Concyclic points in $\mathbb{Z}^2$

Question

So I just came up with this question that I thought would be interesting to share: Consider $\mathbb{Z}^2$ as a subset of $\mathbb{R}^2$ and for a circumference $C$ in $\mathbb{R}^2$, let $f(C)=|C\cap\mathbb{Z}^2|$, i.e. the number of points of the lattice that lie on the circumference. Find the minimum $k\in\mathbb{N}$ such that $f(C)\neq k$ for all circumference $C\in\mathbb{R}^2$

(I hope it's worth a shot)

Let's just say (for convenience of this question) that for $k\in\mathbb{N}$, if there exists a circumference such that $f(C)=k$, then $k$ is admissible (not admissible otherwise). And let's say that a finite set of concyclic points $\mathcal{A}=\{A_1,\ldots,A_n\}$ is comaximal if they lie on a circumference and $f(C)=|\mathcal{A}\cap C|=n$ (for some $C$), i.e. there are no more points in $\mathbb{Z}^2$ that are concyclic with the points in $\mathcal{A}$.

At first it is easy to check that 1,2,3 and 4 are admissible.

Later I thought, "probably more even numbers than odd numbers are admissible, since there is usually plenty of symmetry in a circumference (if the centre lies in $\mathbb{Z}^2$ for example)", which is just a vague idea, because I just haven't thought this through. So I started to wonder "perhaps 5 is not admissible", but then I found that the points $(8,8), (10,0), (0,10), (-1,-5)$ and $(-5,-1)$ are comaximal, so $\min\{k:k\text{ is admissible}\}\geq7$, and that's all I have so far.

It'd be really cool if no such minimum exists though.

Cheers!

EDIT

The set of 5 points I suggested that satisfy the conditions of the problem were incorrect, though I've found now that the set $\mathcal{B}=\{(10,0),(0,10),(-5,-5),(-3,9),(9,-3)\}$ actually works with the relation $$(x-\frac{5}{4})^2+(y-\frac{5}{4})^2=\left(\frac{25\sqrt{2}}{4}\right)^2$$

If I understand your question correctly you can reduce your question to simply:
What is the smallest number $n$ such that there does not exist $n$ concyclic Guassian integers? Or is there no smallest number? — Mason, Jul 06 '18 at 22:02
Can you give the $h,k,r$ such that $(x-h)^2+(y-k)^2=r^2$ is satisfied by the (5) points: $(8,8), (10,0), (0,10), (-1,-5), $and $(-5,-1)$? — Mason, Jul 06 '18 at 22:04
About what you wrote firstly, yes. As for the second one $h=k=7/3$ and $r=17\sqrt{2}/3$ — Cristian Baeza, Jul 06 '18 at 22:20
It is nonetheless an interesting question. You should edit it though... — Mason, Jul 06 '18 at 22:48

score 1 · Answer 1 · answered Jul 09 '18 at 17:24

I just found with maple that for $h=k=\frac{395}{14}$ and $r=\frac{325\sqrt{2}}{14}$, you get 7 comaximal points, namely $$(0,45),(45,0),(20,60),(60,20),(44,57),(57,44),(5,5)$$ and 9 comaximal points show up more often, for example $h=k=\frac{89}{6}$ and $r=\frac{65\sqrt{2}}{6}$ yield $$(0,11),(11,0),(7,28),(28,7),(9,29),(29,9),(17,30),(30,17),(4,4)$$ so $\min\{k:k \text{ is admissible}\}\geq10$.

Concyclic points in $\mathbb{Z}^2$

1 Answers1