Conway Can Divide by Three, but I Can’t [pdf]
math.berkeley.edu
math.berkeley.edu
Or something like that.
This paper is not refuting Conway's, and Conway's paper does not prove the claim that a set can be divided in 3+ parts without relying on AC.
What the Conway's paper proves is that, without assuming AC, if there is a bijection between A×n and B×n for some finite n, then there is a bijection between A and B. Axn can be equivalently written as the union of a×n, as a ranges over the elements of A, similarly B×n can be written as the union over b×n. This paper shows that if instead you take the union over a×N_a, where the sets N_a are pairwise disjoint and have n elements, and similarly instead of considering B×n you consider the union of b×N_b, where the sets N_b are pairwise disjoint and have n elements, then the existence of a bijection between those two unions is not sufficient to construct a bijection between A and B if we're not assuming AC. The main point here is that without choice we cannot order all of the N_a's and N_b's at the same time, while in Conway's paper, since N_a=N_b=n={0,1,...,n-1}, they are already uniformly ordered and no such issue arises.