i am trying to solve RSA algorithm, how i generate public key? in mathematical problem…i can't understand












0















Select two prime no's. Suppose P = 53 and Q = 59.
Now First part of the Public key : n = P*Q = 3127.



We also need a small exponent say e :
But e Must be



An integer.



Not be a factor of n.



1 < e < Φ(n) [Φ(n) is discussed below],
Let us now consider it to be equal to 3.



Our Public Key is made of n and e
Generating Private Key :



We need to calculate Φ(n) :
Such that Φ(n) = (P-1)(Q-1)

so, Φ(n) = 3016



Now calculate Private Key, d :
d = (k*Φ(n) + 1) / e for some integer k
For k = 2, value of d is 2011.










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  • 2





    Probably a better fit in Cryptography, SO is about coding.

    – zaph
    Nov 14 '18 at 8:35






  • 1





    I'm voting to close this question as off-topic because this is about crypto, not programming.

    – Maarten Bodewes
    Nov 14 '18 at 15:55
















0















Select two prime no's. Suppose P = 53 and Q = 59.
Now First part of the Public key : n = P*Q = 3127.



We also need a small exponent say e :
But e Must be



An integer.



Not be a factor of n.



1 < e < Φ(n) [Φ(n) is discussed below],
Let us now consider it to be equal to 3.



Our Public Key is made of n and e
Generating Private Key :



We need to calculate Φ(n) :
Such that Φ(n) = (P-1)(Q-1)

so, Φ(n) = 3016



Now calculate Private Key, d :
d = (k*Φ(n) + 1) / e for some integer k
For k = 2, value of d is 2011.










share|improve this question


















  • 2





    Probably a better fit in Cryptography, SO is about coding.

    – zaph
    Nov 14 '18 at 8:35






  • 1





    I'm voting to close this question as off-topic because this is about crypto, not programming.

    – Maarten Bodewes
    Nov 14 '18 at 15:55














0












0








0








Select two prime no's. Suppose P = 53 and Q = 59.
Now First part of the Public key : n = P*Q = 3127.



We also need a small exponent say e :
But e Must be



An integer.



Not be a factor of n.



1 < e < Φ(n) [Φ(n) is discussed below],
Let us now consider it to be equal to 3.



Our Public Key is made of n and e
Generating Private Key :



We need to calculate Φ(n) :
Such that Φ(n) = (P-1)(Q-1)

so, Φ(n) = 3016



Now calculate Private Key, d :
d = (k*Φ(n) + 1) / e for some integer k
For k = 2, value of d is 2011.










share|improve this question














Select two prime no's. Suppose P = 53 and Q = 59.
Now First part of the Public key : n = P*Q = 3127.



We also need a small exponent say e :
But e Must be



An integer.



Not be a factor of n.



1 < e < Φ(n) [Φ(n) is discussed below],
Let us now consider it to be equal to 3.



Our Public Key is made of n and e
Generating Private Key :



We need to calculate Φ(n) :
Such that Φ(n) = (P-1)(Q-1)

so, Φ(n) = 3016



Now calculate Private Key, d :
d = (k*Φ(n) + 1) / e for some integer k
For k = 2, value of d is 2011.







cryptography






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asked Nov 14 '18 at 7:04









Tara JanwaTara Janwa

1




1








  • 2





    Probably a better fit in Cryptography, SO is about coding.

    – zaph
    Nov 14 '18 at 8:35






  • 1





    I'm voting to close this question as off-topic because this is about crypto, not programming.

    – Maarten Bodewes
    Nov 14 '18 at 15:55














  • 2





    Probably a better fit in Cryptography, SO is about coding.

    – zaph
    Nov 14 '18 at 8:35






  • 1





    I'm voting to close this question as off-topic because this is about crypto, not programming.

    – Maarten Bodewes
    Nov 14 '18 at 15:55








2




2





Probably a better fit in Cryptography, SO is about coding.

– zaph
Nov 14 '18 at 8:35





Probably a better fit in Cryptography, SO is about coding.

– zaph
Nov 14 '18 at 8:35




1




1





I'm voting to close this question as off-topic because this is about crypto, not programming.

– Maarten Bodewes
Nov 14 '18 at 15:55





I'm voting to close this question as off-topic because this is about crypto, not programming.

– Maarten Bodewes
Nov 14 '18 at 15:55












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