> No, that's not a conjecture. If you factor the modulus you can directly calculate the private key from the public one. You have misread my post. The conject

> No, that's not a conjecture. If you factor the modulus you can directly calculate the private key from the public one. You have misread my post. The conjecture is that the RSA problem is as difficult as the factoring problem. In other words, it's possible that solving the RSA problem isn't as difficult as the factoring problem (but most people doubt it). A <= B. Most people think A == B. This is the inverse of your statement. Factoring breaking RSA doesn't tell us if there's a non-factoring solution to the RSA problem. Everyone knows breaking RSA is no more difficult than factoring, but it may be easier than factoring. A break in the Rabin cryptosystem would provably break RSA, but it's not necessarily true that a break in RSA would break the Rabin cryptosystem. Edit: Paragraph 4 of https://en.wikipedia.org/wiki/RSA_problem https://en.wikipedia.org/wiki/RSA_problem at least believes it's still a conjecture that RSA is as difficult as the factoring problem.