Is the integer factorization problem harder than RSA factorization: $n = pq$? This is a cross-post from math.stackexchange.

Let FACT denote the integer factoring problem: given $n \in \mathbb{N},$ find primes $p_i \in \mathbb{N},$ and integers $e_i \in \mathbb{N},$ such that $n = \prod_{i=0}^{k} p_{i}^{e_i}.$
Let RSA denote the special case of factoring problem where $n = pq$ and $p,q$ are primes. That is, given $n$ find primes $p,q$ or NONE if there is no such factorization.
Clearly, RSA is an instance of FACT. Is FACT harder than RSA? Given an oracle that solves RSA in polynomial time, could it be used to solve FACT in polynomial time?
(A pointer to literature is much appreciated.)

Edit 1: Added the restriction on computational power to be polynomial time.

Edit 2: As pointed out in the answer by Dan Brumleve, there are papers arguing for and against RSA being harder (or easier) than FACT. I found the following papers so far:

D. Boneh and R. Venkatesan. Breaking RSA may be easier than factoring. EUROCRYPT 1998. http://crypto.stanford.edu/~dabo/papers/no_rsa_red.pdf
D. Brown: Breaking RSA may be as difficult as factoring. Cryptology ePrint Archive, Report 205/380 (2006) http://eprint.iacr.org/2005/380.pdf
G. Leander and A. Rupp. On the Equivalence of RSA and Factoring regarding Generic Ring Algorithms. ASIACRYPT 2006. http://www.iacr.org/archive/asiacrypt2006/42840239/42840239.pdf
D. Aggarwal and U. Maurer. Breaking RSA Generically Is Equivalent to Factoring. EUROCRYPT 2009. http://eprint.iacr.org/2008/260.pdf

I have to go through them and find a conclusion. Is someone aware of these results, who can provide a summary?
As far as I can see, an efficient algorithm for factoring semiprimes (RSA) does not automatically translate into an efficient algorithm for factoring general integers (FACT). However, in practice, semiprimes are the hardest integers to factor. One reason for this is that the maximum size of the smallest prime is dependent on the number of factors. For an integer $N$ with $f$ prime factors, the maximum size of the smallest prime factor is $\lfloor N^\frac{1}{f} \rfloor$, and so (via the prime number theorem) there are approximately $\frac{f N^\frac{1}{f}}{\log(N)}$ possibilities for this. Thus increasing $f$ decreases the number of possibilities for the smallest prime factor. Any algorithm which works be successively reducing this space of probabilities will then work best for large $f$ and worst for $f=2$. This is borne out in practice, as many classical factoring algorithms are much faster when the number being factored has more than 2 prime factors.

Further the General Number Field Sieve, the fastest known classical factoring algorithm, and Shor's algorithm, the polynomial time quantum factoring algorithm, work equally well for non-semiprimes. In general, it seems much more important that the factors by coprime than that they be prime.

I think part of the reason for this is the decision version of factoring co-primes is most naturally described as a promise problem, and any way of removing the promise of the input being semiprime is to either


introduce an indexing on the semiprimes (which in itself I suspect is as hard as factoring them), or
by generalising the problem to include non-semiprimes. 


It seems likely that in the latter case the most efficient algorithm would solve FACT as well as RSA, though I have no proof of this. However, a proof is a little to much to ask for, since given an oracle for RSA proving that this cannot efficiently solve FACT amounts to proving that $P\neq NP$.

Lastly it is worth pointing out that RSA (the cryptosystem, not the factoring problem you defined above) trivially generalizes beyond semi-primes.