0 = {c : U0(c, y1, . . . , yv) = 0 has a solution} This is our “enumeration of the solvable diophantine equations”. We shall show that H0 is diophantine and Hc

0 is not!

John Lindsay Orr Hilbert’s Tenth Problem

slide-60
SLIDE 60

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Theorem For each k, there exists a universal diophantine equation Uk(a1, . . . , ak, c, y1, . . . , yw) Let H0 = {c : U0(c, y1, . . . , yv) = 0 has a solution} This is our “enumeration of the solvable diophantine equations”. We shall show that H0 is diophantine and Hc

0 is not!

John Lindsay Orr Hilbert’s Tenth Problem

slide-61
SLIDE 61

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H1 = {k : U1(k, k, y1, . . . , yw) = 0 has a solution}

  • Claim. H1 is a diophantine set but Hc

1 is not.

  • Proof. (First part) Write D(k, y1, . . . , yw) = U1(k, k, y1, . . . , yw).

Then k ∈ H1 ⇐ ⇒ D(k, y1, . . . , yw) has a solution Thus H1 is diophantine.

John Lindsay Orr Hilbert’s Tenth Problem

slide-62
SLIDE 62

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H1 = {k : U1(k, k, y1, . . . , yw) = 0 has a solution}

  • Claim. H1 is a diophantine set but Hc

1 is not.

  • Proof. (First part) Write D(k, y1, . . . , yw) = U1(k, k, y1, . . . , yw).

Then k ∈ H1 ⇐ ⇒ D(k, y1, . . . , yw) has a solution Thus H1 is diophantine.

John Lindsay Orr Hilbert’s Tenth Problem

slide-63
SLIDE 63

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H1 = {k : U1(k, k, y1, . . . , yw) = 0 has a solution}

  • Claim. H1 is a diophantine set but Hc

1 is not.

  • Proof. (First part) Write D(k, y1, . . . , yw) = U1(k, k, y1, . . . , yw).

Then k ∈ H1 ⇐ ⇒ D(k, y1, . . . , yw) has a solution Thus H1 is diophantine.

John Lindsay Orr Hilbert’s Tenth Problem

slide-64
SLIDE 64

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H1 = {k : U1(k, k, y1, . . . , yw) = 0 has a solution}

  • Claim. H1 is a diophantine set but Hc

1 is not.

  • Proof. (First part) Write D(k, y1, . . . , yw) = U1(k, k, y1, . . . , yw).

Then k ∈ H1 ⇐ ⇒ D(k, y1, . . . , yw) has a solution Thus H1 is diophantine.

John Lindsay Orr Hilbert’s Tenth Problem

slide-65
SLIDE 65

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H1 = {k : U1(k, k, y1, . . . , yw) = 0 has a solution}

  • Claim. H1 is a diophantine set but Hc

1 is not.

  • Proof. (Second part) If Hc

1 were diophantine there would be a

code, k, for it. But then ask: Does U1(k, k, y1, . . . , yw) = 0 have a solution? If “yes” then k ∈ H1. But k is the code for the set Hc

1 so in

general: U1(k, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒ a ∈ Hc

1

Thus, k ∈ Hc

  • 1. Contradiction!

If “no” then k ∈ Hc

  • 1. But likewise a ∈ Hc
  • 1. Contradiction!

John Lindsay Orr Hilbert’s Tenth Problem

slide-66
SLIDE 66

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H1 = {k : U1(k, k, y1, . . . , yw) = 0 has a solution}

  • Claim. H1 is a diophantine set but Hc

1 is not.

  • Proof. (Second part) If Hc

1 were diophantine there would be a

code, k, for it. But then ask: Does U1(k, k, y1, . . . , yw) = 0 have a solution? If “yes” then k ∈ H1. But k is the code for the set Hc

1 so in

general: U1(k, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒ a ∈ Hc

1

Thus, k ∈ Hc

  • 1. Contradiction!

If “no” then k ∈ Hc

  • 1. But likewise a ∈ Hc
  • 1. Contradiction!

John Lindsay Orr Hilbert’s Tenth Problem

slide-67
SLIDE 67

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H1 = {k : U1(k, k, y1, . . . , yw) = 0 has a solution}

  • Claim. H1 is a diophantine set but Hc

1 is not.

  • Proof. (Second part) If Hc

1 were diophantine there would be a

code, k, for it. But then ask: Does U1(k, k, y1, . . . , yw) = 0 have a solution? If “yes” then k ∈ H1. But k is the code for the set Hc

1 so in

general: U1(k, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒ a ∈ Hc

1

Thus, k ∈ Hc

  • 1. Contradiction!

If “no” then k ∈ Hc

  • 1. But likewise a ∈ Hc
  • 1. Contradiction!

John Lindsay Orr Hilbert’s Tenth Problem

slide-68
SLIDE 68

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H1 = {k : U1(k, k, y1, . . . , yw) = 0 has a solution}

  • Claim. H1 is a diophantine set but Hc

1 is not.

  • Proof. (Second part) If Hc

1 were diophantine there would be a

code, k, for it. But then ask: Does U1(k, k, y1, . . . , yw) = 0 have a solution? If “yes” then k ∈ H1. But k is the code for the set Hc

1 so in

general: U1(k, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒ a ∈ Hc

1

Thus, k ∈ Hc

  • 1. Contradiction!

If “no” then k ∈ Hc

  • 1. But likewise a ∈ Hc
  • 1. Contradiction!

John Lindsay Orr Hilbert’s Tenth Problem

slide-69
SLIDE 69

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H1 = {k : U1(k, k, y1, . . . , yw) = 0 has a solution}

  • Claim. H1 is a diophantine set but Hc

1 is not.

  • Proof. (Second part) If Hc

1 were diophantine there would be a

code, k, for it. But then ask: Does U1(k, k, y1, . . . , yw) = 0 have a solution? If “yes” then k ∈ H1. But k is the code for the set Hc

1 so in

general: U1(k, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒ a ∈ Hc

1

Thus, k ∈ Hc

  • 1. Contradiction!

If “no” then k ∈ Hc

  • 1. But likewise a ∈ Hc
  • 1. Contradiction!

John Lindsay Orr Hilbert’s Tenth Problem

slide-70
SLIDE 70

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H1 = {k : U1(k, k, y1, . . . , yw) = 0 has a solution}

  • Claim. H1 is a diophantine set but Hc

1 is not.

  • Proof. (Second part) If Hc

1 were diophantine there would be a

code, k, for it. But then ask: Does U1(k, k, y1, . . . , yw) = 0 have a solution? If “yes” then k ∈ H1. But k is the code for the set Hc

1 so in

general: U1(k, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒ a ∈ Hc

1

Thus, k ∈ Hc

  • 1. Contradiction!

If “no” then k ∈ Hc

  • 1. But likewise a ∈ Hc
  • 1. Contradiction!

John Lindsay Orr Hilbert’s Tenth Problem

slide-71
SLIDE 71

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} U1(a, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒Dk(a, x1, . . . , xn) = 0 has a solution ⇐ ⇒W(x1, . . . , xn) = 0 has a solution ⇐ ⇒U0(c(a, k), y1, . . . , yv) = 0 has a solution Thus c(k, k) ∈ H0 ⇐ ⇒ k ∈ H1 c(k, k) ∈ Hc

0 ⇐

⇒ k ∈ Hc

1

  • Fact. c(a, k) is a diophantine polynomial ⇒ Hc

0 is not

diophantine!

John Lindsay Orr Hilbert’s Tenth Problem

slide-72
SLIDE 72

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} U1(a, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒Dk(a, x1, . . . , xn) = 0 has a solution ⇐ ⇒W(x1, . . . , xn) = 0 has a solution ⇐ ⇒U0(c(a, k), y1, . . . , yv) = 0 has a solution Thus c(k, k) ∈ H0 ⇐ ⇒ k ∈ H1 c(k, k) ∈ Hc

0 ⇐

⇒ k ∈ Hc

1

  • Fact. c(a, k) is a diophantine polynomial ⇒ Hc

0 is not

diophantine!

John Lindsay Orr Hilbert’s Tenth Problem

slide-73
SLIDE 73

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} U1(a, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒Dk(a, x1, . . . , xn) = 0 has a solution ⇐ ⇒W(x1, . . . , xn) = 0 has a solution ⇐ ⇒U0(c(a, k), y1, . . . , yv) = 0 has a solution Thus c(k, k) ∈ H0 ⇐ ⇒ k ∈ H1 c(k, k) ∈ Hc

0 ⇐

⇒ k ∈ Hc

1

  • Fact. c(a, k) is a diophantine polynomial ⇒ Hc

0 is not

diophantine!

John Lindsay Orr Hilbert’s Tenth Problem

slide-74
SLIDE 74

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} U1(a, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒Dk(a, x1, . . . , xn) = 0 has a solution ⇐ ⇒W(x1, . . . , xn) = 0 has a solution ⇐ ⇒U0(c(a, k), y1, . . . , yv) = 0 has a solution Thus c(k, k) ∈ H0 ⇐ ⇒ k ∈ H1 c(k, k) ∈ Hc

0 ⇐

⇒ k ∈ Hc

1

  • Fact. c(a, k) is a diophantine polynomial ⇒ Hc

0 is not

diophantine!

John Lindsay Orr Hilbert’s Tenth Problem

slide-75
SLIDE 75

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} U1(a, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒Dk(a, x1, . . . , xn) = 0 has a solution ⇐ ⇒W(x1, . . . , xn) = 0 has a solution ⇐ ⇒U0(c(a, k), y1, . . . , yv) = 0 has a solution Thus c(k, k) ∈ H0 ⇐ ⇒ k ∈ H1 c(k, k) ∈ Hc

0 ⇐

⇒ k ∈ Hc

1

  • Fact. c(a, k) is a diophantine polynomial ⇒ Hc

0 is not

diophantine!

John Lindsay Orr Hilbert’s Tenth Problem

slide-76
SLIDE 76

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} U1(a, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒Dk(a, x1, . . . , xn) = 0 has a solution ⇐ ⇒W(x1, . . . , xn) = 0 has a solution ⇐ ⇒U0(c(a, k), y1, . . . , yv) = 0 has a solution Thus c(k, k) ∈ H0 ⇐ ⇒ k ∈ H1 c(k, k) ∈ Hc

0 ⇐

⇒ k ∈ Hc

1

  • Fact. c(a, k) is a diophantine polynomial ⇒ Hc

0 is not

diophantine!

John Lindsay Orr Hilbert’s Tenth Problem

slide-77
SLIDE 77

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} U1(a, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒Dk(a, x1, . . . , xn) = 0 has a solution ⇐ ⇒W(x1, . . . , xn) = 0 has a solution ⇐ ⇒U0(c(a, k), y1, . . . , yv) = 0 has a solution Thus c(k, k) ∈ H0 ⇐ ⇒ k ∈ H1 c(k, k) ∈ Hc

0 ⇐

⇒ k ∈ Hc

1

  • Fact. c(a, k) is a diophantine polynomial ⇒ Hc

0 is not

diophantine!

John Lindsay Orr Hilbert’s Tenth Problem

slide-78
SLIDE 78

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Let H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} U1(a, k, y1, . . . , yw) = 0 has a solution ⇐ ⇒Dk(a, x1, . . . , xn) = 0 has a solution ⇐ ⇒W(x1, . . . , xn) = 0 has a solution ⇐ ⇒U0(c(a, k), y1, . . . , yv) = 0 has a solution Thus c(k, k) ∈ H0 ⇐ ⇒ k ∈ H1 c(k, k) ∈ Hc

0 ⇐

⇒ k ∈ Hc

1

  • Fact. c(a, k) is a diophantine polynomial ⇒ Hc

0 is not

diophantine!

John Lindsay Orr Hilbert’s Tenth Problem

slide-79
SLIDE 79

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Summary

We have seen that H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} is not Turing decidable. The elements of H0 are in one-to-one correspondence with the solvable diophantine equations. Thus, there is no algorithm to decide which diophantine equations are solvable and which are not.

John Lindsay Orr Hilbert’s Tenth Problem

slide-80
SLIDE 80

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Summary

We have seen that H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} is not Turing decidable. The elements of H0 are in one-to-one correspondence with the solvable diophantine equations. Thus, there is no algorithm to decide which diophantine equations are solvable and which are not.

John Lindsay Orr Hilbert’s Tenth Problem

slide-81
SLIDE 81

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Summary

We have seen that H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} is not Turing decidable. The elements of H0 are in one-to-one correspondence with the solvable diophantine equations. Thus, there is no algorithm to decide which diophantine equations are solvable and which are not.

John Lindsay Orr Hilbert’s Tenth Problem

slide-82
SLIDE 82

Introduction Sketch of Proof Going Into the Details Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

Summary

We have seen that H0 = {k : U0(k, y1, . . . , yv) = 0 has a solution} is not Turing decidable. The elements of H0 are in one-to-one correspondence with the solvable diophantine equations. Thus, there is no algorithm to decide which diophantine equations are solvable and which are not.

John Lindsay Orr Hilbert’s Tenth Problem

slide-83
SLIDE 83

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Outline

1

Introduction Disclaimer History and Statement of the Problem

2

Sketch of Proof Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

3

Going Into the Details Working with Diophantine Sets Coding n-tuples

John Lindsay Orr Hilbert’s Tenth Problem

slide-84
SLIDE 84

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Unions and Intersections

Let S1, S2 ⊆ Nk be diophantine sets with representations (a1, . . . , ak) ∈ S1 ⇐ ⇒ D1(a1, . . . , ak, x1, . . . , xm) = 0 has a solution and (a1, . . . , ak) ∈ S1 ⇐ ⇒ D2(a1, . . . , ak, y1, . . . , yn) = 0 has a solution Then S1 ∪ S2 and S1 ∩ S2 are diophantine sets.

  • Proof. Consider

D1(a1, . . . , ak, x1, . . . , xm)D2(a1, . . . , ak, y1, . . . , yn) = 0 and D1(a1, . . . , ak, x1, . . . , xm)2 + D2(a1, . . . , ak, y1, . . . , yn)2 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-85
SLIDE 85

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Unions and Intersections

Let S1, S2 ⊆ Nk be diophantine sets with representations (a1, . . . , ak) ∈ S1 ⇐ ⇒ D1(a1, . . . , ak, x1, . . . , xm) = 0 has a solution and (a1, . . . , ak) ∈ S1 ⇐ ⇒ D2(a1, . . . , ak, y1, . . . , yn) = 0 has a solution Then S1 ∪ S2 and S1 ∩ S2 are diophantine sets.

  • Proof. Consider

D1(a1, . . . , ak, x1, . . . , xm)D2(a1, . . . , ak, y1, . . . , yn) = 0 and D1(a1, . . . , ak, x1, . . . , xm)2 + D2(a1, . . . , ak, y1, . . . , yn)2 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-86
SLIDE 86

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Some Basic Diophantine Sets

The set {(a, b) : aRb} is diophantine when “R” is one of the relations: a = b (consider “∃x s.t. x + (a − b)2 = 0”) a < b (consider “∃x s.t. a + x + 1 = b”) a|b (consider “∃x s.t. ax = b”)

John Lindsay Orr Hilbert’s Tenth Problem

slide-87
SLIDE 87

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Some Basic Diophantine Sets

The set {(a, b) : aRb} is diophantine when “R” is one of the relations: a = b (consider “∃x s.t. x + (a − b)2 = 0”) a < b (consider “∃x s.t. a + x + 1 = b”) a|b (consider “∃x s.t. ax = b”)

John Lindsay Orr Hilbert’s Tenth Problem

slide-88
SLIDE 88

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Some Basic Diophantine Sets

The set {(a, b) : aRb} is diophantine when “R” is one of the relations: a = b (consider “∃x s.t. x + (a − b)2 = 0”) a < b (consider “∃x s.t. a + x + 1 = b”) a|b (consider “∃x s.t. ax = b”)

John Lindsay Orr Hilbert’s Tenth Problem

slide-89
SLIDE 89

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Some Basic Diophantine Sets

The set {(a, b) : aRb} is diophantine when “R” is one of the relations: a = b (consider “∃x s.t. x + (a − b)2 = 0”) a < b (consider “∃x s.t. a + x + 1 = b”) a|b (consider “∃x s.t. ax = b”)

John Lindsay Orr Hilbert’s Tenth Problem

slide-90
SLIDE 90

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Some Basic Diophantine Sets

The set {(a, b) : aRb} is diophantine when “R” is one of the relations: a = b (consider “∃x s.t. x + (a − b)2 = 0”) a < b (consider “∃x s.t. a + x + 1 = b”) a|b (consider “∃x s.t. ax = b”) The set {(a, b, c) : a = rem(b, c)} is diophantine. Proof. a = rem(b, c) ⇐ ⇒a < c & c|b − a ⇐ ⇒∃x, y s.t. (a + x + 1 − b)2 + (cy − (b − a))2 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-91
SLIDE 91

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Some Basic Diophantine Sets

The set {(a, b) : aRb} is diophantine when “R” is one of the relations: a = b (consider “∃x s.t. x + (a − b)2 = 0”) a < b (consider “∃x s.t. a + x + 1 = b”) a|b (consider “∃x s.t. ax = b”) The set {(a, b, c) : a = rem(b, c)} is diophantine. Proof. a = rem(b, c) ⇐ ⇒a < c & c|b − a ⇐ ⇒∃x, y s.t. (a + x + 1 − b)2 + (cy − (b − a))2 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-92
SLIDE 92

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Some Basic Diophantine Sets

The set {(a, b) : aRb} is diophantine when “R” is one of the relations: a = b (consider “∃x s.t. x + (a − b)2 = 0”) a < b (consider “∃x s.t. a + x + 1 = b”) a|b (consider “∃x s.t. ax = b”) The set {(a, b, c) : a = rem(b, c)} is diophantine. Proof. a = rem(b, c) ⇐ ⇒a < c & c|b − a ⇐ ⇒∃x, y s.t. (a + x + 1 − b)2 + (cy − (b − a))2 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-93
SLIDE 93

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Some Basic Diophantine Sets

The set {(a, b) : aRb} is diophantine when “R” is one of the relations: a = b (consider “∃x s.t. x + (a − b)2 = 0”) a < b (consider “∃x s.t. a + x + 1 = b”) a|b (consider “∃x s.t. ax = b”) The set {(a, b, c) : a = rem(b, c)} is diophantine. Proof. a = rem(b, c) ⇐ ⇒a < c & c|b − a ⇐ ⇒∃x, y s.t. (a + x + 1 − b)2 + (cy − (b − a))2 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-94
SLIDE 94

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

The set {(a, b, c) : a ≡ b (mod c)} is diophantine. Proof. a ≡ b (mod c) ⇐ ⇒ rem(a, c) = rem(b, c) ⇐ ⇒∃v, w s.t. v = rem(a, c) & w = rem(b, c) & w = v ⇐ ⇒∃v, w, x, y, x′, y′, z s.t. ((v + x + 1 − a)2 + (cy − (a − v))2)2 + ((w + x′ + 1 − b)2 + (cy′ − (b − w))2)2 + (z + (v − w)2)2 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-95
SLIDE 95

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

The set {(a, b, c) : a ≡ b (mod c)} is diophantine. Proof. a ≡ b (mod c) ⇐ ⇒ rem(a, c) = rem(b, c) ⇐ ⇒∃v, w s.t. v = rem(a, c) & w = rem(b, c) & w = v ⇐ ⇒∃v, w, x, y, x′, y′, z s.t. ((v + x + 1 − a)2 + (cy − (a − v))2)2 + ((w + x′ + 1 − b)2 + (cy′ − (b − w))2)2 + (z + (v − w)2)2 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-96
SLIDE 96

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

The set {(a, b, c) : a ≡ b (mod c)} is diophantine. Proof. a ≡ b (mod c) ⇐ ⇒ rem(a, c) = rem(b, c) ⇐ ⇒∃v, w s.t. v = rem(a, c) & w = rem(b, c) & w = v ⇐ ⇒∃v, w, x, y, x′, y′, z s.t. ((v + x + 1 − a)2 + (cy − (a − v))2)2 + ((w + x′ + 1 − b)2 + (cy′ − (b − w))2)2 + (z + (v − w)2)2 = 0

John Lindsay Orr Hilbert’s Tenth Problem

slide-97
SLIDE 97

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Exponentiation is Diophantine

Theorem (Matiyasevich, 1970) The set {(a, b, c) : a = bc} is diophantine. Corollary The set {(a, n) : a = n!} is diophantine. a is prime ⇐ ⇒ a > 1 & gcd(a, (a − 1)!) = 1

John Lindsay Orr Hilbert’s Tenth Problem

slide-98
SLIDE 98

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Exponentiation is Diophantine

Theorem (Matiyasevich, 1970) The set {(a, b, c) : a = bc} is diophantine. Corollary The set {(a, n) : a = n!} is diophantine. a is prime ⇐ ⇒ a > 1 & gcd(a, (a − 1)!) = 1

John Lindsay Orr Hilbert’s Tenth Problem

slide-99
SLIDE 99

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Exponentiation is Diophantine

Theorem (Matiyasevich, 1970) The set {(a, b, c) : a = bc} is diophantine. Corollary The set {(a, n) : a = n!} is diophantine. a is prime ⇐ ⇒ a > 1 & gcd(a, (a − 1)!) = 1

John Lindsay Orr Hilbert’s Tenth Problem

slide-100
SLIDE 100

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Outline

1

Introduction Disclaimer History and Statement of the Problem

2

Sketch of Proof Turing Machines and Decidability Diophantine Sets Universal Diophantine Equations

3

Going Into the Details Working with Diophantine Sets Coding n-tuples

John Lindsay Orr Hilbert’s Tenth Problem

slide-101
SLIDE 101

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Coding n-tuples

(a0, a1, . . . , an)  

  • a = a0 + a1b + a2b2 + · · ·
  • y

+ akbk

ebk

+ · · · + anbn

  • xbk+1

e = Elem(k, a, b) ⇐ ⇒ ∃x, y s.t. a = y + ebk + xbk+1 & e < b & y < bk

John Lindsay Orr Hilbert’s Tenth Problem

slide-102
SLIDE 102

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Coding n-tuples

(a0, a1, . . . , an)  

  • a = a0 + a1b + a2b2 + · · ·
  • y

+ akbk

ebk

+ · · · + anbn

  • xbk+1

e = Elem(k, a, b) ⇐ ⇒ ∃x, y s.t. a = y + ebk + xbk+1 & e < b & y < bk

John Lindsay Orr Hilbert’s Tenth Problem

slide-103
SLIDE 103

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Primes

(b + 1)n = n

  • +

n 1

  • b + · · · +

n k

  • bk + · · · +

n n

  • bn

a = n k

⇒ a = Elem(k, (b + 1)n, b) & b = 2n a is prime ⇐ ⇒ a > 1 & gcd(a, (a − 1)!) = 1

John Lindsay Orr Hilbert’s Tenth Problem

slide-104
SLIDE 104

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Primes

(b + 1)n = n

  • +

n 1

  • b + · · · +

n k

  • bk + · · · +

n n

  • bn

a = n k

⇒ a = Elem(k, (b + 1)n, b) & b = 2n a is prime ⇐ ⇒ a > 1 & gcd(a, (a − 1)!) = 1

John Lindsay Orr Hilbert’s Tenth Problem

slide-105
SLIDE 105

Introduction Sketch of Proof Going Into the Details Working with Diophantine Sets Coding n-tuples

Primes

(b + 1)n = n

  • +

n 1

  • b + · · · +

n k

  • bk + · · · +

n n

  • bn

a = n k

⇒ a = Elem(k, (b + 1)n, b) & b = 2n a is prime ⇐ ⇒ a > 1 & gcd(a, (a − 1)!) = 1

John Lindsay Orr Hilbert’s Tenth Problem