[openssl-dev] common factors in (p-1) and (q-1)

[mancha]

On Fri, Jul 31, 2015 at 06:46:22PM +0000, Viktor Dukhovni wrote:
> On Fri, Jul 31, 2015 at 11:19:39AM -0700, Bill Cox wrote:
> 
> > Cool observation.  From running a bit of Python code, it looks like
> > the probability that GCD(p-1, p-q) == 4 is a bit higher than 15%, at
> > least for random numbers between 2048 and 4096 bits long.  It looks
> > like putting in a GCD(p-1, q-1) check will slow down finding
> > suitable p and q by around a factor of 6.5.
> 
> A smaller slow-down would be incurred one were to restrict both of p,q
> to 3 mod 4. In that case 2 would be the largest common even factor of
> (p-1) and (q-1), and any appreciably large common odd factor
> (necessarily above 17863 due to how each of p/q is chosen) would be
> very rare.
> 
> Is there a good argument for adding the gcd test?  How big does the
> common factor have to be for any information it might provide to be
> substantially useful in finding 1/e mod phi(m)?
> 
> The larger the common factor is, the smaller the probability of p-1
> and q-1 sharing it (for a given sufficiently large prime factor "r" of
> (p-1), the probability of (q-1) also having that factor is 1/(r-1)).
> If say "r" needs be 80 bits long to be useful in attacking RSA 1024,
> then only ~1 in 2^80 (p-1,q-1) pairs will have such a common factor,
> which is sufficiently rare not warrant any attention.
> 
> Also one still needs to be able to fully factor (n-1).  After tens of
> thousands of trials, I managed to generate a (p,q,n) triple with a
> 1024-bit modulus n in which (p-1,q-1) have a common odd factor.
> 
>     n =
>     123727085863382195696899362818055010267368591819174730632443285012648773223152448218495408371737254282531468855140111723936275062312943433684139231097953508685462994307654703316031424869371422426773001891452680576333954733056995016189880381373567072504551999849596021790801362257131899242011337424119163152403
> 
>     e = F_4 = 65537
> 
>     gcd(p-1,q-1) = 2 * 28559
> 
> What can the OP tell us about d, p or q?  Can anyone produce a full
> factorization of n - 1?

n-1 = 2 * 3^3 * 7 * 13 * 67 * 2399 * 28559 *
5485062554686449262177590194597345407327047899375366044215091312099734701911004226037445837630559113651708968440813791318544450398897628672342337619064712331937685677843283385813601700381667290503026724160750373906990713551823941904482040860073543880341612964100618466865014941425056336718955019

--
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