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Consider the set of coprime integer pairs $\mathcal{C} \subset \mathbb{Z}^2$ and the circle of radius $r$ centered at the origin. The function $$f(r) = \min_{(m,n) \in \mathcal{C}} \bigl| \sqrt{m^2 + n^2} - r \bigr|$$ represents the minimum Euclidean distance between $\mathcal{C}$ and the circle of radius $r$.

I am interested in an explicit upper bound on $f(r)$. An obvious upper bound is $f(r) \leq \frac{1}{2}$ for $r \geq 1$, since circles of radius at least $1$ will cross the line $(1,\mathbb{Z})$ which consists of consecutive coprime pairs. But I need way better than this, ideally a bound that tends to zero as $r \to \infty$.

First of all, is it even true that $\limsup_{r \to \infty} f(r) = 0$?

The classical result on the asymptotic density of coprimes tending to $6/\pi^2$ would indeed suggest this, but the statement seems way too weak to be of any use here. I am also aware of the Jacobsthal function, though bounds thereof do not seem to help either for the question at hand.

NB: in my actual research problem I consider centered ellipses rather than circles, which might make the problem even harder. Specifically, I want to find a good upper bound on $$\min_{(m,n) \in \mathcal{C}} \left| \begin{bmatrix} m & n \end{bmatrix} \mathbf{A} \begin{bmatrix} m \\ n \end{bmatrix} - C \right|$$ for some $C > 0$ and positive definite $\mathbf{A} \in \mathbb{R}^{2 \times 2}$. This bound should tend to zero as $C \to \infty$.

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    $\begingroup$ Google "primitive circle problem." (Will add an answer later if it's not resolved by then.) $\endgroup$
    – Lucia
    Oct 27, 2016 at 21:05
  • $\begingroup$ Thanks for this pointer, it is very related indeed! However, it seems that the primitive circle problem is typically studied in terms of the number $V(r) = \frac{6}{\pi}r^2 + O(r^{1+\epsilon})$ of primitive points inside a circle. However, in my problem I do not need bounds on the $O(r^{1+\epsilon})$ term (the typically studied problem), but lower bounds on increments of this error term. $\endgroup$
    – jens
    Oct 27, 2016 at 21:51
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    $\begingroup$ If you look at $m^2+n^2-r^2$, you're looking at gaps in the sequence of sums of two squares (well, two coprime squares). $\endgroup$ Oct 27, 2016 at 21:56
  • $\begingroup$ Perhaps you might find the following useful : Adhikari and various coauthors (Chen and Granville to name two) have papers on visibility on points of the plane. The points nearest your ellipse , assuming the ellipse is contained in an n by n box, are visible from fewer than C log n points. There may be enough close points to the ellipse that you can see one from the center using this estimate. Gerhard "Maybe Close Is Good Enough?" Paseman, 2016.10.27. $\endgroup$ Oct 28, 2016 at 5:07

1 Answer 1

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Here's an easy way to show that $f(r)$ goes to zero. Indeed it shows that $f(r) \le C r^{-1/2+\epsilon}$ for some constant $C$ and any $\epsilon > 0$.

We start with the observation that given any number $n$, $$ \sum_{\substack{k\le x \\ (k,n) =1}} 1 = \frac{\phi(n)}{n} x + O(2^{\omega(n)}), $$ which follows from inclusion-exclusion (see Section 3.1 of Montgomery and Vaughan's book Multiplicative number theory, for this and other details below). Here $\omega(n)$ is the number of distinct prime factors of $n$. Since $\phi(n)/n$ is never too small (always bounded below by $c/\log \log n$), and $2^{\omega(n)}$ may be bounded by $n^{\epsilon/2}$, it follows that every interval of length $Cn^{\epsilon}$ contains an integer coprime to $n$. One can quantify this better, but this is enough.

Now choose $n$ to be the largest integer below $r$. Therefore $\lceil r^2\rceil -n^2= N$ is an integer of size $\le 3r$. Now choose $m$ to be the largest integer below $\sqrt{N}$ that is also coprime to $n$. From our observation, we may choose $m$ within $Cn^{\epsilon}$ of $\sqrt{N}$, and so $0\le N-m^2 \le C_1 \sqrt{N} n^{\epsilon}$.
Then $$ 0\le r - \sqrt{n^2 +m^2} \le \frac{1}{r} (r^2 -n^2-m^2) \le \frac{1}{r} (C_1 \sqrt{N} n^{\epsilon}) \le C_2 r^{-1/2 +\epsilon}, $$ as claimed.

My earlier answer produced $p^2+m^2$ close to $r^2$ for a prime $p$, but this is overkill. You can also adapt the argument to ellipses $f(x,y)= ax^2+bxy+cy^2$ with discriminant $D=b^2-4ac <0$. One way to do this is to multiply $f(x,y)$ by $4a$ and complete the square, thus getting $(2ax+by)^2 -Dy^2$. Now look for coprime $(X,Y)$ with $X^2-DY^2$ close to $4ar^2$, and arrange for $Y$ to lie in a suitable progression (given $X$) so that you can recover $y=Y$ and $x =(X-bY)/(2a)$.

Alternatively, you can use work toward the primitive circle problem. It $r_0(n)$ denotes the number of primitive representations of $n$ as a sum of two squares then it is known that $$ \sum_{n\le x} r_0(n) = C x +O(x^{1/2} \exp(-C (\log x)^{3/5-\epsilon})), $$ and this is enough to show that $f(r) \to 0$ (you just need to beat $\sqrt{x}$ in the error term). This again is known for ellipses; see for example Nowak and follow the references there.

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  • $\begingroup$ $-3/5$ in the final exponent? $\endgroup$ Oct 28, 2016 at 16:12
  • $\begingroup$ @GregMartin: No just 3/5 -- as in the prime number theorem. Missing a close bracket though. $\endgroup$
    – Lucia
    Oct 28, 2016 at 16:15
  • $\begingroup$ @Lucia: Thanks a lot for your multiple answers. It takes me a while to unravel your answer because I am unfamiliar with number theory. Why do you say your previous version was overkill? Didn't it yield a much better exponent than the $-1/2+\epsilon$ that you have now? I am primarily interested in the best exponent, and to the extent that it is possible, in explicit (constant) factors. For the interest of other readers, may I suggest to restore your previous solution (in addition to the other two approaches)? $\endgroup$
    – jens
    Oct 28, 2016 at 23:46
  • $\begingroup$ It yielded a worse exponent! You want to get the exponent as negative as possible -- the earlier argument had -0.2375 while this has -1/2+epsilon. The exponent here is definitely the best known -- even if you omit the coprimality condition, nothing more is known. $\endgroup$
    – Lucia
    Oct 28, 2016 at 23:50
  • $\begingroup$ Ah, of course! Thanks again for your valuable inputs. Now, for my understanding, how can I derive (or find a reference for) the identity which "follows from inclusion-exclusion"? I presume $\phi$ is the totient function, right? Do you think there is a chance of getting an explicit upper bound on $C_2$? And finally, do you reckon that your approach is extensible to ellipses? $\endgroup$
    – jens
    Oct 29, 2016 at 0:10

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