2
$\begingroup$

In there a finite set of $(a_k,b_k)\in\mathbb{Q}_{\geq 0}\times \mathbb{Q}_{\geq 0}$ so that any $n\in \mathbb{Z}_{>0}$ can be written as $n=a_k x^2+b_k y^2$ for some $k$ ?

This is related to things like universal binary quadratic forms, but in my case I don't allow cross-terms $c_k xy$, and everything is nonnegative.

I imagine this must be well-known, but I can't find a reference.

EDIT: after a comment by Ofir, I'd like to ask about the case of $n$ being a prime. Already half of all primes are of the form $x^2+y^2$, so one can wonder whether finitely many forms would suffice for all primes.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ 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 $\endgroup$
    – Ofir Gorodetsky
    Commented Nov 25 at 18:54
  • $\begingroup$ Would the density be different for $n$ prime? $\endgroup$
    – Dima Pasechnik
    Commented Nov 25 at 19:05
  • $\begingroup$ 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 ). $\endgroup$
    – Ofir Gorodetsky
    Commented Nov 25 at 19:08
  • 1
    $\begingroup$ 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). $\endgroup$
    – Aleksei Kulikov
    Commented Nov 25 at 19:13
  • $\begingroup$ Are $x$ and $y$ supposed to be integers, or allowed to be rationals? $\endgroup$
    – Peter Mueller
    Commented Nov 25 at 19:41

1 Answer 1

Reset to default
5
$\begingroup$

There is no such finite set of pairs $(a_k,b_k)\in\mathbb{Q}_{\geq 0}^2$. Indeed, because the problem concerns rational representations, we can assume without loss of generality that $(a_k,b_k)\in\mathbb{Z}_{\geq 1}^2$. If the prime $p$ is represented by some $a_kx^2+b_ky^2$ over $\mathbb{Q}$, then there exists a positive integer $m$ such that $pm^2$ is primitively represented by $a_kx^2+b_ky^2$ over $\mathbb{Z}$. So there is a quadratic form $pm^2x^2+cxy+dy^2$ which is equivalent to $a_kx^2+b_ky^2$. Taking the discriminants of these quadratic forms, we see that $-4a_kb_k$ is a square modulo $p$. Passing to the fundamental discriminants underlying the discriminants $-4a_kb_k$, we would obtain a finite set $\mathcal{D}$ of negative fundamental discriminants such that for every sufficiently large prime $p$, there exists $d\in\mathcal{D}$ with $\chi_d(p)=1$. This is impossible by Lucia's response to MO question 373900.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Sorry, I don't follow - why is it possible to assume $(a_k,b_k)∈Z_{≥1}$ ? You can assume that denominators of $a_k$ and $b_k$ are square-free, but further than that? $\endgroup$
    – Dima Pasechnik
    Commented Nov 26 at 20:05
  • $\begingroup$ @DimaPasechnik Let $a_k=c_k/q_k$ and $b_k=d_k/q_k$, where $c_k,d_k\in\mathbb{Z}_{\geq 0}$ and $q_k\in\mathbb{Z}_{\geq 1}$. It is easy to see that the range $\{a_kx^2+b_ky^2:(x,y)\in\mathbb{Q}^2\}$ equals $\{c_kq_kx^2+d_kq_k y^2:(x,y)\in\mathbb{Q}^2\}$. So we can assume that $(a_k,b_k)\in\mathbb{Z}_{\geq 0}^2$. If $a_k=0$ (resp. $b_k=0$), then we can replace it by $a_k=1$ (resp. $b_k=1$), because this only increases the range $\{a_kx^2+b_ky^2:(x,y)\in\mathbb{Q}^2\}$. Hence we can also assume that $(a_k,b_k)\in\mathbb{Z}_{\geq 1}^2$. I am not using squarefreeness anywhere. I hope this clarifies. $\endgroup$
    – GH from MO
    Commented Nov 26 at 21:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .