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Simple question, Given c=$b$$e$ mod $m$, we all know finding $e$‎ is equivalent to solving the discrete logarithm.
But what about finding $b$ from c ; $e$ and the semi‑prime $m$ ? Is it something harder than factoring $m$ too ?

If yes as this is different from the ʀꜱᴀ problem, how to compute $b$ when $e$ is more than 128‑bits long (so not small) ? Is the possibility to set c to arbitrary values while changing prime $e$ is making things easier to get at least 1 example where $b$ is found ?

user2284570
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This is equivalent to finding the RSA plaintext $b$ given the public key $(m,e)$. Usually called The RSA Problem. It is no harder than factoring $m$. There is however no evidence that it is easier than factoring.

Edit: As @DanielS points out I was a bit sloppy. For the mapping to be one-to-one, i.e., for the encryption to be reversible, we require $\textrm{gcd}(e,\lambda(m))=1$ and the term RSA problem denotes this case.

kodlu
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