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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 $f(x,y) = x^d -y$.

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  • $\begingroup$ Just out of curiosity, is there a story (similar to classical theta functions, along the lines of Fourier coefficient of modular forms) for series of the shape $\sum_{n \in \mathbb{Z}} \exp(2\pi i f(n)x)$ where $f(x) \in \mathbb{Z}[x]$ is a degree $d > 2$ polynomial? $\endgroup$ – user31415 Mar 2 '16 at 6:48
  • $\begingroup$ A related question is mathoverflow.net/questions/45511. $\endgroup$ – Richard Stanley Mar 2 '16 at 22:05
  • $\begingroup$ The implied constant in the bound you seek can depend on $f$, correct? $\endgroup$ – Bobby Grizzard Mar 2 '16 at 23:20
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    $\begingroup$ @GH, are you saying that because then the genus is at least 2, and boundedness would follow from the Bombieri-Lang Conjecture, or is there something more concrete that would apply to these specific curves? $\endgroup$ – Bobby Grizzard Mar 3 '16 at 2:51
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    $\begingroup$ I assume Bobby Grizzard is referring to the result of Caporaso-Harris-Mazur that shows that Bombieri-Lang gives a uniform bound on the rational points of any curve of a fixed genus greater than or equal to 2: ams.org/mathscinet-getitem?mr=1325796 . This would indeed seem to give GH's claim, conditional of course on Bombieri-Lang. $\endgroup$ – Terry Tao Mar 3 '16 at 5:06
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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.

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  • $\begingroup$ I had also observed this after posting the question. It certainly seems that this can't be generalized in any straightforward way to sums since there are polynomials such that f(x)+f(y) does not have linear factors. $\endgroup$ – Mark Lewko Mar 6 '16 at 23:38
  • $\begingroup$ Do you know when $f(x)+f(y)$ is absolutely irreducible? It's tempting to try and repeat the argument in a number field. $\endgroup$ – Brandon Hanson Mar 6 '16 at 23:48
  • $\begingroup$ I'm not sure. I thought about that briefly but didn't get too far. This quickly leads to questions outside of my expertise. $\endgroup$ – Mark Lewko Mar 6 '16 at 23:54
  • $\begingroup$ The answer at mathoverflow.net/questions/105304/… seems relevant here. $\endgroup$ – Terry Tao Mar 7 '16 at 7:55

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