On 01/24/2014 09:22 PM, Port563 wrote: > "Hlauk" <firstname.lastname@example.org> wrote in message > news:email@example.com... >> On Friday, January 24, 2014 10:50:44 AM UTC-5, David Bernier wrote: > >>> 69176818182079702962583905305944091851396378426614231238000704281 * >>> 68643130470496608225965397098921704892910199186048409746525386899 >>> = >>> 47485133560063190433548259679204355907477810307482663990525310023\ >>> 91821388435849831808696657092215642611599144927466443608010614619 >>> >> Your composite has prime factors that are close to it's square root. >> Should it make it easier to factor? >> >> Dan > > Yes, assuming that the breaker does not only start its check on the > smallest factor and work upwards, but also (or first) starts at > the square root and works down. > > But David wasn't trying to make a good public key. > > If the non-trivial factors are p1, p2; p1 > p2; > we have: > p1 ~ 6.91768181820797E+64 > p2 ~ 6.86431304704966E+64 > (p1-p2)/2 ~ 2.66843855791553E+62 > > So, as David said there was nothing "special" about his choice of composite, > a very crude measure of the factorisation difficulty is to compare > (p1-p2)/2 with p2; this ratio is < 0.4%. > > One where p1/p2 ~ 3, say, this ratio would be ~100%. > > [I am aware that primes thin out as one goes higher, so the two sieve > approaches are working at very different prime densities. As I said, this > difficulty measure is very crude] > >
If we call N the composite, then N = 68909974326288155594274651202432898372153288806331320492263045590^2 - 266843855791547368309254103511193479243089620282910745737658691^2
that is, N as a difference of two squares.
As you say , (p1-p2)/2 ~= 2.7e+62. This is Fermat's method.
I'm not sure it helps a whole lot here, though ...