TITLE: Lower bounds for $|A+A|$ if $A$ contains only perfect squares
QUESTION [11 upvotes]: Let $A$ a set  with $|A|=n$ that contains only perfect squares of integers. 
What lower bounds can we give for $|A+A|$?
I think the lower bound $\gg \frac{n^2}{\sqrt{log \,n}}$ holds (this would be the best bound possible, with "equality" for $A=\{1^2,2^2,...n^2\}$). However, this estimate seems really hard.
I couldn't even prove the bound $|A+A|>Cn$ for all constant $C$ (for big $n$). I think proving this bound would already be a good start. Any idea or technique related to the problem is appreciated.
Thanks in advance.

REPLY [17 votes]: This is a well-known (and difficult) problem.
The current record is $|A+A| \geq \log(|A|)^{c\log\log(|A|)}|A|$ due to Schoen (in 2011), using his near optimal form of Freiman's theorem. Note this is just shy of a power gain.
See Chang's paper On problems of Erdos and Rudin for more discussion of this and related questions.