IMT4551 – Selected topics in cryptology
Exercise 4
1801
1. Evaluate the Jacobi symbol (8191), using the properties of the Jacobi symbol.
The following theorem can be useful in the computation of the Jacobi symbol:
Theorem
Let n be an odd positive integer and let a,b0. Then the following identities hold
1.
 0  1,if n  1
 
 n  0 ,if n  1
2.
 2  1,if n  1 mod 8
 
 n   1,if n  3 mod 8
3.
a b
     if ab (mod n)
n n
4.
 ab   a  b 
     
 n   n  n 
5.
 n
   if a  n  3 mod 4
a  a
 
 n   n 
  , otherwise
 a 
We are going to check those properties in this order: 1, 5, 3, 4, 2 and thus reduce the
complexity of the evaluation problem.
 1801  5  8191  3  987  5  1801  3

 
 
 

 8191   1801   1801   987 
3 814

 4  2  407  2  407  5  987  3

 

  
 

 987   987  987   987   407 
3 173

 5  407  3  61  5  173  3  51  5

 
 
 
   
 407   173   173   61   61 
5 61 3 10 4
     2  5  2  5  5
          
 51   51   51  51   51 
5
 51  3  1 
       1
 5  5
2. Check whether n=8633 is prime, by means of the Solovay-Strassen algorithm. Set a=7019
7019
We first have to compute the Jacobi symbol (8633). We use the properties of the Jacobi
symbol.
 7019  5  8633  3  1614  4  2  807  2

 
 
 


 8633   7019   7019   7019  7019 
2
 807  5  7019  3  563  5  807  3
 
 
 
  

 7019   807   807   563 
2
3
 244  4  2   61  2  61  5  563  3  14  4
 
  
 
  
  
    
 563   563   563   563   61   61 
4
 2  7  2  7  5  61  3  5  5
           
 61  61   61   7   7 
5 7 3 2 2
   
      1
5 5
Then we compare the computed value of the Jacobi symbol with 𝑦 = 𝑎
𝑦 = 7019
8633−1
2
𝑛−1
2
mod 𝑛
mod 8633 = 70194316 mod 8633
We use modular exponentiation. 431610=10000110111002 (use the arrow algorithm).
Thus we have
i
0
1
2
3
4
5
6
7
8
9
10
11
12
2𝑖
20
21
22 *
23 *
24 *
25
6
2 *
27 *
28
29
210
211
12
2 *
𝑖
70192 mod 8633
7019
6463
3915
3650
1781
3650
1781
3650
1781
3650
1781
3650
1781
So we have
70194316 mod 8633 = 3915 ∙ 3650 ∙ 1781 ∙ 1781 ∙ 3650 ∙ 1781 = 5784 (mod 8633)
We conclude that 8633 is not prime.
3. Factorize the integer 8633 using Fermat’s factorization.
Let n=a×b, where a and b are close together. Then
n  ab 
a 2 ab b 2  a 2 ab b 2


 

4
2
4  4
2
4
2
2




 a b   a b 
2
2

 
  t  s  t  s t  s 
 2   2 
As a and b are close, s is small, so t is only slightly larger than √𝑛 . In that case we can
find a and b by trying all values of t starting with ⌊√𝑛 + 1⌋, until we find one for which
𝑠 2 = 𝑡 2 − 𝑛 is a perfect square.
We first compute the square root of 8633
√8633 = 92,91
That means that we should try the integers 92+1, 92+2, …
92+1=93
932-8633=16=42
Since 16 is a perfect square, we have
8633=932-42=(93-4)(93+4)=8997
4. Factorize the integer 82123 using the Pollard’s rho algorithm. Set x0=631 and f(x)=x2+1.
We give the Pollard’s rho algorithm here:
Let x0 be a random integer in Zn
Let y0x0, i0
repeat
ii+1; xif(xi-1) mod n; yif(f(yi-1)) mod n
if 1<(xi-yi,n)<n then return (xi-yi,n)
else if (xi-yi,n)=n then return “failure”
In our case, n=82123, x0=631 and f(x)=x2+1. We set y0=x0=631, i=0.
ii+1=1
x1=f(x0) mod n=6312+1 mod 82123 = 69670
y1=f(f(y0)) mod n = (6312+1)2+1 mod 82123 = 28986
By using Euclidean algorithm, we get that (69670-28986,82123)=(40684,82123)=1
ii+1=2
x2=f(x1) mod n=696702+1 mod 82123 = 28986
y2=f(f(y1)) mod n = (289862+1)2+1 mod 82123 = 13166
By using Euclidean algorithm, we get that (28986-13166,82123)=(15820,82123)=1
ii+1=3
x3=f(x2) mod n=289862+1 mod 82123 = 69907
y3=f(f(y2)) mod n = (131662+1)2+1 mod 82123 = 40816
By using Euclidean algorithm, we get that (69907-40816,82123)=(29091,82123)=1
ii+1=4
x4=f(x3) mod n=699072+1 mod 82123 = 13166
y4=f(f(y3)) mod n = (408162+1)2+1 mod 82123 = 20459
By using Euclidean algorithm, we get that (13166-20459,82123)=(-7923,82123)=
=(-7923+82123,82123)=(74830,82123)=1
ii+1=5
x5=f(x4) mod n=131662+1 mod 82123 = 64027
y5=f(f(y4)) mod n = (204592+1)2+1 mod 82123 = 6685
By using Euclidean algorithm, we get that (64027-6685,82123)=(57342,82123)=1
ii+1=6
x6=f(x5) mod n=640272+1 mod 82123 = 40816
y6=f(f(y5)) mod n = (66852+1)2+1 mod 82123 = 75835
By using Euclidean algorithm, we get that (40816-75835,82123)=(-35019,82123)=
=(-35019+82123,82123)=(47104,82123)=1
ii+1=7
x7=f(x6) mod n=408162+1 mod 82123 = 80802
y7=f(f(y6)) mod n = (758352+1)2+1 mod 82123 = 17539
By using Euclidean algorithm, we get that (80802-17539,82123)=(63263,82123)=41.
The algorithm stops and 41 is a factor of 82123. Thus 82123=412003.
5. Apply the baby-step giant-step algorithm to find the discrete logarithm of 3 and
∗
10 in 𝑍23
using the generator  = 5.
n=23, =5, =3. Then 𝑚 = ⌈√23⌉ = 5
We produce the table of powers of j, 0jm-1 first and then we sort it
j
5j mod 23
0
1
1
5
2
2
3
10
4
4
j
5j mod 23
0
1
2
2
4
4
1
5
3
10
Then we compute -m=-5=(5)-1=(55)-1 mod 23=20-1. We use the extended Euclidean
algorithm.
23=120+3
20=63+2
3=12+1
So we have
1=3-2=3-(20-63)=3-20+63=73-20=7(23-20)-20=723-720-20=723-820
If we take modulo 23 on both sides we get
1=-820=1520
That means -m=15. We set ==3
Now for i=0,1,2,3,4 we try whether -m can be found in the second row of the table.
i=0, =3
i=1, =315 mod 23 = 22
i=2, =2215 mod 23 = 8
i=3, =815 mod 23 = 5
Since 5 can be found in the second row of the sorted table for j=1, we get
log53=im+j=35+1=16
n=23, =5, =10. Then 𝑚 = ⌈√23⌉ = 5
We use the same tables from the first example, since they do not depend on . We see
that ==10 can be found in the second row of the sorted table for i=0, j=3. Thus
log510=im+j=05+3=3.
∗
6. Let q=113, and =3 be a generator of 𝑍113
. Alice and Bob want to establish a common
secret key by means of the Diffie-Hellman secret key exchange algorithm. If Alice
generates a random number x=30 and Bob generates a random number y=59, show how
they establish the common secret key.
Alice computes a=x mod q=330 mod 113=91 and sends a to Bob. Bob computes
b=y mod q =359 mod 113=86 and sends b to Alice.
Then Alice computes bx mod q=(y )x mod q=xy mod q =8630 mod 113=87. Bob computes
ay mod q=(x )y mod q=xy mod q =9159 mod 113=87.