Sujet : Re: Some naive musings about RSA
De : ec1828 (at) *nospam* somewhere.edu (Ethan Carter)
Groupes : comp.miscDate : 08. Jul 2025, 00:45:10
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <87h5znlg55.fsf@somewhere.edu>
References : 1
Sylvia Else <
sylvia@email.invalid> writes:
Sometimes in an idle moment, I find myself wondering about possible
ways to attack RSA. Spoiler alert, I haven't found one.
>
But in passing, I found myself thinking about this:
>
For the uninitiated, RSA is based on the fact that if p and q are
large primes, then the product p*q is a large number that is
computationally infeasible to factorise - which is to say, no one's
found a way, though quantum computers offer a possibility.
>
Let m be p * q, and n be (p-1)(q-1). Then RSA further depends on the
fact that for any integer a, co-prime with m, a^n modulo m is 1. Since
2 will be co-prime with m, that means that 2^n modulo m is also 1. We
can write that equivalently as
>
2^n - 1 = k * m
>
where k is an unknown integer. Actually, there are an infinity of
possible k, but we're interested in the lowest (ignoring the trivial
case of k = 0).
I suggest you should not use the word ``unknown'' here because the
equation is true for all integers k, as you observe on the following
sentence. The equation is saying that 2^n - 1 is a multiple of m.
Mathematical lingo is such that ``an unknown integer'' means a /fixed/
number k (that we don't which it is), but your k here is not fixed.
(You can say ``an arbitrary integer'' instead.)
If we knew k, then that would give us n, or a factor of n (n is never
prime, it is at least divisible by 4).
Something isn't quite right here. We trivially know the smallest
positive k that satisfy the equation---that's k = 1. For example, take
p = 3, q = 5 so that m = 15 and n = 8. Then the equation
2^8 - 1 = k * 15
is trivially satisfied by taking k = 1, which is the smallest /positive/
integer you're looking for.
So I did not really understand the rest of your post. You seemed to be
looking for a factor of n. You're then looking for a factor of
p - 1 or q - 1.
If these numbers would have /small/ factors, you could find them by
trial division, say. That would indeed be a weakness. RSA is not safe
if we pick p, q such that p - 1 or q - 1 have small factors. (A keyword
here would be ``Pollard's p - 1 method''.) However, Ronald Rivest and
Robert Silverman have argued that by just selecting large random primes,
we're unlikely to run into trouble. (In other words, we don't need
``strong primes'', meaning the security gain is negligible.)