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Given a positive integer $n$, consider the diophantine equation $4x^2+y^2+4x+y=2n$ with solutions in non-negative integers $x$ and $y$.

Define the proportion $$\delta_n=\frac{\#\{(x,y)\in\mathbb{Z}^2_{\geq0}: 4x^2+y^2+4x+y=2n\}}n.$$

I would like to ask:

QUESTION. Is this true? $$\lim_{n\rightarrow\infty}\delta_n=0.$$

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    $\begingroup$ Yes, it's less than the proportion of integers of the form $8n+5$ which are sum of two squares, which tends to $0$ like $n/\sqrt{\log(n)}$. $\endgroup$
    – Henri Cohen
    Commented Mar 25, 2022 at 15:07
  • $\begingroup$ @HenriCohen: if you could add details and references, that could be instructional and useful. I may encourage you to write it as an answer. $\endgroup$
    – T. Amdeberhan
    Commented Mar 25, 2022 at 15:13
  • $\begingroup$ @T. Amdeberhan rearranging the terms we get $(4x+2)^2+(2y+1)^2=8n+5$ or $a^2+b^2=8n+5$. $\endgroup$
    – Alapan Das
    Commented Mar 25, 2022 at 15:16
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    $\begingroup$ Your question is answered here (when you combine with Alapan Das's observation): math.stackexchange.com/questions/264069/… $\endgroup$
    – GH from MO
    Commented Mar 25, 2022 at 16:29

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