TITLE: What is the status of the Gauss Circle Problem?
QUESTION [62 upvotes]: For $r > 0$, let $L(r) = \# \{ (x,y) \in \mathbb{Z}^2 \ | \ x^2 + y^2 \leq r^2\}$ be the number of lattice points lying on or inside the standard circle of radius $r$.  It is easy to see that $L(r) \sim \pi r^2$ as $r \rightarrow \infty$.  The Gauss circle problem is to give the best possible error bounds: put
$E(r) = |L(r) - \pi r^2|$.
Gauss himself gave the elementary bound $E(r) = O(r)$.  In 1916 Hardy and Landau showed that it is not the case that $E(r) = O(r^{\frac{1}{2}})$.  It is now believed that this is "almost" true: i.e.:
Gauss Circle Conjecture: For every $\epsilon > 0$, $E(r) = O_{\epsilon}(r^{\frac{1}{2}+\epsilon})$.
So far as I know the best published result is a 1993 theorem of Huxley, who shows one may take $\epsilon > \frac{19}{146}$.
(For a little more information, see here)
In early 2007 I was teaching an elementary number theory class when I noticed that Cappell and Shaneson had uploaded a preprint to the arxiv claiming to prove the Gauss Circle Conjecture:

http://arxiv.org/abs/math/0702613

Two more versions were uploaded, the last in July of 2007.
It is now a little more than three years later, and so far as I know the paper has neither been published nor retracted.  This seems like a strange state of affairs for an important classical problem.  Can someone say what the status of the Gauss Circle Problem is today?  Is the argument of Cappell and Shaneson correct?  Or is there a known flaw?

REPLY [13 votes]: This recent arxiv posting by Shaneson claims that one may take $\epsilon > 1153/9750$, improving on Huxley's bound.  It also includes the passage

In 2007 Cappell and the author posted a paper on the arXiv claiming to obtain
  the estimate [in my notation -- PLC] $O(r^{1/2+\epsilon})$.  Unfortunately we have not been able to produce an error free version. The present paper shares with that paper the Proposition in section 5 and there is also something there akin to what immediately follows the Proposition.

I guess that's that.