- Rumyantsev
- 文章數 : 2
積分 : 8
注冊日期 : 2020-05-20
Assume you are to apply the RSA algorithm. You decided to choose the prime numbers p and q as follows: p= 7 and q= 11. You also selected e= 13. (a) What is the public key that you want to advertise? (Hint: find N) (b) What is the totient value: (N) (c) Fi
- Vibhav G.Contributor
- 文章數 : 12
積分 : 12
注冊日期 : 2020-05-01
a)
p=7, q=11
n=p*q, n=77
e =13
the public key is (n, e) (77,13)
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b)
Φ(n) = (p-1)(q-1)
Φ(n) = (6)*(10) = 60
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c)
e*d mod Φ(n) =1
13*d mod 60 =1
step 1: Euclidean alogrithm
60x+13y=1
60 = 4(13)+8
13 = 1(8)+5
8 = 1(5)+3
5 = 1(3)+2
3 = 1(2)+1
step 2: Back substitution
1= 3 - 1( 2 )
1 = 3 -1 (5 - 1(3))
1 = 2( 3 ) - 1(5)
1 = 2(8 -1(5)) - 1(5)
1= 2(8 ) - 3(5)
1= 2(8) - 3(13-1(8))
1 = 5(8)-3(13)
1= 5(60 - 4(13))-3(13)
1 = 5(60) - 23(13)
d=-23
it is negative so one more step
d= Φ(n)-d
d= 60-23
=37
d will act as the private key. our private key is 37.
p=7, q=11
n=p*q, n=77
e =13
the public key is (n, e) (77,13)
=====================================================================================
b)
Φ(n) = (p-1)(q-1)
Φ(n) = (6)*(10) = 60
=====================================================================================
c)
e*d mod Φ(n) =1
13*d mod 60 =1
step 1: Euclidean alogrithm
60x+13y=1
60 = 4(13)+8
13 = 1(8)+5
8 = 1(5)+3
5 = 1(3)+2
3 = 1(2)+1
step 2: Back substitution
1= 3 - 1( 2 )
1 = 3 -1 (5 - 1(3))
1 = 2( 3 ) - 1(5)
1 = 2(8 -1(5)) - 1(5)
1= 2(8 ) - 3(5)
1= 2(8) - 3(13-1(8))
1 = 5(8)-3(13)
1= 5(60 - 4(13))-3(13)
1 = 5(60) - 23(13)
d=-23
it is negative so one more step
d= Φ(n)-d
d= 60-23
=37
d will act as the private key. our private key is 37.
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