- Hi, all,

I refer to a guide called:

Beginners Guide to NFS factoring using GGNFS and MSIEVE

(http://gilchrist.ca/jeff/factoring/nfs_beginners_guide.html)

AND

MSIEVE and GGNFS in Combination using PYTHON (factmsieve.py)

(http://gladman.plushost.co.uk/oldsite/computing/factoring.php)

and successfully factorize several numbers.

Now I have a projectÂ£Âº

Given a number, like:

1010646509523626909471007838658716275297171202169082755460653497731277859829955377460426014100492764686887

And after calculation, the two prime factors should be :

r1=23998563736897718596683178098390962435026361950437211 (pp53)

r2=42112791440504457422110905563891953193367924242081317 (pp53)

which means the given number can be factorize into two same digit prime. And I am always given numbers like this. What I need to do is only factorize them into two same digit prime.

Do I have any approach or command to do it quickly?

Thank you for your attention! > which means the given number can be factorize into two same digit prime.

When the number is large enough, there is no method known for factoring

> And I am always given numbers like this. What I need to do is only

> factorize them into two same digit prime.

>

> Do I have any approach or command to do it quickly?

integers that

- takes advantage of the factors having the same number of bits

- is faster than the number field sieve

If the factors are extremely close together (i.e. not only the same

size but many of the top bits of each factor are identical), then

possibly a variant of Fermat's or Lehman's method can find them more

quickly, but such numbers make terrible RSA keys for this reason and

a good RSA key will never have this form.

Otherwise, NFS is unfortunately the best we can do.

jasonp