Discrete logarithm with exponentiation oracle Suppose $(n, d)$ is an RSA private keypair. We know the public key only (wlog suppose it's $(n, 2^{16}+1)$ and we are given oracle $E$, decryption oracle for $(n, d)$. Is there any efficient algorithm to recover $d$? This is a bit easier than directly attacking for known pair of plaintext and ciphertext, as for any $m$ we can astablish this relation:

$$
E(m) \equiv m^d \pmod n
$$
No, there is no known way to recover $d$ using an Oracle that, given $m$, returns $m^d \bmod n$; there is also no proof that such a method does not exist.

Such a method would imply that the RSA problem (which is, given $m^e \bmod n$, recover $m$) is equivalent to the factorization problem (which knowledge of $d, e$ would allow you to solve); it is unknown if these two problems are equivalent.