Cryptanalysis of RSA Factorization - Library(ISI Kolkata) - Indian ...

Chapter 4: Cryptanalysis of RSA with more than one Decryption Exponent 70
Calculation of s 1
s 1 =
m+1
∑
i 1 =0
m−i
∑ 1 +1
r=0
( r+n−1
r
)
(r+t+1)·i 1
−
m∑
i 1 =0
m−i
∑ 1
r=0
( r+n−1
r
)
(r+t+1)·i 1
Now, using Lemma 4.1 and neglecting the lower order terms, we obtain
≈
m+1
∑
i 1 =0
m+1
∑
i 1 =0
m+1
∑
≈
i 1 =0
m+1
∑
≈
T=0
m−i
∑ 1 +1
r=0
m−i
∑ 1 +1
r=0
( r +n−1
r
r n−1
(n−1)! (r+t)·i 1
)
(r +t+1)·i 1
(m−i 1 +1) n+1 m+1
(n−1)!·(n+1) ·i ∑
1 +
T n+1 m+1
(m−T +1)
(n−1)!·(n+1) + ∑
T=0
i 1 =0
(m+1) n+3
≈
(n−1)!·(n+3)(n+2)(n+1) +
(m−i 1 +1) n
(n−1)!·n ·ti 1
t· Tn (m−T +1)
(n−1)!·n , where T = m−i 1 +1
t
(n−1)! ·
Thus, the final expression for s 1 is as follows.
(m+1) n+2
n(n+1)(n+2) .
s 1 ≈
≈
(m+1) n+3
(n−1)!·(n+3)(n+2)(n+1) + t
(n−1)! · (m+1) n+2
n(n+1)(n+2)
m n+3
−
(n−1)!·(n+3)(n+2)(n+1) − t
(n−1)! · m n+2
n(n+1)(n+2)
m n+2
(n−1)!·(n+2)(n+1) + tm n+1
(n−1)!·n(n+1)