Adding integers represented by their factorization is as hard as factoring? Reference request I'm looking for a reference for the following result:


  Adding two integers in the factored representation is as hard as factoring two integers in the usual binary representation.


(I'm pretty sure it's out there because this is something I had wondered at some point, and then was excited when I finally saw it in print.)

"Adding two integers in the factored representation" is the problem: given the prime factorizations of two numbers $x$ and $y$, output the prime factorization of $x+y$. Note that the naive algorithm for this problem uses factorization in the standard binary representation as a subroutine.

Update: Thanks Kaveh and Sadeq for the proofs. Obviously the more proofs the merrier, but I would also like to encourage more help in finding a reference, which as I said I'm fairly sure exists. I recall reading it in a paper with other interesting and not-often-discussed ideas in it, but I don't recall what those other ideas were or what the paper was about in general.
Assume that we can solve the problem (lets call it FactSum) in complexity class $C$ and $C$ is closed under $\log$-iteration (aka $\log$-bounded recursion)(e.g. if we can compute $x*y$ where $*$ is a binary function, we can computed $x_1*\ldots*x_{\log{n}}$) and contains $\mathsf{P}$ (this last condition can be made weaker). We show that factoring is in also in $C$.

Note that each number can be written as a sum of $\log n$ powers of $2$. Each of them are easy to factor. 

Now given a number, write it as the sum of powers of it, then write each summand in the factoring representation, and then use the algorithm to sum them in the factoring representation. The result will be the factoring of the input number.

This shows that factoring is reducible to $\log$-iteration of your problem FactSum. Therefore factoring is in $\mathsf{P}^{\text{FactSum}}$ (and I think $\mathsf{P}$ can be replaced with  $\mathsf{NC^1}$ here).