The summary in the Rivest-Kaliski paper does indicate that two specific problems are equivalent to the RSA problem itself.
Due to the algebraic properties of the field $Z_n$, if there exists any small fraction of weak cipher texts (i.e. a particular subset of cipher texts $C$ for which finding the solution $M$ to $C = M^e \pmod N$ is easy), then the RSA Problem is at most as hard as the running time of the adversarial procedure for decrypting the weak cipher texts times a polynomial factor. IOW: The RSA Problem guarantees that all cipher texts are equally hard to invert.
A similar argument applies to each individual bit of a RSA cipher text. If an adversary has a non trivial advantage to determine any bit of a RSA plain text $M$ given a RSA cipher text $C$, then all bits of $M$ are equally easy to determine.