Download Mühendislik Fakültesi Diferansiyel Denklemler Vize Sınav

Fen Edebiyat Fakültesi Matematik Bölümü
Introduction to Cryptography (Şifrelemeye Giriş)
Final Exam (Final Sınav Soruları) Güz Yarıyılı-2013-2014
Instructor: Prof. Dr. İrfan ŞİAP
Question:
Grade:
1
2
3
4
Student’s
5
Total
Name Surname:
Number:
Show your work to get full credit. Time is 90 minutes. Good luck.
1.
a.
Solve the congruence x11  2(mod 41)
b.
Compute  ( (1905))
a. (11, 40)  1 and 11 11  3  40  1
 3 40
x  x1111
 ( x11 )11 ( x40 )3  211  39(mod 41)
b.  ( (1905))   ( (3  5  127))   (2  4  126)   (24  32  7)  8  6  6  288
2.
a.
b.
Assume that Alice publishes her RSA public key: N  33 , e  3 .
Encrypt the message “L&M” and find the ciphertext.
Assume that Oscar obtained the ciphertext “SIL”Help him to find the plaintext.
a)) “L & M”  14 | 29 |15 Therefore
143  5 (mod33), 293  2 (mod33), 153  9 (mod33) Hence the ciphertext is “ECH”
b)) N  3  11 Therefore we can find the decryption key d wchich is ed  1 (mod  ( N )) .
Since  (33)  2  10  20 the decryption key d is 7.
SIL  21|10 |14
217  21 (mod33), 107  10 (mod33), 147  20 (mod33)
Thus the plaintext is “SIR”
3.
Show that 159991 is not a prime number. (Hint: Use difference of squares
method.)
b.
Show that a  2 is a Rabin-Miller witness for (the compositeness of) n  49.
a.
159991  12  15999
159991  22  159995 So 159991  4002  32  (397)(403) is not a prime number.
159991  32  160000
n  1  48  24  3
23  8  1 (mod 49)
22 3  8  1 (mod 49)
0
b))
22 3  15  1 (mod 49)
1
22 3  29  1 (mod 49)
2
22 3  8  1 (mod 49)
Therefore 2 is a Rabin Miller witness for the compositeness of n  49.
3
4. Use the data provided to find values of a and b satisfying a2  b2 (mod N ) by illustrating
linear equation system and then compute gcd( N , a  b) in order to find a nontrivial factor
of N .
N  16873
13002  2700 (mod16873) and 2700  22  33  52
50032  7350 (mod16873) and 7350  2  3  52  7 2
31502  1176 (mod16873) and 1176  23  3  7 2
 0 1 1
 0

  u1   
 1 1 1 u    0
 0 0 0   2   0 

  u3   
 0 0 0
 0
 1 0 0
 0

  u1   
 0 1 1 u    0
 0 0 0   2   0 

  u3   
 0 0 0
 0
(0,1,1) is a nontrivial solution. So it gives a congruence a 2  b2 (mod 16873) that
has the potential to provide a factorization of 16873. It says that if we multiply the
2nd and 3rd numbers in the list we will get a square, indeed
50032  31502  (2  3  52  7 2 )(23  3  7 2 ) (mod 16873)
 (2 2  3  5  7 2 ) 2  2940 2
Next we compute
gcd(16873,5003  3150  2940)  gcd(16873,15756510)  359
Therefore 359 |16873 .
5.
Determine whether the following numbers are quadratic residue in the given
modulus by using Legendre symbol.
a) 102 in modulo 37.
b) 41 in modulo 163.
c) -23 in modulo 83.
d) Write down all quadratic residues and nonresidues in modulo 17.
2
 102   28   2   7   7   37   2 
a)) 
                   1 So 102 is QR
 37   37   37   37   37   7   7 
 41   163   1 
b)) 

     1 So 41 is QR
 163   41   41 
 23   60   2   3   5   3   5 
 83   83 
 2  3
c)) 
                             1
 83   83   83   83   83   83   83 
 3  5 
 3  5 
So -23 is NR
2
12  1
22  4
32  9
4 2  16
d))
52  8
So QR  1, 2, 4,8,9,13,15,16 and NR  3,5,6,7,10,11,12,14
62  2
7 2  15
82  13
Appendix: Correspondece between Turkish alphabet and
33
A
B
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Ç
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Ğ
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L
0
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M
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T
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&
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