Cryptanalysis of RSA Factorization - Library(ISI Kolkata) - Indian ...
Cryptanalysis of RSA Factorization - Library(ISI Kolkata) - Indian ...
Cryptanalysis of RSA Factorization - Library(ISI Kolkata) - Indian ...
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Chapter 7: Approximate Integer Common Divisor Problem 134<br />
Let X 2 = X 3 = ··· = X k = X be the common upper bound on each co-ordinate<br />
<strong>of</strong> the root (˜x 2 ,...,˜x k ). The shift polynomials from Equation (7.20) contribute<br />
P ′ 1 =<br />
m∏<br />
(X r a m−r<br />
r=0<br />
1 ) (k+r−2<br />
r ) = X<br />
η 4<br />
a η 5<br />
with η 4 = ∑ m<br />
r=0 r( )<br />
k+r−2<br />
r , η5 = ∑ m<br />
r=0 (m − r)( )<br />
k+r−2<br />
r , to the determinant <strong>of</strong> L ′ .<br />
(Note that this P 1 ′ is same as P 1 in Corollary 7.9). The shift polynomials from<br />
Equation (7.21) contribute<br />
P ′ 2 =<br />
t∏<br />
i 2 =1<br />
(X i 2<br />
X m ) (k+m−2 m ) = X<br />
η 6<br />
with η 6 = ∑ t<br />
i 2 =1 (i 2+m) ( )<br />
k+m−2<br />
m , to the determinant <strong>of</strong> L ′ . The dimension <strong>of</strong> L ′ is<br />
ω ′ =<br />
m∑<br />
( ) ( )<br />
k +r −2 m+k −2<br />
+t .<br />
r m<br />
r=0<br />
Now, we have ( )<br />
k+r−2<br />
r =<br />
r k−2<br />
(k−2)! +o(rk−2 ). Using Lemma 4.1 and neglecting lower<br />
order terms, we obtain<br />
P ′ 1 ≈ X ∑ m<br />
r=0 r rk−2<br />
∑ m rk−2<br />
r=0 (m−r)<br />
(k−2)! a<br />
(k−2)!<br />
1 ≈ X 1<br />
1<br />
m k 1 m k<br />
(k−2)! k a<br />
(k−2)! k−1 − 1 m k<br />
(k−2)! k<br />
1 ,<br />
P 2 ′ ≈ X ∑ t<br />
i 2 =1 (i 2+m) mk−2<br />
(k−2)!<br />
≈ X 1<br />
(k−2)! (t2 m k−2 +tm k−1) 2<br />
, and<br />
m∑<br />
ω ′ ≈<br />
r=0<br />
r k−2 mk−2<br />
+t<br />
(k −2)! (k −2)!<br />
Following Theorem 2.23, the required condition is<br />
≈<br />
det(L ′ ) = P ′ 1P ′ 2 < g mω′ ,<br />
m k−1 mk−2<br />
+t<br />
(k −1)(k −2)! (k −2)!<br />
where g is the common divisor. Let X = a α+β . Then putting the values <strong>of</strong> g,X<br />
in det(L ′ ) = P ′ 1P ′ 2 < g mω′ , we get,<br />
( m<br />
k<br />
k + mk−2 t 2<br />
2<br />
+m k−1 t<br />
)(α+β)+ mk<br />
k −1 − mk<br />
k<br />
< (1−α)<br />
(m k−1 t+ mk<br />
k −1<br />
)<br />
. (7.22)