Todays Agenda Wrap up of Number Theory (Sec. 3.7) Fermats Little - - PowerPoint PPT Presentation

Today’s Agenda

• Wrap up of Number Theory (Sec. 3.7)
• Fermat’s Little Theorem
• Public Key Cryptography (RSA)
• Strings and Languages (Chap. 12)

1 Based on Rosen and slides by K. Busch

2

Fermat’s little theorem:

For any prime and integer not divisible by ( ):

p

a p

) (mod 1

1

p a p 



Example:

2  a 5  p

) 5 (mod 1 16 24  

1 ) , gcd(  p a

Pierre de Fermat (1601-1665)

a

(We will use FLT in the RSA cryptosystem)

3

Public Key Cryptography (RSA cryptosystem) “MEET YOU IN THE PARK” “9383772909383637467”

n x x f

e mod

) (  n y y f

d mod

) (

1





encryption decryption

q p n  

Large primes

e n,

are public keys

q p,

are private keys for finding d for any e

1 )) 1 )( 1 ( , gcd(    q p e

(with the condition that )

Key Idea: Everyone knows n (= pq) and e, but to find d to decrypt, need to know what p and q are.

4

Practically impossible to factor n into p and q if p and q are chosen to be primes of 200 digits or more.

5

Message to encrypt: “STOP” Encryption example:

43  p 59  q 13  e 2537    q p n

1 ) 58 42 , 13 gcd( )) 1 )( 1 ( , gcd(      q p e Translate to equivalent numbers

“18 19 14 15” “1819 1415”

Group into blocks of two numbers

6

“1819 1415” “2081 2182”

2537 mod mod ) (

13

x n x x f

e

 

2081 2537 mod 1819 ) 1819 (

13

  f 2182 2537 mod 1415 ) 1415 (

13

  f

Apply encryption function to each block

Encrypted message:

Use fast modular exponentiation algorithm:

7

Message decryption

M C

:an original block of the message :respective encrypted block “1819 1415” “2081 2182” We want to recover by knowing

M e q p C , , , ) (modn M C

e



encrypt

8

Let = inverse of modulo

d

e

) 1 )( 1 (   q p

)) 1 )( 1 (mod( 1    q p de

) 1 )( 1 ( 1     q p k de

Inverse exists because

1 )) 1 )( 1 ( , gcd(    q p e

by definition of congruent

) 1 )( 1 mod( 1 i.e., ) 1 )( 1 ( 1 )) 1 )( 1 ( , gcd(           q p se q p t se q p e

s d 

Does inverse d always exist?

9

) (modn M C

e



 

) (modn M C

d e d 

) (mod

) 1 )( 1 ( 1

n M M C

q p k de d   

 

) 1 )( 1 ( 1     q p k de

Encryption Decryption

10

In real-world case,

1 ) , gcd(  p M

(because is a large prime and is small)

p M

1 ) , gcd(  p M

By Fermat’s little theorem

) (mod 1

1

p M p 



Remember me?

11

) (mod 1

1

p M p 



 

) (mod 1 1

) 1 ( ) 1 ( 1

p M

q k q k p

 

  

 

) (mod 1

) 1 ( 1

p M M M

q k p

  

 

) (mod p M M 

) (mod

) 1 )( 1 ( 1

p M M

q p k



  

Multiply under mod

12

) (mod

) 1 )( 1 ( 1

p M M

q p k



  

By symmetry (by replacing with ):

) (mod

) 1 )( 1 ( 1

q M M

q p k



  

q p

We showed: By Exercise 23 (Sec. 3.7):

) (mod ) (mod

) 1 )( 1 ( 1

n M pq M M

q p k

 

  

13

) (mod

) 1 )( 1 ( 1

n M C

q p k d   



) (mod

) 1 )( 1 ( 1

n M M

q p k



  

We showed:

) (modn M Cd 

In other words, the original message:

n C M

d mod



14

Decryption example: “2081 2182”

43  p 59  q 13  e 2537    q p n

1 ) 58 42 , 13 gcd( )) 1 )( 1 ( , gcd(      q p e

937

n C M

d mod



1819 2537 mod 2081937  1415 2537 mod 2182937 

“1819 1415” “18 19 14 15” = “STOP” Compute = inverse of modulo =

e

58 42

d