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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57 3.3 A New Class <strong>of</strong> Weak Keys<br />
8975911152337295179208385986195123683049905106852500834430037397316270<br />
0864249336588337032797599912676619611066951994203277539744143449608166<br />
337312588384073377751.<br />
Then the continued fraction <strong>of</strong> e gives X,Y (respectively) as<br />
N<br />
1684996666696914987166688442938726917102321526408785780068975640579,<br />
1675051614915381290330108388747571693885770140577513454985303531396.<br />
Then we find pu+v as<br />
1455711706935944763346481078909022402000177527246411623972976816582783<br />
4476154028615829072068297700713978442098223807592204425878660324320671<br />
085270850022954001141596160.<br />
From pu+v we get p using the idea <strong>of</strong> Theorem 3.4 as in this case u = 2 52 is<br />
not a multiple <strong>of</strong> q and v = 2 175 is less than N 1 4.<br />
Next we give an example using the LLL technique that cannot be done using<br />
the technique <strong>of</strong> Lemma 3.1.<br />
Example 3.6. The public exponent e is a 1000-bit integer as follows.<br />
6875161700303375704546089777254797601588535353918071259131130001533282<br />
9990657553375889250468602806539681604179662083731636763206733203004319<br />
5254536198882701786034369526609680993429591977911304810695130485689008<br />
4599364131003915783164223278468592950590533634401668968574079388141851<br />
069809468480532614755.<br />
Using Theorem 3.2 we get Y (a 240 bit integer) as follows.<br />
1743981747042138853816214839550693531666858348727675018411981662762169<br />
193.<br />
We also get pu+v (a 552 bit integer) as follows.<br />
1455711706935944763346481078909022402000177527246411623972976816582783<br />
4476154028615829072068297700713978442098223807592204425878660324320671<br />
085270850022954001141596160.<br />
In this case u,v are same as in Example 3.5 and hence p can be found using<br />
the idea <strong>of</strong> Theorem 3.4. It is to note that in this case X is a 241 bit integer<br />
17668470647783843295832975007429185158274838968756189581216062012<br />
92619790.<br />
Note that ⌈ N<br />
pu+v<br />
⌉ is a 448-bit integer. Now 2XY should be less than 448 bit<br />
for a success using the technique <strong>of</strong> Lemma 3.1, which is not possible as X,Y are