Factorization Methods Bernd Schr oder Bernd Schr oder Louisiana - - PowerPoint PPT Presentation

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factorization Methods

Bernd Schr¨

• der

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Introduction

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Introduction

• 1. The prime factorization of integers has many applications.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Introduction

• 1. The prime factorization of integers has many applications.
• 2. But how do we find it?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Introduction

• 1. The prime factorization of integers has many applications.
• 2. But how do we find it?
• 3. How do we find it in acceptable time?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Introduction

• 1. The prime factorization of integers has many applications.
• 2. But how do we find it?
• 3. How do we find it in acceptable time?
• 4. Because the ability to factor numbers connects to the

ability to find large prime numbers (needed for internet security), as well as to the ability to crack internet encryptions (by factoring the numbers used in the code), there is a lot of interest in this question.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Introduction

• 1. The prime factorization of integers has many applications.
• 2. But how do we find it?
• 3. How do we find it in acceptable time?
• 4. Because the ability to factor numbers connects to the

ability to find large prime numbers (needed for internet security), as well as to the ability to crack internet encryptions (by factoring the numbers used in the code), there is a lot of interest in this question.

• 5. So, for the latest, it is best to check the current literature (if

it is accessible).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Introduction

• 1. The prime factorization of integers has many applications.
• 2. But how do we find it?
• 3. How do we find it in acceptable time?
• 4. Because the ability to factor numbers connects to the

ability to find large prime numbers (needed for internet security), as well as to the ability to crack internet encryptions (by factoring the numbers used in the code), there is a lot of interest in this question.

• 5. So, for the latest, it is best to check the current literature (if

it is accessible).

• 6. This first presentation is only the introduction of a

recurrent theme.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factor the number 91

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factor the number 91

• 1. 2 ∤ 91

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factor the number 91

• 1. 2 ∤ 91, 3 ∤ 91

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factor the number 91

• 1. 2 ∤ 91, 3 ∤ 91, 5 ∤ 91

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factor the number 91

• 1. 2 ∤ 91, 3 ∤ 91, 5 ∤ 91, 7 | 91!

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factor the number 91

• 1. 2 ∤ 91, 3 ∤ 91, 5 ∤ 91, 7 | 91!
• 2. So 91 = 7·x and x = 91

7 = 13.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factor the number 91

• 1. 2 ∤ 91, 3 ∤ 91, 5 ∤ 91, 7 | 91!
• 2. So 91 = 7·x and x = 91

7 = 13.

• 3. So 91 = 7·13.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factor the number 91

• 1. 2 ∤ 91, 3 ∤ 91, 5 ∤ 91, 7 | 91!
• 2. So 91 = 7·x and x = 91

7 = 13.

• 3. So 91 = 7·13.

That does not seem so bad.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factor the number 91

• 1. 2 ∤ 91, 3 ∤ 91, 5 ∤ 91, 7 | 91!
• 2. So 91 = 7·x and x = 91

7 = 13.

• 3. So 91 = 7·13.

That does not seem so bad. Trying it with 53,198,462,357

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factor the number 91

• 1. 2 ∤ 91, 3 ∤ 91, 5 ∤ 91, 7 | 91!
• 2. So 91 = 7·x and x = 91

7 = 13.

• 3. So 91 = 7·13.

That does not seem so bad. Trying it with 53,198,462,357 shows that it actually is quite bad.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Factor the number 91

• 1. 2 ∤ 91, 3 ∤ 91, 5 ∤ 91, 7 | 91!
• 2. So 91 = 7·x and x = 91

7 = 13.

• 3. So 91 = 7·13.

That does not seem so bad. Trying it with 53,198,462,357 shows that it actually is quite bad. 53,198,462,357 = 11,113,111·4,787.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Lemma.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let n be an odd positive integer. There is a one-to-one

correspondence between the factorizations of n into two positive integers and differences of two squares that equal n.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let n be an odd positive integer. There is a one-to-one

correspondence between the factorizations of n into two positive integers and differences of two squares that equal n. Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let n be an odd positive integer. There is a one-to-one

correspondence between the factorizations of n into two positive integers and differences of two squares that equal n.

• Proof. To write a factorization of n as ab = n

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let n be an odd positive integer. There is a one-to-one

correspondence between the factorizations of n into two positive integers and differences of two squares that equal n.

• Proof. To write a factorization of n as ab = n = t2 −s2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let n be an odd positive integer. There is a one-to-one

correspondence between the factorizations of n into two positive integers and differences of two squares that equal n.

• Proof. To write a factorization of n as ab = n = t2 −s2, use

t = a+b

2 , s = a−b 2 .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let n be an odd positive integer. There is a one-to-one

correspondence between the factorizations of n into two positive integers and differences of two squares that equal n.

• Proof. To write a factorization of n as ab = n = t2 −s2, use

t = a+b

2 , s = a−b 2 . The one-to-one correspondence (injective

function) is that we consider the factorization itself as input (a,b) and f(a,b) = (t,s)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let n be an odd positive integer. There is a one-to-one

correspondence between the factorizations of n into two positive integers and differences of two squares that equal n.

• Proof. To write a factorization of n as ab = n = t2 −s2, use

t = a+b

2 , s = a−b 2 . The one-to-one correspondence (injective

function) is that we consider the factorization itself as input (a,b) and f(a,b) = (t,s) = a+b

2 , a−b 2

• .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let n be an odd positive integer. There is a one-to-one

correspondence between the factorizations of n into two positive integers and differences of two squares that equal n.

• Proof. To write a factorization of n as ab = n = t2 −s2, use

t = a+b

2 , s = a−b 2 . The one-to-one correspondence (injective

function) is that we consider the factorization itself as input (a,b) and f(a,b) = (t,s) = a+b

2 , a−b 2

• .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Implementation

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Implementation

• 1. Take t to be the smallest integer greater than √n.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Implementation

• 1. Take t to be the smallest integer greater than √n.
• 2. Look for perfect squares s2 in the sequence t2 −n,

(t +1)2 −n, (t +2)2 −n, ....

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Implementation

• 1. Take t to be the smallest integer greater than √n.
• 2. Look for perfect squares s2 in the sequence t2 −n,

(t +1)2 −n, (t +2)2 −n, .... Then s2 = (t +k)2 −n

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Implementation

• 1. Take t to be the smallest integer greater than √n.
• 2. Look for perfect squares s2 in the sequence t2 −n,

(t +1)2 −n, (t +2)2 −n, .... Then s2 = (t +k)2 −n and n = (t +k)2 −s2 = (t +k +s)(t +k −s)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Implementation

• 1. Take t to be the smallest integer greater than √n.
• 2. Look for perfect squares s2 in the sequence t2 −n,

(t +1)2 −n, (t +2)2 −n, .... Then s2 = (t +k)2 −n and n = (t +k)2 −s2 = (t +k +s)(t +k −s)

• 3. Because n =

n+1

2

2 − n−1

2

2 we will find a number that works in at most n

2 −√n steps.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Implementation

• 1. Take t to be the smallest integer greater than √n.
• 2. Look for perfect squares s2 in the sequence t2 −n,

(t +1)2 −n, (t +2)2 −n, .... Then s2 = (t +k)2 −n and n = (t +k)2 −s2 = (t +k +s)(t +k −s)

• 3. Because n =

n+1

2

2 − n−1

2

2 we will find a number that works in at most n

2 −√n steps.

• 4. But we cannot jump directly to the end

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Implementation

• 1. Take t to be the smallest integer greater than √n.
• 2. Look for perfect squares s2 in the sequence t2 −n,

(t +1)2 −n, (t +2)2 −n, .... Then s2 = (t +k)2 −n and n = (t +k)2 −s2 = (t +k +s)(t +k −s)

• 3. Because n =

n+1

2

2 − n−1

2

2 we will find a number that works in at most n

2 −√n steps.

• 4. But we cannot jump directly to the end, because using the

above difference only leads to the factorization n = n·1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Example.

◮ Factoring 1,363 takes one step with this method.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Example.

◮ Factoring 1,363 takes one step with this method. ◮ Factoring 1,463 takes 10 steps.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Example.

◮ Factoring 1,363 takes one step with this method. ◮ Factoring 1,463 takes 10 steps. ◮ Unfortunately, for products of factors that are not

approximately equal, Fermat factorization can take a lot longer:

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Example.

◮ Factoring 1,363 takes one step with this method. ◮ Factoring 1,463 takes 10 steps. ◮ Unfortunately, for products of factors that are not

approximately equal, Fermat factorization can take a lot longer: 1,461 is divisible by 3 and Fermat factorization does not yield a factorization in 100 steps.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Example.

◮ Factoring 1,363 takes one step with this method. ◮ Factoring 1,463 takes 10 steps. ◮ Unfortunately, for products of factors that are not

approximately equal, Fermat factorization can take a lot longer: 1,461 is divisible by 3 and Fermat factorization does not yield a factorization in 100 steps.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1 = 24 +54

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1 = 24 +54 225 +1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1 = 24 +54 225 +1 = 232 +1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1 = 24 +54 225 +1 = 232 +1 = 24 ·228 +1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1 = 24 +54 225 +1 = 232 +1 = 24 ·228 +1 =

• 641−54

228 +1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1 = 24 +54 225 +1 = 232 +1 = 24 ·228 +1 =

• 641−54

228 +1 = 641·228 −

• 5·274 +1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1 = 24 +54 225 +1 = 232 +1 = 24 ·228 +1 =

• 641−54

228 +1 = 641·228 −

• 5·274 +1 = 641·228 −(641−1)4 +1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1 = 24 +54 225 +1 = 232 +1 = 24 ·228 +1 =

• 641−54

228 +1 = 641·228 −

• 5·274 +1 = 641·228 −(641−1)4 +1

= 641·

• 228 −6413 +4·6412 −6·641+4
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1 = 24 +54 225 +1 = 232 +1 = 24 ·228 +1 =

• 641−54

228 +1 = 641·228 −

• 5·274 +1 = 641·228 −(641−1)4 +1

= 641·

• 228 −6413 +4·6412 −6·641+4
• So 641|225 +1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1 = 24 +54 225 +1 = 232 +1 = 24 ·228 +1 =

• 641−54

228 +1 = 641·228 −

• 5·274 +1 = 641·228 −(641−1)4 +1

= 641·

• 228 −6413 +4·6412 −6·641+4
• So 641|225 +1. Now do the division

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Numbers of the form 22n +1 are called Fermat numbers. Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect.

• Example. 225 +1 = 641·6,700,417 (Euler, 1732).

Proof. 641 = 5·27 +1 = 24 +54 225 +1 = 232 +1 = 24 ·228 +1 =

• 641−54

228 +1 = 641·228 −

• 5·274 +1 = 641·228 −(641−1)4 +1

= 641·

• 228 −6413 +4·6412 −6·641+4
• So 641|225 +1. Now do the division

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1 Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1

• Example. 223 +1 is prime, because

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1

• Example. 223 +1 is prime, because 24k +1 = 16k +1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1

• Example. 223 +1 is prime, because 24k +1 = 16k +1 and

223 +1 = 257, which is not divisible by 17

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1

• Example. 223 +1 is prime, because 24k +1 = 16k +1 and

223 +1 = 257, which is not divisible by 17 and all other numbers of the form 16k+1 are too large to be factors

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1

• Example. 223 +1 is prime, because 24k +1 = 16k +1 and

223 +1 = 257, which is not divisible by 17 and all other numbers of the form 16k+1 are too large to be factors, actually, 17 is, too.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1

• Example. 223 +1 is prime, because 24k +1 = 16k +1 and

223 +1 = 257, which is not divisible by 17 and all other numbers of the form 16k+1 are too large to be factors, actually, 17 is, too.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1

• Example. 223 +1 is prime, because 24k +1 = 16k +1 and

223 +1 = 257, which is not divisible by 17 and all other numbers of the form 16k+1 are too large to be factors, actually, 17 is, too. Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1

• Example. 223 +1 is prime, because 24k +1 = 16k +1 and

223 +1 = 257, which is not divisible by 17 and all other numbers of the form 16k+1 are too large to be factors, actually, 17 is, too.

• Example. 226 +1 is not prime

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1

• Example. 223 +1 is prime, because 24k +1 = 16k +1 and

223 +1 = 257, which is not divisible by 17 and all other numbers of the form 16k+1 are too large to be factors, actually, 17 is, too.

• Example. 226 +1 is not prime, but it takes a while until a factor

256·1071+1 is found.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 22n +1 is of the form 2n+1k +1

• Example. 223 +1 is prime, because 24k +1 = 16k +1 and

223 +1 = 257, which is not divisible by 17 and all other numbers of the form 16k+1 are too large to be factors, actually, 17 is, too.

• Example. 226 +1 is not prime, but it takes a while until a factor

256·1071+1 is found.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Facts About Fermat Numbers

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Facts About Fermat Numbers

• 1. Only the first four Fermat numbers are known to be prime:

221 +1 = 5, 222 +1 = 17, 223 +1 = 257, 224 +1 = 65,537.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Facts About Fermat Numbers

• 1. Only the first four Fermat numbers are known to be prime:

221 +1 = 5, 222 +1 = 17, 223 +1 = 257, 224 +1 = 65,537.

• 2. Another 243 are known to be composite.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Facts About Fermat Numbers

• 1. Only the first four Fermat numbers are known to be prime:

221 +1 = 5, 222 +1 = 17, 223 +1 = 257, 224 +1 = 65,537.

• 2. Another 243 are known to be composite.
• 3. So now there also is a conjecture that only the first four

Fermat numbers are prime.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Facts About Fermat Numbers

• 1. Only the first four Fermat numbers are known to be prime:

221 +1 = 5, 222 +1 = 17, 223 +1 = 257, 224 +1 = 65,537.

• 2. Another 243 are known to be composite.
• 3. So now there also is a conjecture that only the first four

Fermat numbers are prime.

• 4. The research in this direction certainly tests the limits of

what is possible: There are only 7 composite Fermat numbers for which we know the complete factorization.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Facts About Fermat Numbers

• 1. Only the first four Fermat numbers are known to be prime:

221 +1 = 5, 222 +1 = 17, 223 +1 = 257, 224 +1 = 65,537.

• 2. Another 243 are known to be composite.
• 3. So now there also is a conjecture that only the first four

Fermat numbers are prime.

• 4. The research in this direction certainly tests the limits of

what is possible: There are only 7 composite Fermat numbers for which we know the complete factorization. (Think about it: 2212 +1 = 24096 +1 > 101365.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Lemma.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2. Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5 = 15

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5 = 15 = 17−2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5 = 15 = 17−2 = F2 −2.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5 = 15 = 17−2 = F2 −2. Induction Step, n → n+1. F0F1F2 ···Fn

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5 = 15 = 17−2 = F2 −2. Induction Step, n → n+1. F0F1F2 ···Fn = (F0F1F2 ···Fn−1)Fn

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5 = 15 = 17−2 = F2 −2. Induction Step, n → n+1. F0F1F2 ···Fn = (F0F1F2 ···Fn−1)Fn = (Fn −2)Fn

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5 = 15 = 17−2 = F2 −2. Induction Step, n → n+1. F0F1F2 ···Fn = (F0F1F2 ···Fn−1)Fn = (Fn −2)Fn =

• 22n −1
• 22n +1
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5 = 15 = 17−2 = F2 −2. Induction Step, n → n+1. F0F1F2 ···Fn = (F0F1F2 ···Fn−1)Fn = (Fn −2)Fn =

• 22n −1
• 22n +1
• =
• 22n+2n −1
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5 = 15 = 17−2 = F2 −2. Induction Step, n → n+1. F0F1F2 ···Fn = (F0F1F2 ···Fn−1)Fn = (Fn −2)Fn =

• 22n −1
• 22n +1
• =
• 22n+2n −1
• =
• 22n+1 +1−2
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5 = 15 = 17−2 = F2 −2. Induction Step, n → n+1. F0F1F2 ···Fn = (F0F1F2 ···Fn−1)Fn = (Fn −2)Fn =

• 22n −1
• 22n +1
• =
• 22n+2n −1
• =
• 22n+1 +1−2
• = Fn+1 −2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Lemma. Let Fk := 22k +1. Then for all positive integers n we

have F0F1F2 ···Fn−1 = Fn −2.

• Proof. Induction on n.

Base Step, n = 2. F0F1 = 3·5 = 15 = 17−2 = F2 −2. Induction Step, n → n+1. F0F1F2 ···Fn = (F0F1F2 ···Fn−1)Fn = (Fn −2)Fn =

• 22n −1
• 22n +1
• =
• 22n+2n −1
• =
• 22n+1 +1−2
• = Fn+1 −2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Theorem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime. Consequently, there must be infinitely many prime numbers.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime. Consequently, there must be infinitely many prime numbers. Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime. Consequently, there must be infinitely many prime numbers.

• Proof. Without loss of generality, let m < n

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime. Consequently, there must be infinitely many prime numbers.

• Proof. Without loss of generality, let m < n and suppose for a

contradiction that d|Fn and d|Fm. Then d|Fn −F0F1F2 ···Fn−1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime. Consequently, there must be infinitely many prime numbers.

• Proof. Without loss of generality, let m < n and suppose for a

contradiction that d|Fn and d|Fm. Then d|Fn −F0F1F2 ···Fn−1 = 2, a contradiction.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime. Consequently, there must be infinitely many prime numbers.

• Proof. Without loss of generality, let m < n and suppose for a

contradiction that d|Fn and d|Fm. Then d|Fn −F0F1F2 ···Fn−1 = 2, a contradiction. So now, if there were only finitely many prime numbers

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime. Consequently, there must be infinitely many prime numbers.

• Proof. Without loss of generality, let m < n and suppose for a

contradiction that d|Fn and d|Fm. Then d|Fn −F0F1F2 ···Fn−1 = 2, a contradiction. So now, if there were only finitely many prime numbers, say, N

• f them

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime. Consequently, there must be infinitely many prime numbers.

• Proof. Without loss of generality, let m < n and suppose for a

contradiction that d|Fn and d|Fm. Then d|Fn −F0F1F2 ···Fn−1 = 2, a contradiction. So now, if there were only finitely many prime numbers, say, N

• f them, then two of F0,...,FN would not be relatively prime

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime. Consequently, there must be infinitely many prime numbers.

• Proof. Without loss of generality, let m < n and suppose for a

contradiction that d|Fn and d|Fm. Then d|Fn −F0F1F2 ···Fn−1 = 2, a contradiction. So now, if there were only finitely many prime numbers, say, N

• f them, then two of F0,...,FN would not be relatively prime,

because each Fk has a prime factor that does not occur in any of the factorizations of the earlier Fermat numbers

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime. Consequently, there must be infinitely many prime numbers.

• Proof. Without loss of generality, let m < n and suppose for a

contradiction that d|Fn and d|Fm. Then d|Fn −F0F1F2 ···Fn−1 = 2, a contradiction. So now, if there were only finitely many prime numbers, say, N

• f them, then two of F0,...,FN would not be relatively prime,

because each Fk has a prime factor that does not occur in any of the factorizations of the earlier Fermat numbers (and F0 itself is prime, too).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. For distinct nonnegative integers n and m, the

Fermat numbers Fn and Fm are relatively prime. Consequently, there must be infinitely many prime numbers.

• Proof. Without loss of generality, let m < n and suppose for a

contradiction that d|Fn and d|Fm. Then d|Fn −F0F1F2 ···Fn−1 = 2, a contradiction. So now, if there were only finitely many prime numbers, say, N

• f them, then two of F0,...,FN would not be relatively prime,

because each Fk has a prime factor that does not occur in any of the factorizations of the earlier Fermat numbers (and F0 itself is prime, too).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Theorem (without proof).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Theorem (without proof). A regular polygon with n sides can be constructed with straightedge and compass iff n is the product of a nonnegative power of 2 and a nonnegative number

• f distinct Fermat primes.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Diophantine Equations

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Diophantine Equations

• 1. An equation is a diophantine equation if we only allow

integral solutions.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Diophantine Equations

• 1. An equation is a diophantine equation if we only allow

integral solutions.

• 2. The diophantine equation ax+by = c asks for all points on

this straight line whose coordinates are integers.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Diophantine Equations

• 1. An equation is a diophantine equation if we only allow

integral solutions.

• 2. The diophantine equation ax+by = c asks for all points on

this straight line whose coordinates are integers.

• 3. Technically, diophantine equations are solved any time we

convert currency (but of course rounding helps if the equation does not work out).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Diophantine Equations

• 1. An equation is a diophantine equation if we only allow

integral solutions.

• 2. The diophantine equation ax+by = c asks for all points on

this straight line whose coordinates are integers.

• 3. Technically, diophantine equations are solved any time we

convert currency (but of course rounding helps if the equation does not work out). For example, if 1000 US Dollars buy 710 Euros (about the exchange rate as these slides are made) and the bank only has 20 and 50 Euro bills, how can the 710 Euros be paid out?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Diophantine Equations

• 1. An equation is a diophantine equation if we only allow

integral solutions.

• 2. The diophantine equation ax+by = c asks for all points on

this straight line whose coordinates are integers.

• 3. Technically, diophantine equations are solved any time we

convert currency (but of course rounding helps if the equation does not work out). For example, if 1000 US Dollars buy 710 Euros (about the exchange rate as these slides are made) and the bank only has 20 and 50 Euro bills, how can the 710 Euros be paid out?

• 4. Could 715 Euros be paid out?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Theorem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions. Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

• Proof. “⇒.”

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

• Proof. “⇒.” When ax+by = c has an integral solution

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

• Proof. “⇒.” When ax+by = c has an integral solution, then,

by earlier result, c is a multiple of (a,b).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

• Proof. “⇒.” When ax+by = c has an integral solution, then,

by earlier result, c is a multiple of (a,b). “⇐.”

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

• Proof. “⇒.” When ax+by = c has an integral solution, then,

by earlier result, c is a multiple of (a,b). “⇐.” If c = z·(a,b), find s and t so that sa+tb = (a,b)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

• Proof. “⇒.” When ax+by = c has an integral solution, then,

by earlier result, c is a multiple of (a,b). “⇐.” If c = z·(a,b), find s and t so that sa+tb = (a,b) and use x := zs and y = zt.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

• Proof. “⇒.” When ax+by = c has an integral solution, then,

by earlier result, c is a multiple of (a,b). “⇐.” If c = z·(a,b), find s and t so that sa+tb = (a,b) and use x := zs and y = zt. Regarding the infinitely many solutions, note that, in the real numbers, the solutions are on the line ax+by = c, which goes through the point (zs,zt) and which has slope −a

b.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

• Proof. “⇒.” When ax+by = c has an integral solution, then,

by earlier result, c is a multiple of (a,b). “⇐.” If c = z·(a,b), find s and t so that sa+tb = (a,b) and use x := zs and y = zt. Regarding the infinitely many solutions, note that, in the real numbers, the solutions are on the line ax+by = c, which goes through the point (zs,zt) and which has slope −a

• b. So all

numbers of the form (zs+kb,zt −ka) with k ∈ Z are solutions, too.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

• Proof. “⇒.” When ax+by = c has an integral solution, then,

by earlier result, c is a multiple of (a,b). “⇐.” If c = z·(a,b), find s and t so that sa+tb = (a,b) and use x := zs and y = zt. Regarding the infinitely many solutions, note that, in the real numbers, the solutions are on the line ax+by = c, which goes through the point (zs,zt) and which has slope −a

• b. So all

numbers of the form (zs+kb,zt −ka) with k ∈ Z are solutions,

• too. (Note that the geometry is not needed for this last part, but

it helps with the idea.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

• Proof. “⇒.” When ax+by = c has an integral solution, then,

by earlier result, c is a multiple of (a,b). “⇐.” If c = z·(a,b), find s and t so that sa+tb = (a,b) and use x := zs and y = zt. Regarding the infinitely many solutions, note that, in the real numbers, the solutions are on the line ax+by = c, which goes through the point (zs,zt) and which has slope −a

• b. So all

numbers of the form (zs+kb,zt −ka) with k ∈ Z are solutions,

• too. (Note that the geometry is not needed for this last part, but

it helps with the idea. Draw the picture, if it does not “materialize” in your mind.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a,b,c be nonzero integers. The equation

ax+by = c has an integral solution (x,y) iff (a,b)|c. In this case, the equation has infinitely many solutions.

• Proof. “⇒.” When ax+by = c has an integral solution, then,

by earlier result, c is a multiple of (a,b). “⇐.” If c = z·(a,b), find s and t so that sa+tb = (a,b) and use x := zs and y = zt. Regarding the infinitely many solutions, note that, in the real numbers, the solutions are on the line ax+by = c, which goes through the point (zs,zt) and which has slope −a

• b. So all

numbers of the form (zs+kb,zt −ka) with k ∈ Z are solutions,

• too. (Note that the geometry is not needed for this last part, but

it helps with the idea. Draw the picture, if it does not “materialize” in your mind.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

Theorem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions. Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an) = (a1,...,an−2,(an−1,an))

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an) = (a1,...,an−2,(an−1,an)) = (a1,...,an−2,sn−1an−1 +snan)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an) = (a1,...,an−2,(an−1,an)) = (a1,...,an−2,sn−1an−1 +snan) =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an) = (a1,...,an−2,(an−1,an)) = (a1,...,an−2,sn−1an−1 +snan) = ···

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an) = (a1,...,an−2,(an−1,an)) = (a1,...,an−2,sn−1an−1 +snan) = ··· = s1a1 +···+sn−1an−1 +snan

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an) = (a1,...,an−2,(an−1,an)) = (a1,...,an−2,sn−1an−1 +snan) = ··· = s1a1 +···+sn−1an−1 +snan and that any integer s1a1 +···+sn−1an−1 +snan is a multiple of (a1,...,an)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an) = (a1,...,an−2,(an−1,an)) = (a1,...,an−2,sn−1an−1 +snan) = ··· = s1a1 +···+sn−1an−1 +snan and that any integer s1a1 +···+sn−1an−1 +snan is a multiple of (a1,...,an) (induction proof

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an) = (a1,...,an−2,(an−1,an)) = (a1,...,an−2,sn−1an−1 +snan) = ··· = s1a1 +···+sn−1an−1 +snan and that any integer s1a1 +···+sn−1an−1 +snan is a multiple of (a1,...,an) (induction proof, do it).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an) = (a1,...,an−2,(an−1,an)) = (a1,...,an−2,sn−1an−1 +snan) = ··· = s1a1 +···+sn−1an−1 +snan and that any integer s1a1 +···+sn−1an−1 +snan is a multiple of (a1,...,an) (induction proof, do it). Now mimic the proof of the preceding result

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an) = (a1,...,an−2,(an−1,an)) = (a1,...,an−2,sn−1an−1 +snan) = ··· = s1a1 +···+sn−1an−1 +snan and that any integer s1a1 +···+sn−1an−1 +snan is a multiple of (a1,...,an) (induction proof, do it). Now mimic the proof of the preceding result (good exercise, too).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations

• Theorem. Let a1,...,an be nonzero integers. The equation

a1x1 +···+anxn = c has an integral solution (x1,...,xn) iff (a1,...,an)|c. In this case, the equation has infinitely many solutions.

• Proof. First note that

(a1,...,an) = (a1,...,an−2,(an−1,an)) = (a1,...,an−2,sn−1an−1 +snan) = ··· = s1a1 +···+sn−1an−1 +snan and that any integer s1a1 +···+sn−1an−1 +snan is a multiple of (a1,...,an) (induction proof, do it). Now mimic the proof of the preceding result (good exercise, too).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Factorization Methods