Yesterday I read the Quanta article [How a Strange Grid Reveals Hidden Connections Between Simple Numbers](https://www.quantamagazine.org/the-sum-product-problem-shows-how-addition-and-multiplication-constrain-each-other-20190206/) about the sum-product problem: 


Let $A$ be a set of integers. Erdös and Szemerédi conjectured that for any $\epsilon>0$, there exists a $c\_{\epsilon}>0$ such that 


$\max\{|A+A|,|A \cdot A| \}\geq c\_{\epsilon}|A|^{2-\epsilon}$.


The Quanta article talks about recent progress in proving this conjecture. While I was reading the article, I was inspired to try to use the identity


$xy=((x+y)^2-x^2-y^2)/2$


to try to prove this conjecture, since I see squaring and adding numbers as more primitive operations than multiplying two numbers. Using this identity and the fact that $|(A+A)^2|=|A+A|$, I found that:


$|A \cdot A|+|A+A| = |A \cdot A|+|(A+A)^2|=|\{x^2+y^2-(x+y)^2:x,y \in A\}|+|\{(x+y)^2:x,y \in A\}| \geq |\{x^2+y^2:x,y \in A\}| = |A^2+A^2|$.


So to prove the conjecture, it suffices to prove that for any $\epsilon>0$, there exists a $c\_{\epsilon}>0$ such that 


$|A^2+A^2|=|\{x^2+y^2: x,y \in A\}|\geq c\_{\epsilon}|A|^{2-\epsilon}$.


A lot is known about the [sum of two squares](https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem). Not every number can be expressed as the sum of two squares, but many can. My question is is there a known number $n \leq 2$ such that for any $\epsilon>0$, there exists a $c\_{\epsilon}>0$ such that 


$|A^2+A^2|\geq c\_{\epsilon}|A|^{n-\epsilon}$?


Has this strategy been tried before?



Trick for the sum-product problem
Your question is a well-known and difficult open problem. See [Lower bounds for $|A+A|$ if $A$ contains only perfect squares](https://mathoverflow.net/questions/150142/lower-bounds-for-aa-if-a-contains-only-perfect-squares/150147#150147). To repeat my answer from that question, the best lower bound to date is:
$$|A^2+A^2| \geq |A| (\log |A| )^{c \log \log |A|}$$
due to Schoen in 2011 using his bounds on Freiman's theorem.


I think one drawback to your strategy is that the problem you end up with is considerably harder than the original problem. Indeed there are non-trivial polynomial-type lower bounds on the sum-product problem via rather elementary arguments. On the other hand, there isn't any non-trivial polynomial-type lower bound known on the sum of squares problem and Schoen's theorem above relies on two deep results: Freiman's theorem and a deep theorem about squares in arithmetic progressions due to Bombieri, Granville, Pintz using arguments related to Faltings's theorem.


One explanation for this is Boris' first comment / "counterexample". To make progress on the sum of squares problem you need to exploit the fact that the numbers involved are integers, where most of the sum-product technology doesn't distinguish between real numbers and integers (which, generally, is a feature).


