Problem Set #5
Brandon DeAlmeida
1)
For the past two weeks we
’
ve been continuing learning about the RSA algorithm. While
covering more about the cryptosystem we
’
ve learned new techniques for factorization
that are relevant to RSA. These, such as Sieving, have allowed us to learn about other
topics such as B-smooth.
2)
We wish to factor
N = 493
by sieving prime powers up to
B = 11
on the values of
F(23)
to
F(38)
where
F(t) = t
2
− N
. Starting with 23 we get:
23
2
≡
N
36 ≡
N
2
2
⋅ 3
2
24
2
≡
N
83
25
2
≡
N
132 ≡
N
2
2
⋅ 3 ⋅ 11
26
2
≡
N
183 ≡
N
3 ⋅ 61
27
2
≡
N
236 ≡
N
2
2
⋅ 59
28
2
≡
N
291 ≡
N
3 ⋅ 97
29
2
≡
N
348 ≡
N
2
2
⋅ 3 ⋅ 29
30
2
≡
N
407 ≡
N
11 ⋅ 37
31
2
≡
N
468 ≡
N
2
2
⋅ 3
2
⋅ 13
32
2
≡
N
38 ≡
N
2 ⋅ 19
33
2
≡
N
103
34
2
≡
N
170 ≡
N
2 ⋅ 5 ⋅ 17
35
2
≡
N
239
36
2
≡
N
310 ≡
N
2
2
⋅ 5 ⋅ 31
37
2
≡
N
383
38
2
≡
N
458 ≡
N
2 ⋅ 229
39
2
≡
N
42 ≡
N
2 ⋅ 3 ⋅ 7
40
2
≡
N
121 ≡
N
11
2
Using the fact that
40
2
≡
N
11
2
we can write
40
2
− 11
2
≡
N
(40 + 11)(40 − 11) ≡
N
0
.
Finally, the GCD of
40 ± 11
and N will give us a factor of N:
493 − 9 ⋅ 51 = 34
51 − 34 = 17
34 − 2 ⋅ 17 = 0
∴ 17 divides 493

3)
We wish to compute the following values of
Ψ(X, B)
which is the number of B-smooth
numbers from 2 through X:
3.1)
Ψ(25,3)
2 = 2
3 = 3
4 = 2
2
5 = 5
6 = 2 ⋅ 3
7 = 7
8 = 2
3
9 = 3
2
10 = 2 ⋅ 5
11 = 11
12 = 2
2
⋅ 3
13 = 13
14 = 2 ⋅ 7
15 = 3 ⋅ 5
16 = 2
4
17 = 17
18 = 2 ⋅ 3
2
19 = 19
20 = 2
2
⋅ 5
21 = 3 ⋅ 7
22 = 2 ⋅ 11
23 = 23
24 = 2
3
⋅ 3
25 = 5
2
∴ Ψ(25,3) = 10
3.2)
Ψ(35,5)