Decrypting small integers under RSA Let $(n,e)$ be an RSA public key. Suppose $c = m^e \pmod n$, where $c&gt;1$ is a very small integer. For concreteness, say $c=2$ or $c=4$.


  Is it hard to find $m$ under the RSA assumption (or any of its variants)?
Since the RSA problem is assumed hard, we do not know and can't find the factorization of $n$.
We know (from standard RSA) that $m=c^{\left(e^{-1}\bmod\varphi(n)\right)}\bmod n$ fulfills the requirement $c\equiv m^e\pmod n$, but we do not know how to compute $m$ without some extra info or oracle.

Is it hard to find $m$ under the RSA assumption (or any of its variants)?

Yes, but I have no better argument than: among integers $c&gt;1$ independent of $n$, only exact $e^\text{th}$ powers are known to make it easy to solve the RSA problem for arbitrary $n$ too large to factor and odd $e&gt;1$ making $(n,e)$ a valid RSA public key. In addition, there in no such $c$ in the interval $[2,2^e)$,  which with $e=65537$ as in a comment by the OP includes about all commonly used values of $n$, thus of $c$.
Solving $c\equiv m^e\pmod n$ for small $c$ is not necessarily hard for other $c$ and whenever $n$ is hard to factor. Proof by counterexample: $c=2$, $e=65537$, $n=(3^{65537}-2)/29$. I can't factor that $n$, yet it's easy to find the solution $m=3$.
More formally: I'm about as confident about the hardness of the RSA problem than I am about that problem restricted to small $c&gt;1$, provided that $c$ is not a power of $e$, and that factors of $n$ are random distinct primes large enough to make factoring $n$ unfeasible. Yet I'd be quite surprised if we could prove the hardness of that restricted RSA problem on the basis of the hardness of the normal RSA problem.