TITLE: Is the number of representations as the sum of two elements of a polynomial sequence always small?
QUESTION [15 upvotes]: Let $f(x) \in \mathbb{Z}[x]$ be a degree $d>1$ polynomial with integer coefficients. Define 
$$r(n) := | \{x,y \in \mathbb{Z} : f(x)+f(y) = n  \}|. $$
My question is:


Is it true that $r(n) \ll_{\epsilon} n^{\epsilon} $ for $\epsilon >0$? 


In certain cases (such as $ f(x)=x^{2k}$), one can "factor" the problem and deduce the desired result from the divisor bound. However, I do not see how to approach the general case in this manner. I am aware that there is a weaker but more general result of Bombieri and Pila which states that 
$$r'(n,M) := | \{x,y \in \mathbb{Z} : f(x,y) = n, |x|,|y| < M  \}| $$
satisfies $r'(n,M) \ll_{\epsilon} M^{1/d + \epsilon} $when $f(x,y)$ is an absolutely irreducible polynomial of degree $d$. In this greater level of generality this is nearly best possible as can be seen by taking $f(x,y) = x^d -y$.

REPLY [2 votes]: I don't have the rep to comment so I will just mention something here, though I would guess you might have recognized it already. 
You can certainly bound the number of solutions to $f(x)-f(y)=n$ by writing $$f(x)-f(y)=(x-y)H(x,y)$$ for some polynomial $H$ that depends only on $f$. Then $x-y=d$ divides $n$ and $H(x,x-d)=n/d$ so there are $O_f(1)$ choices for $x$. Going from differences to sums is probably hard.