11.07.2014 Views

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

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

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

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

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

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!