Here's a quote from Douglas Stinson:

“[i]f a cryptosystem can be ‘broken’ in some specific way, then it would be possible to efficiently solve some well-studied problem that is thought to be difficult. For example, it may be possible to prove a statement of the type “a given cryptosystem is secure if a given integer n cannot be factored.” [...] [B]ut it must be understood that this approach only provides a proof of security relative to some other problem, not an absolute proof of security. This is a similar situation to proving that a problem is NP-complete [...].”

Source: Stinson, D. R. (2006). Cryptography, Theory and Practice. Chapman and Hall, CRC, 3rd edition. Chapter 2, section 2.1, page 45.

Does this apply to RSA? Suppose I find the private exponent by some other way other than factoring $n$. Then I would have broken RSA without factoring.

So although Stinson's example mentions factoring, we can't immediately think of RSA here because there's no proof that factoring is the only way to recover a private key. What do you say?

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    $\begingroup$ Well, the security of RSA (with a secure padding scheme) is provably reducible to the RSA problem. But I'm not sure if that's really the kind of answer you're looking for. $\endgroup$ Commented Dec 8, 2018 at 18:40
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    $\begingroup$ @IlmariKaronen Technically the RSA problem is a "well-studied problem that is thought to be difficult". $\endgroup$
    – Maeher
    Commented Dec 8, 2018 at 18:42
  • $\begingroup$ @IlmariKaronen Perhaps I could put my question this way: if I have $n$, the composite, $e$ the public exponent and $d$, the private exponent, is there a known method to find $\phi(n)$ (or to factor $n$)? $\endgroup$
    – user45491
    Commented Dec 8, 2018 at 18:48
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    $\begingroup$ Yes, there is. $\endgroup$ Commented Dec 8, 2018 at 18:53
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    $\begingroup$ @user45491 no, finding a private key ($d$) is not the necessary condition for broking PKE scheme. E.g., if an attacker is able to find just a forgery - a signature of some message, the scheme is already broken, although the attacker can't find the private key. $\endgroup$ Commented Dec 8, 2018 at 20:02

1 Answer 1


Every cryptosystem is "provably secure" under at least one hardness assumption: the assumption that it cannot be broken. Hence, the only question which matters is whether a cryptosystem is provably secure under a well-known and well-studied assumption.

This is kind of the case for RSA, but in a somewhat unsatisfying way: the IND-CPA security of RSA with appropriate padding can be reduced to the RSA problem (extracting $e$-th roots mod $n$), which is a well-studied assumption... But only because this is the one underlying the RSA cryptosystem, and we have been using the cryptosystem a lot in the past decades.

Something much more satisfying would be to reduce it to a problem which had long been studied before, in different context; the most natural such candidate is indeed factoring. Unfortunately, we do not know of a reduction from RSA to the hardness of factoring integers - in fact, we even have proofs that it will be hard to find such a reduction (in a well defined sense).

We have, however, many other cryptosystems which can be reduced to factoring; the earliest example of this kind is the cryptosystem of Rabin.

  • $\begingroup$ The RSA problem asks us to recover the plain text m from (N, e). If I find a way to get d without factoring, I'll be able to recover any m. If I get d, I also factor N. This is not a reduction of RSA to factoring because there may some way to get m without getting d (and so without factoring N). $\endgroup$
    – user45491
    Commented Dec 10, 2018 at 23:46
  • $\begingroup$ Your first paragraph is very well put. Thank you. Your second paragraph is equally good. Your third paragraph I'm still grasping --- but I think I understood it, as expressed in my previous comment. Your last paragraph makes your answer the best thing I could get --- although I had to think about the whole thing for a couple of days. Thank you so much for sharing your very insightful knowledge. Very grateful. $\endgroup$
    – user45491
    Commented Dec 10, 2018 at 23:47

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