1. Sample Solution Cryptology Design Fundamentals
Grundlagen des kryptographischen Systementwurfs
Lecture ID: ET-IDA-28
Final Examination
Closed-book short question part
Prof. W. Adi
Date :
Duration : 20 Minutes , Maximum marks is 30%
Sample Solution
Vorname ………………………………………..
Nachname ………………………………………..
Matrikel-Nr. ………………………………………..

2. Marks: ∑ 2

3. gcd ( a 7 t – a t , a 9 t – a t ) = a t gcd ( a 6 t – 1 , a 8 t – 1 )
Q1: Compute gcd(910,280).
(1 P) n1 n2 q r
910
280 3 70 4
 gcd ( 910 , 280 ) = …70..
Q2: Compute gcd [ (a7 t - at) , (a9 t - at) ], where a and t are non-zero positive integers and gcd (t,9)=gcd(t,7)=1.
(2 P)
gcd ( a 7 t – a t , a 9 t – a t ) = a t gcd ( a 6 t – 1 , a 8 t – 1 )
= a t (a gcd ( 6 t, 8 t ) – 1) = a t (a 2t–1)
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?
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4. Q3: On which claimed unsolvable problems are the securities of the following cryptosystems based?
1. RSA Signature System
Integer factoring problem
2. Omura Proof of Identity Protocol
Discrete Logarithm Problem
3. Omura-Massey Lock for Shamir’s 3-Pass Protocol over GF(p)
Discrete Logarithm Problem over GF(p)
4. Fiat Shamir proof of identity protocol
(4 P)

5. Since ord(3) = 17-1 = 16 as a primitive element
Q4: Compute the multiplicative order of 34 in GF(17) assuming that 3 is a primitive element in GF(17).
Since ord(3) = 17-1 = 16 as a primitive element
Q5: How many elements are there in the group of units Z*m for m =187= 11· 17.
Compute the highest possible multiplicative order for a unit in Z*m
(*)Compute the multiplicative order of the element 2 modulo 187 (optional question +4p)
(3 P)
# of elements in the group is ( ) = (17-1) (11-1) = 160
Highest possible order is: ( ) = lcm [(17) , (11)]
= lcm [(17) , (11)] = lcm [16, 10]
= / gcd(16,10) = 160/2 = 80
(5 P)
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?
Order divides 80 that is 1, 2, 4, 5, 8, 10, 16, 20, 40, or 80
21 = 2 ≠ 1, 22 = 4 ≠ 1, 24 = 16 ≠ 1, 25 = 32 ≠ 1, 28 = 69 ≠ 1, 210 = 89 ≠ 1, 216 = 86 ≠ 1
220 = 67 ≠ 1, 240 = 1 order of 2 is = 40
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6. Q6: Reduce the following expressions to the smallest positive integers in the corresponding deployed algebra: 1. R13 ( – ( 27 ) 5 · 28 2 ) =
R13 ( – (1 ) 5 (2)2 ) = R13 (-1 - 4) = R13 (- 5) =8
2. ( 1 – 2x 2 ) ( 2 + 3x2) over GF(5) =
2 + 3 x2 – 4 x2 - 6x4
= 2 – x2–x4
= 2 + 4x2 + 4 x4
(3 P)

7. and the order of a group’s element?
Q7: Which relationship do exist between the number of the group elements
and the order of a group’s element?
Q8: What is a mathematical „Involution“ function?
Is a function which is equal to its inverse function. ( F= F-1 )
(2 P)
The order of the group element divides the number of its elements
Or the order of each element divides the group’s order (Lagrange Theorem)
(2 P)
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?
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8. + Shamir Perfect Secret Sharing RAND Z
Q9: Sketch Shamir perfect secret sharing scheme for two users
(3 P)
Shamir Perfect Secret Sharing
Shamir
RAND
Random
BSS Z
Secret
Give User A
Common Secret
Between A and B +
Exchange to generate
Common secret
RAND + Z
Give User B

9. Compute the possible multiplicative orders for elements in GF(29)?
Q10: In GF(29).
Compute the possible multiplicative orders for elements in GF(29)?
Possible multiplicative orders are the divisors of (29) = 29-1=28= 2 x 2 x7.
These are: 1, 2, 4, 7,14,28
Compute the number of primitive elements in GF(29).
# of primitive elements (28) = (22.7) = 28 (1- 1/2) . (1- 1/7) = 12
3. Which are the minimum number of tests required to find out weather a given element β in GF(29) is primitive?
β1 ≠ 1, β2 ≠ 1, β4 ≠ 1, β7 ≠ 1, β14 ≠ 1
(6 P)
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?

10. 2-6 = 2t = 2-6 mod 28= 2-6 + 28= 2 22 => order of 2 is 28
Compute the multiplicative order of 2 in GF(29) ≠ 1, 22 ≠ 1, 24 =16≠ 1, 27 =128=12≠ 1, 214 =144=28≠ 1
=> order of 2 is 28
Compute the smallest positive integer t for which 2-6 = 2t holds.
2-6 = 2t = 2-6 mod 28= = 2 22
=> t = 22
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11. Compute a8 and give the corresponding binary vector for a8 .
Q11: GF(26) is generated by the irreducible and primitive polynom P(x)= x6 + x + 1. The element a = = x + 1 is selected from GF(26).
Compute a8 and give the corresponding binary vector for a8 .
a = (x +1), a 2 = (x 2 + 1), a 4 = (x 2 + 1) 2 = x 4 + 1,
a 8 = (x 4 + 1) 2 = x = x 3 + x =
as x 6 + x + 1 = 0  x 6 = x  x 7 = x 2 + x  x 8 = x 3 + x 2
2. Compute the multiplicative order of a (Hint: a= 1+x = x6 )
As P(x) is primitive, the order of x is 26-1=63, as x6=1+x => ord (1+x) = ord( x6)
3. Compute the smallest positive integer t for which a-1 = at holds.
The modulus in the exponent for a is the order of a=63
ord (a) = ord (x6) = (ord x) / gcd (ord x , 6) = (2 6 – 1) / gcd (2 6 – 1 , 6) = 63/3 = 21
 a -1 mod 21 = a = a 20  t = 20
(10 P)
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?

12. Annex: Euler Function (m) (m) = m ( 1 - ) ( 1 - ) ……
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 ) ]
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