Re: Truelly Random Numbers

[tcmay@got.net (Timothy C. May)]

At 8:11 PM 3/2/96, Adam Shostack wrote:
>A. Padgett Peterson P.E. Information Security wrote:

>| True however the current mechanism of generating PGP keys which consists
>| primarily of pseudo-randomly pounding on a keyboard is hardly "truely random.
>|
>| Have no idea of the true number but expect it to be significantly less than
>| that quoted above, even for a 1024 bit key like mine.
>
>        Accroding to Stephan Neuhaus's 'Statistical Properties of IDEA
>session keys in PGP,' the session keys are very well distributed, when
>tested for equidistribution and serial correlation.
>
>        This does not demonstrate that the RSA keys are as well
>distributed, but it does generate some confidence that the key
>generation methods of PGP are not very broken.  Testing for RSA
>generation would be more difficult, since there are some practical
>difficulties in getting a large sample of RSA private keys.

In some PK code I did several years ago in Mathematica, the primes for the
RSA modulus were found by picking a "random" (more on this later) starting
point and then counting up from there, testing for primality (actually,
pseudoprimality, technically). As one would expect, primes are found fairly
quickly.

The "randomness" of the resulting primes--and hence the randomness of the
modulus and hence the "RSA key"--is set by the randomness of the starting
point. With a reasonable amount of entropy, such as picking the next digit
from several keyboard timings, I expect the 150-decimal-digit number to be
*very* random!

(In fact, I'd venture that merely asking people to type in digits would
produce starting points that essentially would be very random...maybe some
clustering here and there, or an unequal number of digits, or too equal a
distribution, but adequate. An since an attacker could not know what the
sources of randomness were for some particular person, I doubt strongly
that factoring the modulus would be any easier.)

Suppose a =
4801747274372727828487361830183561393615106551195496693610351528409257572926
659
2027575902673957001560102249600798767153757681546836352857811107361291541511

(which is about 140-150 digits, "randomly" entered by me)

and p is computed as the first prime larger than this.

q found the same way

Now, is the modulus, n = pq, any more factorable than if a "more random"
source of p and q were used?

(I am actually asking this as a real question. Does anyone know if
factoring is significantly easier for such not-completely-random numbers? I
would expect that in theory it is, but in practice this is not a useful
point of entry into factoring n. Just a hunch.)

--Tim


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