Introduction to Cryptography

Brian Veitch July 2, 2013

Contents

1 Introduction 3 1.1 Alice, Bob, and Eve ...... 3 1.2 Basic Terminology ...... 4 1.3 Brief History of Cryptography ...... 4 1.4 Kerckhoff’s Principles of Cryptographic Secuity ...... 11

2 Substitution Cipher 13 2.1 Caesar Cipher Activity ...... 13 2.2 Keyword Cipher Activity ...... 17 2.3 Vigenere Cipher Activity ...... 20

3 Transposition 22 3.1 Easy Example ...... 22 3.2 Railfence Cipher ...... 22 3.3 Transposition Activity ...... 23

4 Modular Arithmetic 24 4.1 Examples of Congruences ...... 26 4.2 Arithmetic with Congruences ...... 27 4.3 Multipication Tables ...... 27 4.4 Finding the Inverse ...... 29 4.5 Solving Equations (mod n) ...... 32 4.6 The Extended Euclidean Algorithm ...... 34

1 CONTENTS 2

5 Diffie-Hellman Key Exhange 37 5.1 What is it? ...... 37 5.2 The Algorithm ...... 37 5.3 Example of a Key Exhange ...... 38 5.4 Why is Cracking this Exchange Difficult ...... 39 5.5 Key Exhange Activity ...... 40

6 RSA Encryption 42 6.1 Introduction ...... 42 6.2 Example of RSA Encryption ...... 42 6.3 RSA Algorithm ...... 45 6.4 Using The Extended Euclidean Algorithm ...... 46 6.5 ASCII Table ...... 50 6.6 Cracking RSA by finding the private key d...... 52 6.7 Your turn to crack an RSA encrypted message ...... 53

7 Using RSA for Authentication 54 7.1 Algorithm ...... 54 7.2 Authentication Activity ...... 56

8 Important Information 57 8.1 Caesar Shift Wheel ...... 57 8.2 Letter Frequency Tables ...... 58 8.3 Vigenere’s Table ...... 59 8.4 ASCII Table ...... 60 8.5 Useful Websites ...... 60 1 INTRODUCTION 3

1 Introduction

You don’t have to be an expert in cryptogrphy to appreciate the field. Cryptography is all around us. We use it to store senstive data, send secret messages, send financial information over the internet, hide messages from parents, etc. The methods we will cover in this paper can be easily picked up with some knowledge of basic arithmetic and some modular algebra.

There are two basic types of encryption: substitution and transposition. We will cover the basics of each with some easy examples. Since this is an introduction, the techniques will be easier. After reading a section, try to improve on the method by coming up with your own versions.

1.1 Alice, Bob, and Eve Suppose Alice wants to send Bob a message. She doesn’t want Eve to intercept it and read it, so she decides to disguise it. Here’s a diagram that outlines the what’s happening.

Your hope is that Eve doesn’t know how to figure out your message or alter it in some way. 1 INTRODUCTION 4

1.2 Basic Terminology 1. Plaintext - this is the original message to be sent. It can letters, numbers, characters, symbols, etc.

2. Encryption - disgusing the plaintext using some method. These methods usually have a key or algorithm to generate the ciphertext.

3. Ciphertext - the result of the encryption. It can be letters, numbers, characters, etc., but unreadable. 4. Decryption - the process of undoing the encryption. You should get the original plaintext when finished.

5. Key - something that is used to encrypt the message. Think of it as a password that disguises and un-disguises your message.

6. Keyspace - the set of all possible keys. For example, a four digit lock has 10000 possible keys. (0000, 0001, 0002, ..., 9998, 9999)

7. Cryptography - the science of the enciphering and deciphering of messages in secret code or cipher. Also referred to as cryptology.

8. Cryptosystem - a system for encrypting information. 9. Cryptanalysis - the science (and art) of recovering information from ciphers without knowledge of the key.

1.3 Brief History of Cryptography Cryptography is an ancient science. The term comes from the Greek language meaning ”the study of hidden or secret writing.” Ever since there were multiple tribes (or countries), we needed a way to communicate to one another without the other tribe (country) understanding it. Today it’s vital to communicatations, authentication, sending sensative data over the internet, purchasing items from 1 INTRODUCTION 5 eBay, Amazon, and more.

One of the earliest examples of cryptography was in Egypt around 2000 B.C. where they used hieroglyphics to decorate their tombs. Even though they weren’t trying to completely hide the meaning, it still wasn’t easy to interpret.

The Greek writer Polybius used a 5x5 or a 6x6 (with out alphabet) to disguise the message. Below is an example of how it would work,

If you wanted to write the message ”SAVE ME”, it would be disguised as

S-A-V-E-M-E 41-11-44-15-31-15

Another Greek tool for disguising a message was called the Scytale. It is a tool used to perform a transposition cipher, consisting of a cylinder with a strip of parchment wound around it on which is written a message. In other words, it was a way to rearrange the letters.

You would wrap a long thin parchment around a rod. You then write your message across the rod, ”SEND HELP!” 1 INTRODUCTION 6

When you unwrap the parchment the letters appear rearranged. To understand the message, the receiver would need a rod of the same size.

Julius Caesar used his own way of disguising messages to send to his military. We know call it the Caesar Cipher. If he had anything important to send, he would write out his message and then replace each letter of his message with another letter down the alphabet. So A = D, B = E, C = F, D = G, etc. When it needed to be read, the receiver of the message would reverse this process, changing D back to A, E back to B, F back to C. We will go over this method in great detail in the next chapter. The tool we will use will look something like this, 1 INTRODUCTION 7

This method is a form of substitution, where you replace one letter for another. Over the years, the substitution method evolved. In this next chapter, we discuss the Keyword Cipher, which is another form of substitution.

The Vigenere cipher was a substitution cipher where each letter of the message used its own caesar shift. This means the first letter might be shifted by 3 letters, the second letter of the message could be shifted by 5 letters, the third letter by 3, the fourth letter by 5, the fifth letter by 3, and so on. This particular pattern would be two Caesar shifts with 3 and 5. For 300 years it was believed to be indecipherable. It can now be broken by determining the number of different caesar shifts and then do a frequency analysis on the letters that use the same Caesar shift.

To give you an idea, suppose your keyword is ”DOG” and you want to disguise the message ”HELP ME”, you use the Vigenere Table below to determine the ciphertext. 1 INTRODUCTION 8

Message: H E L P M E Keyword: D O G D O G Cipher: K S R S A K

Why was this cipher so strong? Consider the ciphertext. There are two ’K’s but each corresonds to a different letter in the plaintext. One ’K’ decrypts back to H while the other ’K’ decrypts to E.

In 1917 U.S. Army Major Joseph Mauborgne developed a technique known as the One-Time Pad. It works the same way as a Vigenere Cipher. The only 1 INTRODUCTION 9 difference is the keyword is a random string of letters that is the same length as the message. That way the keyword never repeats which makes frequency analysis useless. If this method is done correctly, it is the only provably unbreakable encryption algorithm.

Here’s an example. Suppose the ciphertext is ”KNEXYLTCOW.” Now suppose you used two different one-time pad keys, ”AFTMRDARKF” and ”DJWMRDARKF”. Let’s see what happens. Use the Vigenere Table to convert back to plaintext.

Cipher: K N E X Y L T C O W Key: D J W M R D A R K F Plaintext: H E I L H I T L E R

Cipher: K N E X Y L T C O W Key: A F T M R D A R K F Plaintext: K I L L H I T L E R

This means you can make the plaintext say anything you want. But there are two main weaknessess to this method.

1. Making sure the key is random and of sufficient length. 2. Distributing the keys. How is the receiver suppose to know which key to use?

One way to accompish the distribution of keys is to have an agreed upon book. Each day the random key starts with the first letter of a page in that book. Keep reading the letters from the page until it’s the same length as the message.

Another interesting use of crytography was to send messages through newspapers. Before the telegraph, people needed to send messages through the mail. Postage could be expensive, plus someone could intercept the letter. Newspapers traveled freely and without charge. So someone could publish an article in the newspaper and put a dot above the letter that was part of the plaintext. For example, 1 INTRODUCTION 10

Ia ˙m excit˙ed t˙oa ˙nnounce the wedding of my brother. He is ba˙ck˙ from his journey overseas and is looking forward to a new life with his beautiful bride.

Excluding i’s, which would have two dots, the letters with dots above them spell

attack World War II

The Germans adopted the use of a family of electro-mechanical encryption devices invented by the German engineer Arthur Scherbuis. They became known as the Enigma Machines, and were extremely effective during the war.

The details about how the Engima machine works is a bit complicated. The Enigma looks roughly like a typewriter.

Here’s an excerpt from www.bbc.co.uk/history/tpics/enigma ”Enigma allowed an operator to type in a message, then scramble it by using three to five notched wheels, which displayed different letters of the alphabet. The receiver needed to know the exact settings of these rotors in order to reconstitute the coded text. Over the years the basic machines became more complicated as German code experts added plugs with electronic circuits.”

When a key was pressed, a set of electrical contacts in the disks would complete a circuit lighting up the ciphertext letter. The disks would rotate after each letter is pressed providing a different encryption.

In 1932, Polish cryptologists broke the Enigma by reverse engineering the machine. As Germany added more complexity to their machine, the initial 1 INTRODUCTION 11 decryption techniques became unsuccessful. Poland presented their initial decryption technqiues to the British military intelligence. The cryptanalysts at Bletchley Park used this information to greatly increase their success of decrypting Enigma.

Modern Era

In 1972, the U.S. identified a need for a government-wide data encryption standard. They went with the encryption technique designed by cryptographers from IBM. It became known as DES. AES, the advanced encrpytion standard replaced DES in 2001.

In 1977, the RSA algorithm was presented by its creators Ron Rivest, Adi Shamir, and Leonard Adleman from MIT. With many of the classical cryptosystems, the key used to encrypt a message was also used to decrypt the message. Suppose 300 people send financial information to a bank. Each person could encrypt it and have the bank decrypt it. This would, however, require the bank to know 300 separate keys. Not only is it inefficient, how would the bank know each person’s key?

RSA is called public key encrpytion (asymmetric key encription). To encrypt a message, someone uses a public key known to everyone to encrypt their data. Only the receiver of the message knows the private key to decipher the message. This means 300 people can send their personal info using the bank’s public key to encrypt it. Only the bank can decrypt it with their private key. We will discuss more of RSA later on.

Now this was only a taste of the more famous works in cryptography. More information can be found through reserach in your libraries or online.

1.4 Kerckhoff’s Principles of Cryptographic Secuity You might think that the power behind an encryption technique is keeping the technique secret. This is not true. Kerckhoff’s principles (some of them) state

1. The system must be practically indecipherable. In other words, if the data can be deciphered, it must take a very long time. 1 INTRODUCTION 12

2. The system must not be secret. If you encrypted a message using Caesar’s cipher, then your evesdropper Eve must also know that. 3. Any evesdroppers or attackers must be able to get their hands on the ciphertext. 4. The security of the system should depend solely on the secrecy of the key.

To summarize, a good cryptosystem is one where everyone knows what it is and how it works, everyone can get their hands on the ciphertext, and its strength lies in the secrecy of the key. 2 SUBSTITUTION CIPHER 13

2 Substitution Cipher

2.1 Caesar Cipher Activity With a Caesar Cipher, you replace each letter in a message with a letter along the alphabet. A Caesar cipher shifts the alphabet and is therefore also called a shift cipher. The key is the number of letters you shift. The Caesar Cipher is named after Julius Caesar, who is said to have used it to send messages to his generals over 2,000 years ago.

You should have a pdf of a Caesar Shift Wheel. Follow the directions to create the wheel.

Example: Key: Shift 3

Message: M A T H E Q U A L S M O N E Y Cipher: P D W K H T X D O V P R Q H B

1. Use a key shift of 7, encrypt the following message:

Message: W E L C O M E T O M A T H C A M P

Cipher:

2. Use a key shift of 13, encrypt the following message:

Message: B O W L I N G T O U R N A M E N T T O N I G H T

Cipher: 2 SUBSTITUTION CIPHER 14

3. Use a key shift of 20, encrypt the following message:

Message: T H E S E C R E T P A S S W O R D I S B O O G E R

Cipher:

4.A KEY SPACE is the set of all possible keys. What is the size of the KEY SPACE for the shift cipher? 2 SUBSTITUTION CIPHER 15

To decrypt a shift cipher, you just reverse the shift. If you shifted RIGHT 3, then to decrypt the message we shift LEFT 3.

Example: Key shift: -3

Cipher: P D W K H T X D O V P R Q H B Plaintext : M A T H E Q U A L S M O N E Y

1. Use a key shift of −4, decrypt the following ciphertext:

Ciphertext: G S H I R E Q I: K S S R M I W

Plaintext:

2. Choose your own key shift. Encrypt your name and high school grade level

Plaintext:

Ciphertext:

3. Pass your ciphertext to your neighbor. DO NOT TELL THEM YOUR KEY. Have your neighbor decrypt your message.

4. Go to http://www.brianveitch.com/cryptography/caesar_friendly.php. Use this page to figure out the answer to this riddle. What did the one eye say to the other?

Ciphertext: TWLOWWF QGM SFV EW

Plaintext:

Ciphertext: KGEWLZAFY KEWDDK!

Plaintext: 2 SUBSTITUTION CIPHER 16

5. Why can’t you tell an egg a joke? Key = 23

Ciphertext: FQ JFDEQ ZOXZH RM.

Plaintext:

6. Now that you got a chance to play around with the shift cipher, can you think of any weaknesses in this encryption technique? 2 SUBSTITUTION CIPHER 17

2.2 Keyword Cipher Activity With a Keyword Shift, you place a word underneath the original alphabet. The Keyletter is the letter where you start your keyword. In this example below, the keyword is KINGDOM and the keyletter is G

Starting after the letter M, you fill in the rest of the alphabet. If the letter was already used in the keyword, then you skip it. Also, your keyword cannot have repeated letters.

Example: With the keyword: KINGDOM and keyletter M

Message: G R A V I T Y I S A D O W N E R!

Cipher: K F U P N J S N H U X B Q A Y F!

1. Use the keyword: BACON, keyletter: C 2 SUBSTITUTION CIPHER 18

Message: C R Y P T O I S C O O L

Ciphertext:

2. Use the keyword: BACON, keyletter A to decrypt the following ciphertext.

Ciphertext: T R Y S T B P J G L E H N J J M T M B T R N N

Plaintext:

3. Go to http://www.brianveitch.com/cryptography/keyword.php. Use this page to decrypt the following statement.

Keyword: THUNDERCLAP, Keyletter: E

ZPA’J SPK NWJT DJ ONTA SPK’GT JTQJDAU WAZ CSDAU PA SPKG XWYR ONTA SPKG BNPAT ZTYDZTI JP XT W ADAEW, ICDBI JNGPKUN SPKG HDAUTGI, WAZ WJJWYRI SPKG HWYT! 2 SUBSTITUTION CIPHER 19

4. The following are quotes from famous movies. Each quote has been encrypted with the same keyword and keyletter. Your job is to decrypt them and determine its origin. Identify the keyword and keyletter.

Use the letter frequency calculator at http://www.brianveitch.com/cryptography/relative_frequency.php to help determine the plaintext.

WSC ZE RCVMTD, IV VEFVC M ISCOR FYMF ZD VEFZCVOK SGC SIE.

TMK FYV WSCPV NV IZFY KSG.

FYVCV’D ES AOMPV OZUV YSTV.

KSG PME’F YMEROV FYV FCGFY!

KSG’CV XSEEM EVVR M NZXXVC NSMF.

YVCV’D QSYEEK!

TK ACVPZSGD.

KSGC TSFYVC IMD M YMTDFVC MER KSGC WMFYVC DTVOF SW VORVCNVCCZVD.

Z DYMOO PMOO YZT DBGZDYK. MER YV DYMOO NV TZEV. MER YV DYMOO NV TK DBGZDYK.

FYVKHV RSEV DFGRZVD, KSG UESI. DZJFK AVCPVEF SW FYV FZTV, ZF ISCUD VHVCK FZTV.

CMK, EVJF FZTV DSTVSEV MDUD KSG ZW KSGCV M XSR, KSG DMK KVD! 2 SUBSTITUTION CIPHER 20

2.3 Vigenere Cipher Activity The Vigenre Cipher is a method of encrypting alphabetic text by using a series of different Caesar ciphers based on the letters of a keyword. Instead of shifting every letter of your plaintext by C = 2, each letter of the plaintext will have a different shift.

Though the cipher is easy to understand and implement, for three centuries it resisted all attempts to break it; this earned it the description le chiffre indchiffrable (French for ’the indecipherable cipher’) (wikipedia).

Each letter of the alphabet represents a shift. For example, A = 0 (no shift), B = 1 (shift 1 to the right), C = 2 (shift 2 to the right), D = 3 (shift 3 to the right), E = 4 (shift 4 to the right), ...

If the keyword is ’BED’, it means you shift the first letter of your cipher by B = 1, the second letter of the cipher by E = 4, third letter by D = 3. Repeat this pattern. The keyword ’BED’ converts to the numeric key 143. Consider the following example.

You can use your Caesar Shift Wheel to do this. Since you will have to keep changing the shift for each letter, the wheel will help you do it quickly.

Example: Using the keyword: NEMO, encrypt the following message.

NEMO = (corresponding numeric key) 2 SUBSTITUTION CIPHER 21

Example: Using the keyword: CRYPTO, encrypt the following message.

CRYPTO = (corresponding numeric key)

Decrypting the Ciphertext

Decrypting a Vigenere message requires you to reverse the shift from the keyword. You would line up your ciphertext with the numeric key and reverse the shift, just like you reversed the shift with the Caesar Cipher.

Example: Using the keyword: AVENGERS, decrypt the following message.

Decryping by hand can be tedious. Go to http://www.brianveitch.com/cryptography/vigenere.php. This allows you to encrypt a message with the Vigenere keyword and also decrypt messages very quickly. 3 TRANSPOSITION 22

3 Transposition

There are two basic ways of encrypting a message. One way is through some sort of substitution, which was covered in the previous section. The other way is through transposition. Transposition ciphers are formed by changing the normal position of the letters that make up the message.

3.1 Easy Example A common transposition can be simply writing out the letters of a word backwards.

ATTACK AT DAWN translates into KCATTA TA NWAD

You could also read the entire message right to left, which would encrypt to

NWAD TA KCATTA

3.2 Railfence Cipher Suppose you have the message ”STOP THE ATTACK WE ARE OUTNUMBERED”

1. Write the message vertical. You must choose how many letters you go down. When you reach the bottom of the column, you start the next column.

Some examples,

(a) Vertical: 5 SHAATE TECRNR OAKEUE PTWOMD TTEUB 3 TRANSPOSITION 23

(b) Vertical: 2 SOTETAKEROTUBRD TPHATCWAEUNMEE

2. Next, you would write out the message from left to right like you were reading a book. The cipher text would be

(a) Vertical 5 by Horizontal 6:

SHAATE TECRNR OAKEUE PTWOMD TTEUB or (without spaces)

SHAATETECRNROAKEUEPTWOMDTTEUB

(b) Vertical 2 by Horizontal 15:

SOTETAKEROTUBRD TPHATCWAEUNMEE or (without spaces)

SOTETAKEROTUBRDTPHATCWAEUNMEE

3. You can keep changing the vertical length to get a new cipher for the exact same message. You just need to agree upon a length with the receiver beforehand. 4. To decrypt this message, you’ll need the vertical and horizontal length. It’s possible to do it without, but it will take a lot of trial and error.

3.3 Transposition Activity Choose a vertical length and encrypt a message. Give your encrypted message to your partner. Try to decipher your partner’s message without knowledge of the vertical and horizontal length. After 5 minutes, if you have not deciphered the message, have your partner tell you the vertical and horizontal lengths. 4 MODULAR ARITHMETIC 24

4 Modular Arithmetic

Clock Example

What if we kept counting after 12 o’clock? Where would 13 o’clock be? Turn the time back into our normal 12 hour time.

13 = o’clock 17 = o’clock 23 = o’clock 24 = o’clock

The 24 hour clock face: 4 MODULAR ARITHMETIC 25

With modular arithmetic, we would use the following notation,

13 ≡ 1 (mod 12) 17 ≡ 5 (mod 12) 23 ≡ 11 (mod 12) 24 ≡ 0 (mod 12)

Formal Definition of a ≡ b (mod n)

≡ means congruent a is congruent to b mod n if b − a is a integer multiple of n

That’s great for a math class, but a bit confusing for your first exposure to mod. Another way to look at it is,

Informal Definition of a ≡ b (mod n)

b is the remainder when you divide a by n Examples:

13 ÷ 12 = 1 R 1

17 ÷ 12 = 1 R 5

23 ÷ 12 = 1 R 11

24 ÷ 12 = 2 R 0 4 MODULAR ARITHMETIC 26

So when we talk about (mod n), we really only care about the remainder when you divide by n. Your answer must be a number between

0 and n − 1

4.1 Examples of Congruences 1. 27 ≡ (mod 12) - answer must be between 0 and 11 2. 145 ≡ (mod 12) 3. 214 ≡ (mod 12) 4. 28 ≡ (mod 5) - answer must be between 0 and 4 5. 16 ≡ (mod 3) 6. 1563 ≡ (mod 27) 7. 10625 ≡ (mod 101) - answer must be between 0 and 100 8. 783 ≡ (mod 27)

If the number is negative, you must add multiples of n until you get a number between 0 and n

−3 ≡ 5 (mod 8) because -3 + 8 = 5

−28 ≡ 4 (mod 8) because -28 + 8 + 8 + 8 + 8 = 4

9. −8 ≡ (mod 12) 10. −15 ≡ (mod 4) 11. −126 ≡ (mod 18) 12. −54 ≡ (mod 13) 4 MODULAR ARITHMETIC 27

We can do addition, subtraction, multipication, and exponentiation exactly the same way. Give them a try,

4.2 Arithmetic with Congruences 1. 11 + 4 ≡ (mod 12) 2. 23 − 17 ≡ (mod 12) 3. 16 − 2 ≡ (mod 12) 4.4 · 8 ≡ (mod 5) 5. 16 · 3 ≡ (mod 3) 6. 154 + 27 ≡ (mod 27) 7.2 4 ≡ (mod 27) 8.3 6 ≡ (mod 13) 9.7 3 − 1 ≡ (mod 7) 10.9 4 ≡ (mod 5)

4.3 Multipication Tables Suppose you multiply each number (mod 5) with another number from (mod 5). We can create a table to keep track of the results.

* mod 5 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 3 1 4 2 4 0 4 3 2 1

Notice how 2 · 3 ≡ 1 (mod 5). This will be important very soon. 4 MODULAR ARITHMETIC 28

Example:

Now you try creating a multiplication table for (mod 4), (mod 7) and (mod 11).

* mod 4

* mod 7

* mod 11 4 MODULAR ARITHMETIC 29

Division is very different. What is 4/3 (mod 7)?

Division like this doesn’t make sense. When we added, subtracted, and multiplied we still got an integer. We then divided that integer by n. But 4/3 is not an integer! 1 1 You might recall that 4/3 is the same thing as 4 · And is the inverse of 3. So, 3 3 if we want to compute 4/3 we need to do,

4/3 = 4 · 3−1

So what’s the inverse of 3 (mod 7)? Recall that the inverse of 3 is a number than when you multiply it to 3 you get 1.

3 · x = 1 Let’s look at an example.

4.4 Finding the Inverse 1.3 −1 (mod 7)

We need to solve

3x ≡ 1 (mod 7)

There are three ways we can do this. Let’s first do it by trial and error,

3 · 1 ≡ 3 (mod 7) 3 · 2 ≡ 6 (mod 7) 3 · 3 ≡ 9 ≡ 2 (mod 7) 3 · 4 ≡ 12 ≡ 5 (mod 7) 3 · 5 ≡ 15 ≡ 1 (mod 7) 4 MODULAR ARITHMETIC 30

From the multiplication table, you should have

3 · 5 ≡ 1 (mod 7) So if an inverse exists, you can find it from the multiplication table. The third way will be covered in the section The Extended Euclidean Algorithm.

So when we are in (mod 7), 3−1 is 5 (mod 7).

So 4/3 (mod 7) can be calculated as

4 · 3−1 ≡ 4 · 5 = 20 ≡ 6 (mod 7)

2.3 −1 (mod 4)

3.7 −1 (mod 11) 4 MODULAR ARITHMETIC 31

Please note that some inverses and divisions do not exist. For example, 2−1 (mod 4) did not exist. Therefore, you cannot divide by 2 in (mod 4). However, every number (except 0) in (mod 7) and (mod 11) did have inverses. There is a reason for this. Take some time to think about it. Try creating a multipication table for (mod 12) and see what happens.

* mod 12

11 Example: Find (mod 12) 5 4 MODULAR ARITHMETIC 32

4.5 Solving Equations (mod n) How did we solve an equation like 3x = 5 from basic algebra? We divided both sides by 3.

3x = 5

3x 5 = 3 3 5 x = 3

So how would we solve the equation

3x ≡ 5 (mod 11)

Instead of dividing both sides by 3 (we don’t do division like this in (mod n)), we multiply both sides by the inverse of 3, 3−1.

3x ≡ 5 (mod 11)

3−1 · 3x ≡ 3−1 · 5 (mod 11)

x ≡ 3−1 · 5 (mod 11) So what’s 3−1 (mod 11)?

From the multiplication table, 3−1 ≡ 4 (mod 11) since 3 · 4 ≡ 1 (mod 11).

Therefore, our solution to 3x ≡ 5 (mod 11) is

x ≡ 3−1 · 5 (mod 11)

x ≡ 4 · 5 (mod 11) 4 MODULAR ARITHMETIC 33

x ≡ 9 mod 11 Now let’s verify.

3 · 9 = 27 ≡ 5 (mod 11) Your turn.

1.5 x ≡ 7 (mod 12) 2.7 x ≡ 9 (mod 12) 3. x2 ≡ 1 (mod 12) - use the multiplication table for (mod 12) 4 MODULAR ARITHMETIC 34

4.6 The Extended Euclidean Algorithm A very useful tool to use when trying to find the inverse of a number (mod n) is called the Extended Euclidean Algorithm. As you probably figured out by now, a number will have an inverse (mod n) as long as that number is relativly prime to n.

This method requires the division algorithm on the left hand side, and a recursive formula on the right. Please follow the example below.

We need the recursive formula: pi = pi−2 − pi−1 · qi−2 where p0 = 0 and p1 = 1.

When you move onto the next step, you use ai = bi−1 and bi = ri−1

3 · d ≡ 1 (mod 20)

ai = qi · bi + ri pi = pi−2 − pi−1 · qi−2

Step 0: 20 = (6) · 3 + 2 p0 = 0

Step 1: 3 = (1) · 2 + 1 p1 = 1

Step 2: 2 = (2) · 1 + 0 p2 = p0 − p1 · q0 = 0 − 1(6) = −6

p3 = p1 − p2 · q1 = 1 − (−6) · 1 = 7

Therefore, the solution to 3 · d ≡ 1 (mod 20) is d = 7 (mod 20).

This means 3−1 ≡ 7 (mod 20)

Here’s another example:

7 · d ≡ 1 (mod 40)

Remember, ai = bi−1 and bi = ri−1 4 MODULAR ARITHMETIC 35

ai = qi · bi + ri pi = pi−2 − pi−1 · qi−2

Step 0: 40 = (5) · 7 + 5 p0 = 0

Step 1: 7 = (1) · 5 + 2 p1 = 1

Step 2: 5 = (2) · 2 + 1 p2 = p0 − p1 · q0 = 0 − 1(5) = −5

Step 3: 2 = (2) · 1 + 0 p3 = p1 − p2 · q1 = 1 − (−5)(1) = 6

p4 = p2 − p3 · q2 = −5 − 6 · 2 = −17

Therefore, the solution to 7 · d ≡ 1 (mod 40) is

d = −17 ≡ 23 (mod 40) Just to check,

7 · 23 = 161 ≡ 1 (mod 40)

Therefore, 7−1 ≡ 23 (mod 40).

Your turn! Find 9−1 (mod 16) by solving the following equation.

9 · d ≡ 1 (mod 16)

You’ll find a table on the next page to complete the Extended Euclidean Algorithm. 4 MODULAR ARITHMETIC 36

ai = qi · bi + ri pi = pi−2 − pi−1 · qi−2

Step 0:

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Your inverse is the last pi value in the table. Make sure it’s (mod n).

The inverse of 9−1 (mod 16) is

9−1 ≡ (mod 16)

Use the website http://www.brianveitch.com/cryptography/modinverse.php to confirm the inverse. 5 DIFFIE-HELLMAN KEY EXHANGE 37

5 Diffie-Hellman Key Exhange

5.1 What is it? Suppose Alice sends Bob an encrypted message using the Vigenere Cipher. Bob and Alice need to also agree upon the key, otherwise Bob would have to decipher Alice’s message the hard way. If Eve, our evesdropper, is aware of the key then she can decipher all of Alice’s messages to Bob. Diffie-Hellman Key Exchange allows the sender, Alice, and the receiver, Bob, to come up with a key with public information. The upside to this method is Eve, our evesdropper, is aware of the exchange the whole time but will have a difficult time cracking it and finding the secret key. Here’s how it works.

5.2 The Algorithm 1. Alice and Bob agree, publicly, on a prime number P , and another number N, where N < P . Everyone, including Eve, will know these two numbers. 2. Alice chooses her own personal (private) number A. She does not tell anyone this number. This number should be relatively prime to N. 3. Bob chooses his own personal (private) number B. He does not tell anyone this number. This number should be relatively prime to N. 4. Alice calculates the number

S = N A (mod P ) and sends this number S to Bob. 5. Bob calculates

R = N B (mod P )

and sends this number R to Alice. 6. So now Bob and Alice have two numbers. Eve, who is watching this exchange, is fully aware of the numbers S and R. But since she doesn’t know A and B, she doesn’t know how Alice and Bob calculated the R and S. 5 DIFFIE-HELLMAN KEY EXHANGE 38

7. Alice takes the number she received from Bob and calculates

C = RA (mod P )

and Bob takes the number he received from Alice and calculates

C = SB (mod P )

8. C is their shared key. Since the last step is done in private, Eve does not know C.

5.3 Example of a Key Exhange Alice wants to send the message ”ATTACK AT DAWN” to Bob. Before encrypting it with the Vigenere Cipher, her and Bob need to agree upon the keyword (in numeric form). Alice calls Bob, with Eve evesdropping on this conversation, and come up with a public key

P = 983 and N = 20 You’ll probably want to use the exponentiation calculator at http://www.brianveitch.com/cryptography/modexp.php

1. Alice chooses the number A = 7 and Bob chooses the number B = 9. Remember, these must be relatively prime to N = 20. 2. Alice calculates

S = 207 = 1280000000 ≡ 312 (mod 983) and sends this number to Bob. 3. Bob calculates

R = 209 = 512000000000 ≡ 942 (mod 983) and sends this number to Alice. 5 DIFFIE-HELLMAN KEY EXHANGE 39

4. Alice takes 942 (mod 983) and raises it to her number A = 7.

9427 ≡ 101 (mod 983)

5. Bob takes 312 (mod 983) and raises it to his number B = 9.

3129 ≡ 101 (mod 983)

6. So Alice encrypts the message ”ATTACK AT DAWN” with the Vigenere numeric key 101, or its equivalent alphabetic key BAB. The cipher text is

5.4 Why is Cracking this Exchange Difficult In the previous example, here is all the information Eve has during this key exchange.

P = 983,N = 20,,S = 312, and R = 942

In order for Eve, our evesdropper, to determine the keyword, she needs to solve the following equations

312 ≡ 20A (mod 983) or 942 ≡ 20B (mod 983) This is difficult to do, especially when P and N are extremely large. Our example, they aren’t large. But even for us, our best bet is probably trial and error. We would try to plug in all the values (mod 983) for A and B and see which work. 5 DIFFIE-HELLMAN KEY EXHANGE 40

5.5 Key Exhange Activity Get in groups of 3. One of you will be Alice, another will be Bob, and the third will be Alice. In this activity, Alice and Bob will come up with their Vigenere Cipher key right in front of Eve.

1. Alice writes a message. It doesn’t matter how long since you will be using the online Vigenere Cipher. You want to send this message to Bob. Do not show this message to anyone. 2. Alice and Bob now speak to each other publicly. Eve is there listening and taking notes. Alice and Bob agree on two numbers. Choose a prime number P and another number N less than P . You may use the website http://www.brianveitch.com/cryptography/prime_generator to generate the prime number.

P = and N =

3. Alice and Bob choose their private number. If you’re Alice write it below

A = and calculate

N A ≡ (mod P)

If you are Bob, write it below

B = and calculate

N B ≡ (mod P)

4. Alice and Bob exchange numbers. Eve receives these numbers as well. 5. Alice and Bob finish the calculations to come with the key. Alice uses the Vigenere Cipher at http://www.brianveitch.com/cryptography/vigenere.php to create the ciphertext. 5 DIFFIE-HELLMAN KEY EXHANGE 41

6. Alice sends the message to Bob for him to decipher. 7. Eve should write down the equations she must solve in order to calculate their private key. She may try to crack it by using trial and error.

Change positions and redo this until everyone gets a chance to be Alice. 6 RSA ENCRYPTION 42

6 RSA Encryption

6.1 Introduction I can think of two uses for RSA encryption that should make sense to us. In the last chapter, we discussed a key exhange. This is a method for Alice and Bob to commuicate publicly to create a secret key. If done correctly, Eve should have a very hard time figuring this key out.

RSA encryption can also be used to send the key. It’s very similar to the Diffie-Hellman Key Exhange. RSA is called public key encryption (or asymmetric-key encryption). It means that there is a public key to encrypt the data and a secret key (only the receiver of the message knows) that will decrypt the data. So Alice chooses the keyword for the Vigenere Cipher. She uses the public key to encrypt her keyword and sends the ciphertext to Bob. Bob uses his secret key, known only to him, to decrypt it. With the keyword now agreed upon, Alice can use a different method to encrypt the data and send it to Bob.

Another use of RSA encryption is called Authentication. Suppose Bob recevies a message, supposedly from Alice. How does he actually know this message is from Alice? It’s possible Eve intercepted the message and created her own to send to Bob.

6.2 Example of RSA Encryption Let’s do an example using RSA. Public key is (e, n) = (3, 33)

Message: A T T A C K N O W

First, let’s convert the letters to numbers using A = 0, B = 1, C = 2, D = 3, ..., X = 23, Y = 24, Z = 25. Later on we will convert letters to their ASCII number.

Message: A T T A C K N O W

Pre-Cipher: 0 19 19 0 2 10 13 14 22 e = 3 is called the encryption key. Raise each number in the pre-cipher to e = 3, (mod 33). 6 RSA ENCRYPTION 43

A: 03 ≡ 0 (mod 33) T: 193 = 6859 ≡ 28 (mod 33) T: 193 = 6859 ≡ 28 (mod 33) A: 03 ≡ 0 (mod 33) C: 23 ≡ 8 (mod 33) K: 103 = 1000 ≡ 10 (mod 33) N: 133 = 2197 ≡ 19 (mod 33) O: 143 = 2744 ≡ 5 (mod 33) W: 223 = 10648 ≡ 22 (mod 33)

Message: A T T A C K N O W

Cipher: 0 28 28 0 8 10 19 5 22

Everyone has access to the public key. Only the receiver of the encrypted data has the private key. This is called asymmetric key encryption. It means there is one key to encrypt and a completely different key to decrypt.

The private key for our example is d = 7. Let’s see what happens when we raise our ciphertext numbers to the power of d = 7.

07 ≡ 0 (mod 33) 287 ≡ 19 (mod 33) 287 ≡ 19 (mod 33) 07 ≡ 0 (mod 33) 87 ≡ 2 (mod 33) 107 ≡ 10 (mod 33) 197 ≡ 13 (mod 33) 57 ≡ 14 (mod 33) 227 ≡ 22 (mod 33) 6 RSA ENCRYPTION 44

We get back the original set of numbers

Plaintext: 0 19 19 0 2 10 13 14 22

Message: A T T A C K N O W

Suppose you want to send the number M. The ciphertext is

C ≡ M 3 (mod 33)

To decrypt the ciphertext, you compute

M ≡ C7 (mod 33)

So to encrypt a number, we raised it to the 3rd power. To undo it, we raise it to the 7th power. The question now is how do you come up with the private key d = 7?

Warning:

Notice the same letter encrypted to the same number. Those numbers carry the same statistical information as the Ceasar method. Once I figure out T = 28, I can replace all 28s with T. This is no good to us. So let’s go back to

Message: A T T A C K N O W

Cipher: 0 19 19 0 2 10 13 14 22

Instead of enrypting 0, 19, 19, 0, 2, 10, 13, 14, and 22, we can regroup the numbers. One example would be,

01, 91, 90, 21, 01, 31, 42, 2 6 RSA ENCRYPTION 45

The problem is our modulus is 33, which is too small (since we have numbers larger than 33 - 91, 90, 42) We need to choose a larger modulus and figure out a new public and private key. Let’s figure out how to do this now.

6.3 RSA Algorithm Suppose Alice wants to send Bob a message. She uses Bob’s public key (e, n). Let’s see how Bob came up with his public/private key by coming up with our own public and private key.

1. Bob chooses two large primes. Call them p and q. He must keep these secret!

Now you choose two primes (preferably between 10 and 20).

p = q =

2. He then computes n = p · q. This is part of the public key.

n =

3. Bob computes φ(n) = (p − 1) · (q − 1)

φ(n) =

4. Bob chooses an integer e, 1 < e < φ(n) where e does not share any factors with φ(n). This is called copime (or relatively prime).

Example: 8 and 15 are coprime because they share no factors.

Example: 9 and 24 are NOT coprime because they share a common factor 3. 6 RSA ENCRYPTION 46

e =

5. Bob now has his public key (e, n). Your public key is

(e, n) = ( , )

6. Bob then computes his private key d. To find d, you must solve the following equation

e · d ≡ 1 (mod n)

Your equation is

· d ≡ 1 (mod )

Let’s go over how Bob found d to the following equation. Then follow the pattern. This may take a few tries to get right.

6.4 Using The Extended Euclidean Algorithm

We need the recursive formula: pi = pi−2 − pi−1 · qi−2 where p0 = 0 and p1 = 1.

When you move onto the next step, you use ai = bi−1 and bi = ri−1

3 · d ≡ 1 (mod 20) 6 RSA ENCRYPTION 47

ai = qi · bi + ri pi = pi−2 − pi−1 · qi−2

Step 0: 20 = (6) · 3 + 2 p0 = 0

Step 1: 3 = (1) · 2 + 1 p1 = 1

Step 2: 2 = (2) · 1 + 0 p2 = p0 − p1 · q0 = 0 − 1(6) = −6

p3 = p1 − p2 · q1 = 1 − (−6) · 1 = 7

Therefore, the solution to 3 · d ≡ 1 (mod 20) is d = 7 (mod 20). This is how Bob got his private key.

To view another example, please check out the Extended Euclidean Algorithm section in Modular Arithmetic.

Your turn!

· d ≡ 1 (mod )

ai = qi · bi + ri pi = pi−2 − pi−1 · qi−2

Step 0:

Step 1:

Step 2:

Step 3:

Step 4:

Step 5: 6 RSA ENCRYPTION 48

Your d is the last pi value in the table. Make sure it’s (mod n).

Your private key is

d =

Use the website http://www.brianveitch.com/cryptography/modinverse.php to confirm the inverse.

Example: Use your public key to encrypt the following. First, convert the letters to numbers. Message: M A T H E Q U A L S M O N E Y Pre-Cipher:

To encrypt each number N into the cipher C

C ≡ N e (mod n)

Use my expontentiation calculator at http://www.brianveitch.com/cryptography/modexp.php 6 RSA ENCRYPTION 49

M: ≡ (mod ) A: ≡ (mod ) T: ≡ (mod ) H: ≡ (mod ) E: ≡ (mod ) Q: ≡ (mod ) U: ≡ (mod ) A: ≡ (mod ) L: ≡ (mod ) S: ≡ (mod ) M: ≡ (mod ) O: ≡ (mod ) N: ≡ (mod ) E: ≡ (mod ) Y: ≡ (mod )

To practice RSA encryption, work with a partner. Write a short message (no more than 20 letters) and encrypt it using your partner’s public key. Do not tell your partner your plaintext message. Give them only the ciphertext.

Your partner then decrypts the ciphertext using his or her private key to reveal your message. Remember, to decrypt the message do the following

N ≡ Cd (mod n)

Using the website http://www.brianveitch.com/cryptography/generate_rsa_keys.php, genereate your own public and private key.

Exchange public keys with your partner. Use your partner’s public key to encrypt a word (at most 7 letters). Have your partner decrypt the word. Since the modular exponentiation will be large, use the following websites to help 6 RSA ENCRYPTION 50 http://www.brianveitch.com/cryptography/modexp.php and http://www.brianveitch.com/cryptography/modinverse.php

How do we make this harder to crack?

Let me return to our previous example of encrypting

ATTACK NOW Use the ASCII table provided below to convert every character (including the space, periods, etc.) into its correponding ASCII number.

6.5 ASCII Table 6 RSA ENCRYPTION 51

After converting the characters to their corresponding ASCII numbers, we get the following,

Message: A T T A C K N O W

Plaintext: 65 84 84 65 67 75 32 78 79 87

Let’s use the public key (e = 709, n = 923). To make this harder to crack, we regroup the numbers into blocks of 3. There is nothing special about the size of the block. You just have to make sure the numbers you’re encrypting are less than n = 923.

658|484|656|775|327|879|87 Since the numbers will get large, use the calculator at http://www.brianveitch.com/cryptography/modexp.php

658709 ≡ (mod 923) 484709 ≡ (mod 923) 656709 ≡ (mod 923) 775709 ≡ (mod 923) 327709 ≡ (mod 923) 879709 ≡ (mod 923) 87709 ≡ (mod 923)

Therefore, the message

ATTACK NOW encrypts to 697 419 188 541 288 333 854 6 RSA ENCRYPTION 52

Suppose you’re Eve and you come across this encrypted message. You know the method used and you know Bob’s public key (709, 923). How do we crack it?

6.6 Cracking RSA by finding the private key d. To decrypt the ciphertext, we need the private key d. So let’s try to crack it.

1. Factor n = 923 into two prime numbers p and q.

n = ×

2. Calculte φ(n) = (p − 1) · (q − 1)

φ(923) =

3. Now solve the equation using the Extended Euclidean Algorithm.

709 · d ≡ 1 (mod 840)

qi pi = pi−2 − pi−1 · qi−2

Step 0:

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Step 6: 6 RSA ENCRYPTION 53

Good! You now have the private key d = Use http://www.brianveitch.com/cryptography/modexp.php to complete the exponentiation. Raise each block of the ciphertext by the power of d.

≡ (mod 923) ≡ (mod 923) ≡ (mod 923) ≡ (mod 923 ≡ (mod 923) ≡ (mod 923) ≡ (mod 923)

4. After decrypting the numbers, you should have

658 484 656 775 327 879 87 5. Regroup the numbers back into blocks of 2 (starting from the left)

6. Finally rewrite the ASCII number back into its characters.

6.7 Your turn to crack an RSA encrypted message You are Eve and you intercept a message sent from Alice to Bob. It was encrypted using Bob’s public key (n = 1224825487, e = 1118107197).

The ciphertext is

71218284 237148343 889555017 473479370 649324334 6660643 260396109 637251342 924334351 935096349 271621588 366934580 271621588 576453518 423052470 618017834 332813753 397822860 7 USING RSA FOR AUTHENTICATION 54

To crack the code, find the private key d. Follow the method we used to crack the RSA example from before. Start by factoring n.

You can decrypt it using the private key d using the website http://www.brianveitch.com/cryptography/modexp.php. If you’re doing it by hand, the original block size (before encryption) was 4. or the website http://www.brianveitch.com/cryptography/rsa.php. Use a block size of 2 on this website.

There you go. This is the general idea behind the RSA encryption. Now before we finish with our RSA introduction, I’d like to show you the size of a typical modulus n.

16152174667064029642647365822885998430666314431815268152405470907824573 65903662972483772980826569393306732864932303362619914669385966910731129 68626710792148904239628873374506302653492009810626437582587089465395941 37549600473991849827667633423824146549803003658606392990236819200423317 2032080188726965600617167

Can you see why trying to factor n to crack RSA is hard? Even if you can factor it, you still need to find the private key d.

7 Using RSA for Authentication

7.1 Algorithm As noted earlier, how does Bob actually know the message is sent from Alice. Here’s what she does.

1. Alice pubicly announces her RSA public key, say (n = 2263, e = 917). 2. She converts her name to ASCII numbers: 65 76 73 67 69. She then encrypts her name with her private key d = 1133. Remember that only Alice knows the private key. You may need the calculator at http://www.brianveitch.com/cryptography/modexp.php 7 USING RSA FOR AUTHENTICATION 55

651133 ≡ 228 (mod 2263) 761133 ≡ 1630 (mod 2263) 731133 ≡ 1314 (mod 2263) 671133 ≡ 645 (mod 2263) 691133 ≡ 1405 (mod 2263)

3. She sends her encrypted name along with the encrypted message meant for Bob. ALICE Encrypted: → 228 1630 1314 645 1405

4. Bob receives two ciphers. One is Alice’s encrypted name. The other is the actual encrypted message. Bob decrypts her message, but isn’t convinced it was sent from Alice. 5. Bob uses Alice’s public key, which is accessible by everyone, to decrypt the name.

228917 ≡ 65 (mod 2263) 1630917 ≡ 76 (mod 2263) 1314917 ≡ 73 (mod 2263) 645917 ≡ 67 (mod 2263) 1405917 ≡ 69 (mod 2263)

65 76 73 67 69 → ALICE

6. Since Alice sent her name by encrypting it with her private key, no one can fake this. The only way Eve can fake her name is by encrypting ”ALICE” with her private key. And the only way Eve can do that is by cracking her RSA public key. 7 USING RSA FOR AUTHENTICATION 56

7.2 Authentication Activity 1. Get in groups of 3. One person is Alice, another is Bob, and the third is Alice. 2. Alice needs to generate a public key. She also needs her find her private key. She can do this by hand or use the RSA Key Generator at http://www.brianveitch.com/cryptography/generate_rsa_keys. Alice will create a message and encrypt it. She may use any method she wants. 3. Eve will create her own message (something that may contradict Alice’s message).

Example: (a) Alice may encrypt the message ”ATTACK AT DAWN” using the Caesar Cipher. (b) Eve may encrypt ”DO NOT ATTACK AT DAWN” using the Caesar Cipher. 4. Using Alice’s public key, she encrypts her name by using her private key. 5. Eve will also encrypt Alice’s name, but she must guess on the private key. 6. Both Alice and Eve send Bob their messages with the encrypted name. 7. Bob decrypts both message. Not sure which message came from Alice, use Alice’s public key to decrypt the names. 8. Unless Eve guessed correctly on Alice’s private key, Alice’s encrypted name should decrypt back to Alice and Eve’s should decrypt to jibberish. 8 IMPORTANT INFORMATION 57

8 Important Information

8.1 Caesar Shift Wheel 8 IMPORTANT INFORMATION 58

8.2 Letter Frequency Tables 8 IMPORTANT INFORMATION 59

8.3 Vigenere’s Table 8 IMPORTANT INFORMATION 60

8.4 ASCII Table

8.5 Useful Websites Random Substitution Generator http://www.brianveitch.com/cryptography/sub_shift.php

Easier Caesar Cipher http://www.brianveitch.com/cryptography/caesar_friendly.php

Caesar Cipher http://www.brianveitch.com/cryptography/caesar.php

Keyword Substitution Cipher http://www.brianveitch.com/cryptography/keyword.php 8 IMPORTANT INFORMATION 61

Vigenere Ciphere http://www.brianveitch.com/cryptography/vigenere.php

Letter Frequency http://www.brianveitch.com/cryptography/relative_frequency.php

Modular Exponentiation http://www.brianveitch.com/cryptography/modexp.php

Modular Inverse http://www.brianveitch.com/cryptography/modinverse.php

String to ASCII http://www.brianveitch.com/cryptography/string_to_number.php

ASCII to String http://www.brianveitch.com/cryptography/number_to_string.php

Easier RSA - No Blocking Before Encrypting http://www.brianveitch.com/cryptography/rsa_easy.php

RSA - Blocking Before Encrypting http://www.brianveitch.com/cryptography/rsa.php

Generate RSA Public and Private Keys http://www.brianveitch.com/cryptography/generate_rsa_keys.php

Generate Prime Numbers http://www.brianveitch.com/cryptography/generate_primes.php