1. Network Security Sample Solution Short questions (Closed book)
Final Examination Lecture ID: ET-IDA ( )
Short questions (Closed book)
Duration: Minutes v11
Please write your answer on this question sheet.
Name: ……………………………………………..
Matr. Nr.: ………………………….………………
Sample Solution
Prof. W. Adi

2. Andre Zierfuß, Arther Strasser
Many thanks to :
Andre Zierfuß, Arther Strasser
For their valuable contribution to the sample solution
Seite 1/8

3. Number of invertible elements modulo 63 is  ( 63 )
K1: Compute the gcd(627,494).
(1 P) n1 n2 r
627
494
133 95 38 19
 gcd ( 627,494 ) = 19
K2: Compute the gcd( , ).
(1 P)
gcd ( 233 – 1 , 221 – 1 ) = 2 gcd ( 33,21 ) – 1 = – 1 = 7
gcd ( 33,21 ) = 3
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?
K3: Compute the number of integers smaller than 63 having multiplicative inverses modulo 63.
(1 P)
Number of invertible elements modulo 63 is  ( 63 )
 ( 63 ) = ( 32 x7 ) = 63 (1 – 1/3) (1 – 1/7) = 36 elements

4. 1. Diffie-Hellman key-exchange system Discrete logarithm problem
K4: On which claimed unsolved mathematical problems are the following cryptosystems based?
1. Diffie-Hellman key-exchange system
Discrete logarithm problem
2. Rabin lock
Integer factoring (computing square root in a ring)
3. „Blind Signature“
As RSA system, that is integer factoring
(5 P)
Seite 2/8

5. 2. Compute the highest multiplicative order of an element in Z*m?
K5: 1. How many elements are contained in the group of units Z*m, if m = 51· 32 ?
2. Compute the highest multiplicative order of an element in Z*m?
3. Compute the possible multiplicative orders in Z*m?
4. Compute the multiplicative order of the element 7
(7 P)
Number of elements in Z*m =  (51· 32 ) = 51· 32 ( 1-1/5)( 1-1/3 ) = 24
The highest multiplicative order is  (51· 32 ) called Carmichael’s function  ( m ).
 (51· 32 ) = lcm [  (5 ) ,  (32 ) ]
= lcm [ ( 4 ), (32 (1-1/3))]
= lcm ( 4,6 ) = 4x6/ gcd(4,6) = 12
Divisors of  ( m ) = 12. That is: 1,2,3,4,6, 12 (optimum answer)
Or more general the divisors of  (51· 32 ) = 24
As 71 ≠ 1, 72 =49= 4 ≠ 1, 73 = 28 ≠ 1, = 16 ≠ 1, = 4x16=64=19 ≠ 1
 the order of 7 is 12

6. K7: What is a mathematical „Involutions“ function?
K6: Reduce the following statements to their simplest positive value: 1. R41 ( (83)2 · 432 ) = R41 ( (1)2 · 22 ) = R41 ( · 4 )
= R41 ( -3 ) = = 38
(3 P)
K7: What is a mathematical „Involutions“ function?
(1 P)
Involution: is a self-inverting mathematical function F
that is F = F -1 holds.
Seite 3/8

7. K8: The following questions refer to GF(43).
Compute the possible multiplicative orders of elements in GF(43)?
Compute the number of primitive elements in GF(43).
(6 P)
The divisors of  (43 ) = 43-1= 42 = 2 x 3 x 7. These are: 1,2,3,6,7,14,21,42
Primitive elements have the highest possible order. That is 42
Number of elements having the order 42 =  (42 ) =  (2x3x7) = (2-1)(3-1)(7-1) = 12
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?

8. Compute the multiplicative order of the element 2 in GF(43).
Which minimum number of tests are required to check if an element β in GF(43) is primitive?
Compute the multiplicative order of the element 2 in GF(43).
Compute the smallest t, for which 2-1 = 2t holds.
Possible orders are the divisors of 42, that is: If all the following tests are true
β1 ≠ 1 and β 2 ≠ 1 and β3 ≠ 1 and β6 ≠ 1 and β7 ≠ 1 and β14 ≠ 1 and β21 ≠ 1
then β is primitive!
As 21 ≠ 1, 22 = 4 ≠ 1, 23 = 8 ≠ 1, = 64=21 ≠ 1, 27 = 21x2=42 =-1 ≠ 1, 214 = -1x-1= 1
 the order of 2 is 14
2 -1 mod 14 = = 2 13  t = 13
As the order of 2 is 14
Seite 4/8

9. Compute a8 and give its binary vector presentation.
K9: GF(26) is generated using the irreducible and primitive polynomial P(x)= x6 + x + 1 as modulus. An element a = = x + 1 of GF(26) is chosen.
Compute a8 and give its binary vector presentation.
Compute the multiplicative order of a3 (Hint: a= 1+x = x6 )
Compute the smallest t for which a-1 = at holds.
(5 P)
a = (x +1), a 2 = (x2 + 1), a 4 = (x4 + 1), a 8 = (x8 + 1) = x3 + x2 + 1 =
as x6 + x + 1 = 0  x6 = x + 1  x7 = x2 + x x8 = x3 + x2
a 3 = (x + 1)3 = (x6 )3 = x18
As P(x) is a primitive polynomial, the order of x is = = 63
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?
a = (x + 1) = (x6 )
a -1 mod 21 = a = a 20  t = 20
As the order of a is 21. Using the modulus 63 in the exponent would work,
however the solution would not deliver the required minimum!.

10. On the application layer
K10: On which ISO-OSI layer is SSL located?
(2 P)
On the application layer
K11: Sketch the principle of „Message Authentication Code“ MAC for a message M with a private key Z and a Hash function H.
(10 P)
Source
Destination Z H H
MAC
Message Authentication Code
Z : shared secret key
Seite 5/8

11. K12: For which network security functions is Kerberos used?
Does Kerberos use open or secret key security system?
How many secret-keys are needed for n users in Kerberos system?
(8 P)
Key distribution, Authentication and Confidentiality
Secret-key based concepts
n-keys
K13: Sketch the „Public-Key“ certification scheme in X.509
(6P)
Message
digest CA
Private Key
Verification by CA’s Public Open Key which has to be assumed as somehow initially trustable!
Document
to be signed
Signature

12. K16: What for is „Blind Signature“ in e-Cash Systems used?
K14: What security tasks are attained after a one-way authentication in X.509?
(4 P)
- Authenticates the identity of sender and that the message was generated by the sender
- The message was intended for receiver
- Establishes the integrity and originality of the message; presents credentials
K15: Name two differences between account-based electronic payment systems and the electronic cash systems
(4 P)
Account based system Cash-system
Needs a bank to transfer value no bank required to transfer value
Needs online TTP access Not necessarily consulting online TTP
Can not directly respent can directly re-spent
Single hop multiple hop
No peer to peer transfer offline peer to peer transfer possible
K16: What for is „Blind Signature“ in e-Cash Systems used?
(2 P)
To make coins and users anonymous.
K17: What for is “Secret Splitting technique” used in e-Cash systems?
(2 P)
To detect double spending and disclose identify of the user who double spent the-cash coins
Seite 6/8

13. Passwords pi ‘s are recalled in the reversed order
K18: Sketch the concept of „One-Time Password“ procedure.
(10 P)
One-time password scheme based on idea of Lamport (1981)
h is a one-way hash function (MD5 or SHA-1, for example)
User chooses initial seed k0
System calculates:
h(k0) = k1
h(k1) = k2
h(k2) = k3 …
h(kn–2) = kn–1
h(kn–1) = kn pn
pn-1
pn-2 … p2 p1 h
ki+1 ki
Initialize with k0 at t=0
h(p2) = p1
h(pi) = pi-1
Passwords pi ‘s are recalled in the reversed order

14. (Authentication Request) Shared secret ciphering key
K19: Sketch the private key generation scheme for the payload encryption in GSM System!
(8 P)
Mobile station
Network
128 Bit
Random
Generator Ki
(Authentication Request) Ki
RAND A8 A8
64 Bit
64 Bit
Shared secret ciphering key Kc Kc
Seite 7/8

15. K20: Sketch the concept of authentication in Bluetooth Systems
Authentication through challenge-response technique. The device (unit B below) is challenged by a random sequence AU-RAND. The response SRES = E1( AU-RAND, Link key ,…) is compared. The „Link key“ is the shared secret used for authentication.
A sketch of the used technique is illustrated below

16. Carmicheal´s function (m) :
Annex:
Euler Function (m)
For m = p1 p2 p pt
e1 e2 e et
(m) = m ( ) ( ) …… P1 1 P2 1
Carmicheal´s function (m) :
 (2)= 1, (22) = 2, (2e) = 2e for e  3:
(pe)= (pe) = (p - 1)pe-1 for p odd prim.
for m = p1e1 p2e2 p3e pnen
(m) = lcm [ (p1e1 ), (p2e2 ), … (pnen ) ]
Seite 8/8