IMT4531 Introduction to Cryptology
Exercise 4 – supplement - Asymmetric ciphers, hash functions and digital signatures
1. Use the data from the task 9, exercise 4, to compute the following: Alice wants to digitally
sign the message YES with the RSA system. Bob wants to verify that digital signature.
Alice’s public key is (eA,nA)=(39423,46927) and Alice’s secret key is dA=26767 (see the task
9b). In the RSA digital signature scheme, Alice’s signature is obtained by deciphering the
message. Since the message is YES and its integer encoding is m=16346 (see the task 9a),
Alice has to compute the following
𝑐 = 1634626767 mod 46927 and then c will be the Alice’s digital signature of the message
m=16346. To compute c, we can reuse the pre-computation performed in the task 9a, i.e.
𝑖
i
2𝑖
163462 mod 46927
0
20 *
163461 mod 46927 = 16346
1
1
2 *
163462 mod 46927 = 36305
2
22 *
163464 mod 46927 = 363052 mod 46927 = 14376
3
3
2 *
163468 mod 46927 = 143762 mod 46927 = 2868
4
24
1634616 mod 46927 = 28682 mod 46927 = 13199
5
5
1634632 mod 46927 = 131992 mod 46927 = 20577
2
6
26
1634664 mod 46927 = 205772 mod 46927 = 37535
7
7
2 *
16346128 mod 46927 = 375352 mod 46927 = 33831
8
28
16346256 mod 46927 = 338312 mod 46927 = 33958
9
9
2
16346512 mod 46927 = 339582 mod 46927 = 8593
10
210
163461024 mod 46927 = 85932 mod 46927 = 23478
11
11 2 *
163462048 mod 46927 = 234782 mod 46927 = 11942
12
212
163464096 mod 46927 = 119422 mod 46927 = 211
13
13 2 *
163468192 mod 46927 = 2112 mod 46927 = 44521
14 214 *
1634616384 mod 46927 = 445212 mod 46927 = 16815
15
15
1634632768 mod 46927 = 168152 mod 46927 = 9050
2
Since the binary representation of the exponent 26767 is 110100010001111 (see the task
9b), i.e. 15 bits are used for it, we do not need the last row from the table above (marked
red). The rows that are relevant for the computation of c are marked with * (the
corresponding bits in the binary representation of 26767 are equal 1). So we have
1634626767 mod 46927 = 16346 ∙ 36305 ∙ 14376 ∙ 2868 ∙ 33831 ∙ 11942 ∙ 44521 ∙ 16815
= 997 (mod 46927)
So, Alice sends to Bob the pair (m,c)=(16346,997).
Now Bob wants to verify the signature c, upon receiving (m,c). He needs the Alice’s public
key for it, i.e. the pair (eA,nA)=(39423,46927). Then Bob computes
𝑞 = 99739423 mod 46927
The binary representation of 39423 is 1001100111111111 (see the task 9a). So Bob pre𝑖
computes the powers 9972 mod 46927, i=0,…,15 (since binary representation of 39423 is
performed with 16 bits). He generates the table
i
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
2𝑖
20 *
21 *
22 *
23 *
24 *
25 *
26 *
27 *
28 *
29
210
211 *
212 *
213
214
15
2 *
𝑖
9972 mod 46927
9971 mod 46927 = 997
9972 mod 46927 = 8542
9974 mod 46927 = 85422 mod 46927 = 41206
9978 mod 46927 = 412062 mod 46927 = 21722
99716 mod 46927 = 217222 mod 46927 = 41226
99732 mod 46927 = 412262 mod 46927 = 27917
99764 mod 46927 = 279172 mod 46927 = 42200
997128 mod 46927 = 422002 mod 46927 = 7277
997256 mod 46927 = 72772 mod 46927 = 21073
997512 mod 46927 = 210732 mod 46927 = 1128
9971024 mod 46927 = 11282 mod 46927 = 5355
9972048 mod 46927 = 53552 mod 46927 = 3628
9974096 mod 46927 = 36282 mod 46927 = 22824
9978192 mod 46927 = 228242 mod 46927 = 45276
99716384 mod 46927 = 452762 mod 46927 = 4035
99732768 mod 46927 = 40352 mod 46927 = 44483
Powers of 2 in the table that will be used in the computation are marked with * (they
correspond to the non-zero bits in the binary representation of 39423. Thus we get
𝑞 = 99739423 mod 46927
= 997 ∙ 8542 ∙ 41206 ∙ 21722 ∙ 41226 ∙ 27917 ∙ 42200 ∙ 7277 ∙ 21073
∙ 3628 ∙ 22824 ∙ 44483 = 16346
Since q=m, Alice’s signature has been verified.