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It is known from for example Representations of Integers as Sums of Squares by Grosswald, E. that $$|\{(n_1,n_2,\ldots,n_k)\in\mathbb{Z}^k: \ n_1^2+n_2^2+\cdots+n_k^2=N\}|\leq C_\varepsilon N^{\frac{k-2}{2}+\varepsilon}.$$ More generally, suppose we have a positive definite quadratic form of integer coefficients with possible linear terms $p(n_1,\ldots,n_k)$, say $p(n_1,n_2)=n_1^2+(n_2+1)^2+n_1(n_2+1)$, do we still have $$|\{(n_1,n_2,\ldots,n_k)\in\mathbb{Z}^k: \ p(n_1,\ldots,n_k)=N\}|\leq C_\varepsilon N^{\frac{k-2}{2}+\varepsilon}\ \ \ \ \ \ \ \ \ \ ?$$ I also know of another related fact (The number of integral points on arcs and ovals by E. BOMBIERI and J. PILA): if $\Gamma\subset\mathbb{R}^2$ is a real analytic image of the circle, then $|t\Gamma\cap \mathbb{Z}^2|\leq C_\varepsilon t^\varepsilon$. This could be used to prove special cases of the above inequality, say when $\Gamma=\{X^2+5Y^2=1\}$.

Edit: Also, is it possible to remove the $\varepsilon$ for $k\geq 5$? This is the case for counting representations of integers as sums of squares.

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  • $\begingroup$ Check out my answer to MO question 256576. It answers your question as well. $\endgroup$
    – GH from MO
    Commented Apr 9, 2020 at 3:16
  • $\begingroup$ @GHfromMO Thanks! $\endgroup$ Commented Apr 9, 2020 at 6:56
  • $\begingroup$ @GHfromMO Is it possible to remove the $\varepsilon$ for $k\geq 5$? This is the case for counting the representations as sums of squares. $\endgroup$ Commented Apr 28, 2020 at 1:15
  • $\begingroup$ I don't know, but probably the answer is yes. If you want to prove this, I suggest that you study the "Added 1" section in the link above, and try to generalize it for $k\geq 5$. The Eisenstein component $a(n)$ can be calculated explicitly by Siegel's maass formula, and I am sure one gets the required bound there. The cuspidal component $b(n)$ can probably be estimated similarly as in that post, but the details need to be worked out carefully. $\endgroup$
    – GH from MO
    Commented Apr 28, 2020 at 7:53

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