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Let $C(r)$ be the origin-centered circle of radius $r$, and let $\beta(r)$ be the exterior buffer around $C(r)$: the distance from $C(r)$ to the closest lattice point exterior to $C(r)$:
For example, $\beta(2) = \sqrt{5}-2 \approx 0.24$ because there are no lattice points strictly between $C(2)$ and $C(\sqrt{5})$, and this is the largest buffer around $C(2)$.

I am interested in the behavior of $\beta(r)$ for large $r \in \mathbb{R}$, as I believe understanding that behavior will answer my question concerning ratchet spirals, Lattice radial-step (ratchet) spirals.

I'll pose a specific question before formulating the general question.

Q1. Is there an $R$ such that, for all $r > R$, $\beta(r) < \frac{1}{2}$ ?

If so, then, for example, the spiral $S(3,\frac{1}{2})$ depicted in that question is unbounded.

Q2. Is there an $R(\epsilon)$ such that, for all $r > R(\epsilon)$, $\beta(r) < \epsilon$, where $0 < \epsilon < 1$ ?

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2 Answers 2

up vote 5 down vote accepted

Bambah and Chowla ("On numbers which can be expressed as a sum of two squares") proved that there is a constant $C$ such that for any $x > 0$, there is an integer between $x$ and $x+ Cx^{1/4}$ which is the sum of two squares. From this it easily follows that the answer to the more general Q2 is in the affirmative. (in fact $x + Cx^{1/2 - \epsilon}$ would do for any $\epsilon$).

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I don't understand. Doesn't this give only a bound $\beta(r) \leq Cr^{\frac{1}{4}}$? –  Vít Tuček Oct 30 '13 at 15:53
You apply it with $x = r^2$. It gives a lattice point $(a,b)$ with $n = a^2 + b^2$, where $n$ is between $r$ and $\sqrt{r^2 + C\sqrt{r}}$. Therefore the distance to the circle is $\sqrt{n} - r$, but $r = \sqrt{r^2} \leq \sqrt{n} \leq \sqrt{r^2 + Cr^{1/2}} = r(1 + r^{-3/2}/2 ) = r + r^{-1/2}/2$. –  Abhinav Kumar Oct 30 '13 at 16:13
I see. Much obliged. –  Vít Tuček Oct 30 '13 at 16:39
Great---Thanks, Abhinav! This resolves my other question as well. –  Joseph O'Rourke Oct 30 '13 at 17:20

This is really just a comment, not a complete answer.

Suppose $r$ is integral. The distance from the point $[1,r]$ to $C(r)$ is $$\sqrt{1+r^2}-r$$ which has asymptotics $\frac{1}{2r}$ and hence for integral diameters the answers are affirmative.

If $r$ is nonintegral, then the point $[0,\lceil r \rceil]$ is closer to $C(r)$ than $[0,r]$ and its distance to $C(r)$ $$ \lceil r \rceil - r $$ gets arbitrarily close to 1 (as Abhinav Kumar pointed out in the comments). So for nonintegral diameters one really has to pick a good lattice point inside the first quadrant.

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But for nonintegral $r$, the point $(0,r)$ is not integral ... –  Abhinav Kumar Oct 30 '13 at 12:38
Sorry, I meant to take the ceiling of $r$. Should be fixed now. Thanks! –  Vít Tuček Oct 30 '13 at 13:48
That doesn't quite work - the distance from that point to $C(r)$ is $\lceil r \rceil - r$ which can be arbitrarily close to $1$ (and in particular is not $O(1/r)$). –  Abhinav Kumar Oct 30 '13 at 14:07
Right. There's the stupid mistake I've been talking about. :) I'll edit my "answer" accordingly. Thanks again. –  Vít Tuček Oct 30 '13 at 15:07

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