real valued function with "irregular" forced roots

Question

I'm looking for some techniques or ideas for constructing a real-valued function $ f $ with the following properties:

1. $ f $ is smooth, say $ C^\infty $, but at the very least, continuous on $ \mathbb{R} $.
2. $ f $ is even
3. $ f $ has roots at only the points $ \sqrt{k^2+kl+l^2} $, $ (k,l)\in{\mathbb{Z}}^2 $

The quadratic $ k^2+kl+l^2 $ takes nonzero values $ 1,3,4,7,12,13,19,21,... $ and appears to be sequence OEIS A244819.

A simpler example is $ \sin(\pi x^2) $, which vanishes when $ x = \sqrt{k} $, $ k \in{\mathbb{Z}} $.

Any pointers to references is much appreciated. Thanks!

Answer 1

First of all, let's denote $E = \{\sqrt{k^2 + kl + l^2} : k, l \in \mathbb Z\}$. Note that $E \subseteq [0, \infty)$. Furthermore, let $f: \mathbb R \longrightarrow\mathbb R$ be even. Then $f^{-1}[0]$ is symmetric about the origin, i.e. $-f^{-1}[0] = f^{-1}[0]$. Hence, $E \neq f^{-1}[0]$ as $E$ is wholly nonnegative. Thus, conditions 2 and 3 contradict each other. Regardless, let's proceed.

It's a well known fact that for any closed subset $F \subseteq \mathbb R$ there is a smooth function $f: \mathbb R \longrightarrow \mathbb R$ such that $f^{-1}[0] = F$. Here's a proof. I believe it also works when the domain is an arbitrary smooth manifold, but I digress. Now all we have to do is show that $E$ is closed and apply this result. Indeed, the set $\{k^2 + kl + l^2 : k, l \in \mathbb Z \} \subseteq \mathbb Z$ is discrete, hence closed in $\mathbb R$. $\sqrt: [0, \infty) \longrightarrow [0, \infty)$ is a homeomorphism with inverse $x \mapsto x^2$. Hence, $E$ is closed so there exists a smooth $f: \mathbb R \longrightarrow \mathbb R$ with roots precisely $E$.