Why the course?

Introduction to Number Theory

The make-up of the seminar

Interdisciplinary collaboration

Introduction

1

Enrichment

The needs of an advanced student

Further benefits

What’s in it for the Student

2

Student ownership of the course

Student-teaching-students

Team oriented

Presentations, discussions, assignments,

mini-projects

Seminar Structure

3

Technology

◦Excel

◦C++ programming

An example: Application of the

Euclidean Algorithm

Hands-on and Collaboration

4

Encryption/Decryption Examples

4.2

To illustrate the encoding and decoding procedure with the primes p = 12554 and q = 13007. We multiply them together to get the modulus m = pq = 163276871, and record for future use:

  (m) = (p – 1) (q – 1) = 163251312  Choose a k that is relative prime to (m) so take

k = 79921. 

Encryption:

4.3

In summary,

p=12553, q = 13006, m = pq = 163276871, and k = 79921.

Now to send the message “To be or not to be”, set the cipher shift A = 11, B = 12, and so on, and this message becomes the string of digits:

 30251215252824253030251215

 

4.4

The number m is 9 digits long, so we break the message up in to 8 digit numbers:

 30251215, 25282425, 30302512, 15

  Next we use the method of successive squaring to

raise each of these numbers to the nth power modulo m.

 3025121579921 149419241 (mod 163276871)

2528242579921 62721998 (mod 163276871)

3030251279921 118084566 (mod 163276871)

1579921 40481382 (mod 163276871)

4.5

The encoded message is the list of numbers:

 148419241, 62721998, 118084566, 40481382

 

4.6

Now let’s try decoding a new message. It’s midnight, there’s a knock at our door, and a mysterious messenger delivers the following cryptic missive:

 145387828, 47164891,

152020614, 27279275, 35356191

Use the kth modulo m computing method to solve the congruences.

Decryption:

4.7

 x78821 = 145387828 (mod 163276871)

x = 30182523

x78821 = 47164891 (mod 163276871) x = 26292524

x78821 = 152020614 (mod 163276871)x = 19291924

x78821 = 27279275 (mod 163276871) x = 30282531

x78821 = 35356191 (mod 163276871) x = 122215

4.8

This gives you the string of digits: 301525232629252419291924302825311122215

And now you use the number to letter substitution table for the final decoding step.

  THOMPSONISINTROUBLE 

4.9

Supplying the obvious word breaks and punctuation, you read:

  “Thompson is in trouble”  And off you go to the rescue

4.10

Increases student involvement

Learning by exploration

Getting back to formal mathematics

Broadening the student’s mathematical

horizon

Benefits

5

The Idea Behind the Proof: p2 – 1 = (p + 1) (p – 1) p + 1 and p – 1 are consecutive even

numbers (p + 1) (p – 1) = 2(i) * 2 (i + 1) i is either even or odd integer

Theorem

6.1

Suppose p > 3 is a prime number, Then 24 l p2 – 1

All prime numbers > 3 are odd, so p + 1 and p – 1 are consecutive even numbers and can be written as: (p – 1) (p + 1) = 2 (i) * 2 (i + 1)

p-1, p and p+1 are three consecutive integers. One of three consecutive integers must be divisible by 3.

p is not divisible by 3, since p is a prime

number, therefore either p + 1, or p – 1 is divisible by 3.

6.2

Case 1: (p + 1) (p – 1) = 2(i) * 2 (i + 1)

If i is even, let i = 2j, j = integer, then(p – 1)(p + 1) = 2(2j) *2((2j) + 1) = 8j(2j + 1)

Either j or 2j+ 1 is divisible by 3

If j is divisible by 3, then j = 3k p2 – 1= 8j(2j + 1) = 8(3k)(2j + 1) = 24(k(2j + 1))

If 2j + 1 is divisible by 3, then 2j + 1 = 3kp2 – 1= 8j(2j + 1) = 8(j)(3k) = 24(jk)

Thus, if i is even then p2 – 1 is divisible by 24.

6.3

Case 2: (p + 1) (p – 1) = 2(i) * 2 (i + 1)

If i is odd, let i = 2j + 1, j = integer, then(p – 1) (p + 1) = 2(2j + 1) *2((2j + 1) + 1) = 2(2j + 1) *4(j + 1) = 8(2j + 1)(j + 1)

Either 2j + 1 or j + 1 is divisible by 3

If 2j + 1 is divisible by 3, then 2j + 1 = 3k p2 – 1= 8(3k)(2j + 1) = 24(k(2j + 1)) If j + 1 is divisible by 3, then j + 1 = 3k

p2 – 1= 8(2j + 1)(3k) = 24(2j + 1)(k)

Thus, if i is odd then p2 – 1 is divisible by 24.

6.4

Summary:

Case 1: ◦p2 – 1= 24(k(2j + 1))or◦p2 – 1= 24(jk)

Case 2: ◦p2 – 1= 24(k(2j + 1)) or ◦p2 – 1= 24(2j + 1)(k)

Therefore,

24 l p2 – 1 6.5

Topics are interchangeable

Disciplines are interchangeable

Modular Nature of the Seminar

7

Questions?