The University of Babylon
Department of Software
LECTURE NOTES ON PUBLIC‐ KEY CRYPTOGRAPHY (RSA and ElGamal)
By
Dr. Samaher Hussein Ali
College of Information Technology, University of Babylon, Iraq
Samaher_hussein@yahoo.com
4/11/2012
Public‐key cryptography
Called also asymmetric cryptography The keys used to encrypt and decrypt are different.
Anyone who wants to be a sender needs to “publish” an encryption key, which is known as the public key, KB.
Anyone who wants to be a receiver needs a unique decryption key, which is known as the private key, kv.
If B wants to send an enciphered text to A, B should knows the the encryption algorithm and A’s public key.
4/11/2012
Dr. Samaher Hussein Ali
Notes of Lecture 8
Confidentiality via Public key cryptography
 Samaher wants to send a secret message m to Eman
 Eman should have 2 keys: public KB and private kv
 Prior to message encryption, samaher gets by some means an authentic copy
of Eman’s public key (i.e., the encryption key)
4/11/2012
Dr. Samaher Hussein Ali
Notes of Lecture 8
Public‐key Systems

 It provides confidentiality by depend on digital signatures.
 It reduce number of using keys , example; if we need communication among 100 users in tradition methods, we need
(n(n-1)/2)= 100(99)/2=4950 keys
while, when we use the public key system need only 100 keys
RSA

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It is named after it inventors Ron Rivest, Adi Shamir and Len Adleman.
Published in 1978‫۔‬
It is the most widely used public‐key encryption algorithm today.
It provides confidentiality and digital signatures.
Its security is based on the difficulty of integer factorization
4/11/2012
Dr. Samaher Hussein Ali
Notes of Lecture 8
RSA algorithm (key generation for RSA public‐key encryption)
 Each entity A creates a public key and a corresponding private key by doing
the following
 Who is capable of computing public key?
 Generate three prime numbers p and q, e
 Compute n=p*q
 A’s Public key: (e, n) // to be published
 Who is capable of computing d?
d 
gcd(  ( n )) *  ( n )  1
e
‐‐ (n)=(p‐1)(q‐1)
 To verification d : we must [e*d mod (n)  1]
 A’s private key: (d, n)) // to be kept secretly by A
4/11/2012
Dr. Samaher Hussein Ali
Notes of Lecture 8
RSA Encryption/Decryption
 Summary: A encrypts a message m for B. Upon reception, B decrypts it using its private key.
 Encryption: A should do the following
 Obtain B’s authentic public key (n,e).
 Represent the message as an integer in the interval [0,n‐1]
 Compute c = me mod n // Encryption
 Send the ciphertext c to B
 Decryption: to recover plaintext m from c, B does the following
 Use the private key d to recover m = cd mod n // Decryption
4/11/2012
Dr. Samaher Hussein Ali
Notes of Lecture 8
RSA Encryption/Decryption
4/11/2012
Dr. Samaher Hussein Ali
Notes of Lecture 8
Example 1: Let, p=3, q=17, e=5: Find the public and Secret key then cipher the message ABC n=p*q=3*17=51
Public key ‐‐‐‐‐‐<e,n>=<5,51>
d=(GCD((n))* (n)+1)/e
(n)= (p‐1)(q‐1)=(3‐1)(17‐1)=(2*16)=32
GCD((n))= GCD[(P‐1)(q‐1)]=GCD(2,16)=2
d=((2*32+1)/5)= (64+1) /5 =65/5=13
Secret Key‐‐‐‐‐‐‐<d,n>=<13,51>
To verification d :
e.d mod (n)=1
5*13 mod 32
65 mod 32 =1
4/11/2012
Dr. Samaher Hussein Ali
Notes of Lecture 8
Example 1: Let, p=3, q=17, e=5: Find the public and Secret key then cipher the message ABC 2. Find the cipher text using the public key <e,n>=<5,51>
Ci=Mie mod n
C1=(1)5 Mod 51=1 mod 51 =1
C2=(2)5 Mod 51=32 mod 51 =32
C1=(3)5 Mod 51=243 mod 51 =39
3. Find the cipher text using the Secret Key‐‐‐‐‐‐‐<d,n>=<13,51>
Mi=Cid mod n
M1=(1)13Mod 51=1
M2=(32)13 Mod 51=2
M1=(39)13 Mod 51=3
4/11/2012
Dr. Samaher Hussein Ali
Notes of Lecture 8
Example 2: Confidentiality
 Take p = 7, q = 11, so n =p*q= 77  (n) =(p‐1)(q‐1) =(7‐1)(11‐1)=60
 Let e = 17 ‐‐‐‐‐‐‐‐ public key(17,77)
 Compute d =[gcd(6,10)*60+1]/17= 53
 17 x 53 mod 60 = 1 ‐‐‐‐‐‐‐‐ secret key(53,77)
 samaher wants to secretly send Eman the message HELLO [07 04 11 11 14]
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0717 mod 77 = 28
0417 mod 77 = 16
1117 mod 77 = 44
1117 mod 77 = 44
1417 mod 77 = 42
 samaher sends ciphertext [28 16 44 44 42]
4/11/2012
Dr. Samaher Hussein Ali
Notes of Lecture 8
Example 2: Confidentiality
 Eman receives [28 16 44 44 42]
 Eman uses private key < 53,77>, to decrypt the message:

 2853 mod 77 = 07
H
 1653 mod 77 = 04
E
 4453 mod 77 = 11
L
 4453 mod 77 = 11
L
 4253 mod 77 = 14
O
No one else could read it, as only Eman knows his private key and that is needed for decryption
4/11/2012
Dr. Samaher Hussein Ali
Notes of Lecture 8