This exercise exhibits what is called a protocol failure. It provides an example

Question

This exercise exhibits what is called a protocol failure. It provides an example where ciphertext can be decrypted by an eavesdropper, without determining the key, if a cryptosystem is used in a careless way. Now, suppose that Bob has an RSA Cryptosystem with a large modulus n for which the factorization cannot be found in a reasonable amount of time. Suppose Alice sends a message to Bob by representing each alphabetic character as an integer between 0 and 25 (i.i. A0,B1,C2,...,Z25), and then encrypting each residue modulo 26 as a separate plaintext character.
(a) Describe how Eve can easily decrypt a message which is encrypted this way.

(b) Illustrate this attack by decrypting the following ciphertext (which was encrypted using an RSA cryptosystem with n = 18721 and e = 25) without factoring the modulus: 365,0,4845,14930,2608,2608,0

Explanation / Answer

(a) Eve can simply raise each number from 0 - 25 to the encryption exponent modulo n and record the results.

Using the table table created this way, he can read the message without breaking the system (i.e factoring n).

(b) For n = 18721 and the encryption exponent e = 25

we get the following values for the numbers 0-25 (i.e the letter equivalents):

A(=0)=0 B(=1) =1 C(=2) = 6400 D(=3) = 18718 E (=4) = 17173

F = 1759 G = 18242 H = 12359 I =14930 J = 9

K = 6279 L = 2608 M = 4644 N = 4845 O = 1375

P = 13444 Q =16 R = 13663 S = 1437 T =2940

U =10344 V =365 W =10789 X =8945 Y =11373

Z = 5116.

Therefore the plaintext is Vanilla.

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