[PDF] Logique et calcul : La cryptographie RSA vingt ans après PDF RSA.pdf

Le système de cryptage RSA a été inventé en L'algorithme RSA est moins rapide que les algorithmes lorsque, par exemple, vous souhaitez mener des

Enfin, elle publie dans un annuaire, par exemple sur le web, sa clé publique : Alice public (RSA,p,q) sk = d Exercice 1 On consid`ere les valeurs p = 53,q = 11 et e = 3 a) Calculez la Appliquez cet algorithme pour factoriser 899, 110417

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[PDF] Logique et calcul : La cryptographie RSA vingt ans après

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The RSA Encryption Scheme

Suppose Alice wants her friends to encrypt email messages before sending them to her. Computers represent text as long numbers (01 for \A", 02 for \B" and so on), so an email message is just a very big number. The RSA Encryption Scheme is often used to encrypt and then decrypt electronic communications.

General

Alice's Setup:

Chooses two prime numbers.

Calculates the productn=pq.

Calculatesm= (p1)(q1):

Chooses numberseanddso thatedhas a

remainder of 1 when divided bym.

Publishes her public key (n;e).Example

Alice's Setup:

p= 11 andq= 3. n=pq= 113 = 33: m= (p1)(q1) = 102 = 20:

Ife= 3 andd= 7, thened= 21 has a

remainder of 1 when divided bym= 20.

Publish (n;e) = (33;3).

Bob encrypts a messageMfor Alice:

Finds Alice's public key (n;e).

Finds the remainderCwhenMeis divided

byn. Sends ciphertextCto Alice.Bob encrypts messageM= 14: (n;e) = (33;3).

When 143= 2744 is divided by 33, the re-

mainder isC= 5.

Sends ciphertextC= 5 to Alice.

Alice receives and decrypts ciphertextC:

Uses her private key (n;d).

Finds remainderRwhenCdis divided byn.

Rmatches the messageMthat Bob wanted

to send to Alice!Alice decrypts ciphertextC= 5: (n;d) = (33;7).

When 57= 78125 is divided by 33, the re-

mainder isR= 14.

R= 14 =M, the original message from Bob!

Questions

1. Callie wants to send the messageM= 13 to Alice. Using Alice's public and private keys,

calculate the ciphertextC, and the value forRwhen Alice recovers the message.

2. Dexter wants to set up his own public and private keys. He choosesp= 23 andq= 19 with

e= 283. Finddso thatedhas a remainder of 1 when divided by (p1)(q1).

Connection to the Real World

When your internet browser shows a URL beginning with https, the RSA Encryption Scheme is being used to protect your privacy. For

example, if you log in to Facebook, your computer plays the role of Alice and the Facebook server plays the role of Bob, encrypting and

decrypting the information passed back and forth. In practice, the primespandqare chosen to be very big numbers.

Mathematics is the foundation of modern encryption.

For more Real-World Problems Being Solved by Mathematics, visit http://www.cemc.uwaterloo.ca/resources/real-world.html.

Solution:

1.Callie encrypts messageM= 13:

Alice's public key is (n;e) = (33;3).

WhenMe= 133= 2197 is divided by 33, the remainder isC= 19.

Callie sends to Alice ciphertextC= 19.

Alice receives and decrypts ciphertextC= 19:

Alice uses her private key (n;d) = (33;7).

When 197= 893;871;739 is divided by 33, the remainder isR= 13.

R= 13 =M, the original message from Callie!

2. Withp= 23;q= 19, we havem= (p1)(q1) = 22(18) = 396.

We want to nddso thated= 283dhas a remainder of 1 when divided bym= 396. One way to do this is by simple trial and error, increasing the value ofduntil 283ddivided by 396 leaves a remainder of 1.Remainder when d283d283dis divided by 3961283283

2566170

384957

41132340

51415227

61698114

719811

We see thatd= 7 works; that ised= 2837 = 1981 leaves a remainder of 1 when divided by 396.
In general, trial and error could take a very long time, as the value ofdcould be a big number. Instead, an ancient technique called Euclid's Algorithm can be used to nddin the linear

Diophantine equation 283d+ 396y= 1.

For more Real-World Problems Being Solved by Mathematics, visit http://www.cemc.uwaterloo.ca/resources/real-world.html.

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[PDF] [PDF] Logique et calcul : La cryptographie RSA vingt ans après

The RSA Encryption Scheme

General

Alice's Setup:

Chooses two prime numbers.

Calculates the productn=pq.

Calculatesm= (p1)(q1):

Chooses numberseanddso thatedhas a

Publishes her public key (n;e).Example

Alice's Setup:

Ife= 3 andd= 7, thened= 21 has a

Publish (n;e) = (33;3).

Bob encrypts a messageMfor Alice:

Finds Alice's public key (n;e).

Finds the remainderCwhenMeis divided

When 143= 2744 is divided by 33, the re-

Sends ciphertextC= 5 to Alice.

Alice receives and decrypts ciphertextC:

Uses her private key (n;d).

Finds remainderRwhenCdis divided byn.

Rmatches the messageMthat Bob wanted

When 57= 78125 is divided by 33, the re-

R= 14 =M, the original message from Bob!

Questions

1. Callie wants to send the messageM= 13 to Alice. Using Alice's public and private keys,

2. Dexter wants to set up his own public and private keys. He choosesp= 23 andq= 19 with

Connection to the Real World

Solution:

1.Callie encrypts messageM= 13:

Alice's public key is (n;e) = (33;3).

Callie sends to Alice ciphertextC= 19.

Alice receives and decrypts ciphertextC= 19:

Alice uses her private key (n;d) = (33;7).

R= 13 =M, the original message from Callie!

2. Withp= 23;q= 19, we havem= (p1)(q1) = 22(18) = 396.

2566170

384957

41132340

51415227

61698114

719811

Diophantine equation 283d+ 396y= 1.