🚀 go-pugleaf

In article <1993Jan21.205045.13877@netcom.com> strnlght@netcom.com (David Sternlight) writes:
:In article <1993Jan21.011510.24294@linus.mitre.org> bs@gauss.mitre.org (Robert D. Silverman) writes:
:>27392450308603031423410234291674686281194364367580914627947367941608692026226993634332118404582438634929548737283992369758487974306317730580753883429460344956410077034761330476016739454649828385541500213920807
:>:
:>:Can you factor this number [in your lifetime]?  If so, you may be
:> 
:>What is the source for this number? Is it a random integer? Is it
:>an RSA number, constructed as the product of nearly equal primes?
:>If the latter it is out of current computational range.
:
:(Correcting a previous post)
:Couldn't you take the approximate square root of the number, truncate the
:result, and then start searching up and down from there if the number is the
:product of two nearly equal primes?
 
The above number is 209 digits. Its square root is 105 digits. Let's
say it is the product of primes, the first 10 digits of which are the
same. Thus, they are nearly equal. That only leaves about 10^95 numbers
to search by the proposal you make.

The method you suggest is known as Fermat's method. It is still
only O(sqrt(N)).

--
Bob Silverman
These are my opinions and not MITRE's.
Mitre Corporation, Bedford, MA 01730
"You can lead a horse's ass to knowledge, but you can't make him think"