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.



Lower bounds for $|A+A|$ if $A$ contains only perfect squares
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](http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.dmj/1304429491) (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](http://math.ucr.edu/~mcc/paper/ON%20PROBLEMS%20OF%20ERD%22OS%20AND%20RUDIN.pdf) for more discussion of this and related questions.


