From today\'s standpoint, most people would claim RSA to be secure. However, to
ID: 653109 • Letter: F
Question
From today's standpoint, most people would claim RSA to be secure. However, to my knowledge, this is purely based on the speculation that no one knows a computational feasible way to find a d for given n and e, s.t. ed?1mod?(n) - which can be reduced to the factorization problem.
However, it is still an assumption that the factorization problem is hard to solve but nothing that has been proven mathematically. Worse, as far as I understand it, it is not even clear to which complexity class integer factorization belongs. And if the past year has thought us anything then it is that often the "paranoia argument" is not that paranoid at all.
So I always wondered how probable it is that indeed no one has found an efficient solution to the factorization problem yet. How would one know? Or, in short: How do you convince a paranoid that RSA indeed is secure?
Explanation / Answer
Practical Perspective
While look up tables appear nearly impossible in theory, one will mount attacks on the Pseudo random variety generators.
Also, a nasty implementation of RSA will continuously be fatal. (Especially side channels attacks!)
To add it up, it's laborious to even convert anyone concerning the safety of RSA.
Theoretical Perspective:
The RSA cryptosystem relies on the RSA Assumptions,that may be a stronger assumption than factorization.
Factorization is in NP, but quite know what complexness category the RSA Assumption is in it i.e. all know if got a economical factorization formula, we will use that to interrupt the RSA, however do not know if the finding RSA would imply factorization.
Factoring itself isn't NP complete, thus do not really know if it is very very laborious within the first place!!
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