5.15 This exercise exhibit what is called a protocol. It provides an example whe
ID: 3815072 • Letter: 5
Question
5.15 This exercise exhibit what is called a protocol. It provides an example where ciphertext can be decrypted by an opponent, without determining the key, if a cryptosystem is used in a careless way. The moral is that it is not sufficient to use a "secure" cryptosystem in order to guarantee "secure" communion. TABLE RSA dphertext 8309 14010 8936 27358 25023 16481 25809 7135 24996 30590 27570 26486 30388 9395 6340 23614 14999 4517 12146 29421 26439 1606 17881 27584 7372 18436 12056 13547 25774 7647 23901 25774 1304 1908 8635 2149 1908 22076 7372 8686 4082 11803 5314 107 7359 22470 7372 22827 15698 30317 4685 14696 30388 8671 29956 15705 1417 26905 25809 28347 26277 7897 20240 21519 12437 1108 27106 18743 24144 10685 25234 30155 23005 8267 9917 7994 9694 2149 10042 27705 15930 29748 8635 23645 11738 24591 20240 27212 27486 9741 2149 29329 2149 5501 14015 30155 18154 22319 27705 20321 23254 13624 3249 5443 2149 16975 16087 14600 27705 19386 7325 26277 19554 23614 7553 4734 8091 23973 14015 107 3183 17347 25234 4595 21498 6360 19837 8463 6000 31280 29413 2066 369 23204 8425 7792 25973 4477 30989 Suppose 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 (ie., A 0, B 1. etc.), and then encrypting each residue modulo 26 as a separate plaintext character (a) Describe how oscar can easily decrypt a message which is encrypted in this way. (b) Illustrate this attack by decrypting the following ciphertext (which was en- crypted using an RSA Cryptosystem with n 18721 and b 25) without factoring the modulus: 365, 0,4845, 14930, 2608, 2608, 0. 516 his exercise illustrates another examnlp of a n
Explanation / Answer
a.
It is an insecure method of encryption. Oscar can compute (mb mod n) for all 26 possible values of m. And then create a decryption table where the decryption of (mb mod n) is m.
b.
m = 0 ==> (025 mod 18721) = 0
m = 1 ==> (125 mod 18721) = 1
m = 2 ==> (225 mod 18721) = 6400
m = 3 ==> (325 mod 18721) = 18718
m = 4 ==> (425 mod 18721) = 17173
m = 5 ==> (525 mod 18721) = 1759
m = 6 ==> (625 mod 18721) = 18242
m = 7 ==> (725 mod 18721) = 12359
m = 8 ==> (825 mod 18721) = 14930
m = 9 ==> (925 mod 18721) = 9
m = 10 ==> (1025 mod 18721) = 6279
m = 11 ==> (1125 mod 18721) = 2608
m = 12 ==> (1225 mod 18721) = 4644
m = 13 ==> (1325 mod 18721) = 4845
m = 14 ==> (1425 mod 18721) = 1375
m = 15 ==> (1525 mod 18721) = 13444
m = 16 ==> (1625 mod 18721) = 16
m = 17 ==> (1725 mod 18721) = 13663
m = 18 ==> (1825 mod 18721) = 1437
m = 19 ==> (1925 mod 18721) = 2940
m = 20 ==> (2025 mod 18721) = 10334
m = 21 ==> (2125 mod 18721) = 365
m = 22 ==> (2225 mod 18721) = 10789
m = 23 ==> (2325 mod 18721) = 8945
m = 24 ==> (2425 mod 18721) = 11373
m = 25 ==> (2525 mod 18721) = 5116
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