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Nov 26 at 22:08 vote accept Dima Pasechnik
Nov 26 at 1:16 answer added GH from MO timeline score: 5
Nov 26 at 0:15 history edited LSpice CC BY-SA 4.0
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Nov 25 at 20:50 comment added Dima Pasechnik x and y are allowed to be rationals.
Nov 25 at 19:41 comment added Peter Mueller Are $x$ and $y$ supposed to be integers, or allowed to be rationals?
Nov 25 at 19:13 comment added Aleksei Kulikov Most likely the answer is no, see this MO question mathoverflow.net/questions/373900/… (it deals with the case of integer coefficients, but I would assume it is not much different for rational).
Nov 25 at 19:12 history edited Ofir Gorodetsky CC BY-SA 4.0
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Nov 25 at 19:08 history edited Dima Pasechnik CC BY-SA 4.0
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Nov 25 at 19:08 comment added Ofir Gorodetsky Yes, things are different if you restrict $n$ to be a prime. For each $a,b\ge 0$, the representable set has positive density within the primes (e.g. for $a=b=1$, this is due to Dirichlet). The precise density is related to class field theory, see David A. Cox's book "Primes of the form x² + ny²". So it is likely that in this case finitely many $(a_k,b_k)$-s are sufficient -- but I don't think this was worked out (a version of this has been asked as Problem 4 here: math.purdue.edu/~sahay5/frg-problem-session.pdf ).
Nov 25 at 19:05 comment added Dima Pasechnik Would the density be different for $n$ prime?
Nov 25 at 18:54 comment added Ofir Gorodetsky For fixed $a,b \in \mathbb{Q}_{\ge 0}$ the representable set has density $0$ (for $a=b=1$ this is due to Landau; for general $a,b\ge 0$ this is due to Bernays, at least for integers $a,b$). So a finite set of $(a_k,b_k)$-s is not sufficient. However, see Green and Soundararajan's preprint "Covering integers by x^2+dy^2" for a positive result: arxiv.org/abs/2401.04817
Nov 25 at 18:49 history asked Dima Pasechnik CC BY-SA 4.0