Classic Geometry Problem -- Place a circle on a (2-d) lattice so that n points of the lattice are on the circumference

[mathman]

TL;DR Summary

Place circle on lattice so that n points are on circumference.

In a book (1984) with an interview of Coxeter, an old geometry question was described. Place a circle on a (2-d) lattice so that n points of the lattice are on the circumference. The answer for n=7 was given. Center is $(\frac{1}{3},0)$ and radius is $\frac{5^8}{3}$. Has it been solved for higher values of n since?

 

[jambaugh]

n=8 should be obvious by symmetry. Center (0,0) through $r^2 = 5$. It would pass through $(\pm 2, \pm 1)$ and $(\pm 1, \pm 2)$.

If you insist on an integral radius, try center (0,0) radius 5, passing through $(\pm 3,\pm 4)$ and $(\pm 4, \pm 3)$. Likewise with any other pythagorean triple. I believe there are some distinct (not reduced) pythagorean triples with common hypotenuses so I'm guessing there are some n=16 solutions.

(Will search and edit post...)

[Edit] Ah yes, with radius 25 centered at origin, you get (7,24) and (15,20) and their transposes and sign flips. A simple matter to search through and find more from composite c values.

 

[mfb]

Radius 5 also passes through (0,+-5) and (+-5,0), so we get 12 points. I think OP is looking for exactly n points. Exactly 12 is easy. Exactly 4 is trivial. Exactly 8 has an example. But for everything that's not a multiple of 4 you need to break the symmetry.

Common hypothenuses are easy to make. Multiply the first one by the hypotenuse of the second and vice versa.

 

[jambaugh]

Oops, @mfb, missed those obvious ones, also my n=16 example becomes n=20.

Yes, I agree, the hard problems are for specific n, not just "some bigger value of n". I think that by my scheme it's possible to prove no upper limit on how many by generating an ever increasing sequence using the formula for P-Trips.

 

[jambaugh]

Addendum: I was just thinking about how to tackle the general question. I would guess one would start with the fact that three non-linear points determine a circle uniquely, so characterize cases with triples of grid points then hit it with Galois Theory.

 

[mfb]

Yes, there is no upper limit.
Let a_n,b_n,c_n be N Pythagorean triples. Consider a circle centered at zero with radius C=prod(c_n). It passes through (0,C), (Ca_n/c_n,Cb_n/c_n) and all other symmetric points.

Edit: Obviously you want the N triples to be independent, otherwise the ratios a_n/c_n will be the same for different triples.

 

[mfb]

Circle Lattice Points

Schinzel's theorem shows that for every positive integer n, there exists a circle in the plane having exactly n lattice points on its circumference. The theorem also explicitly identifies such "Schinzel circles" as
$(x-\frac 1 2 )^2 + y^2 = \frac 1 4 5^{k-1}$ for n=2k
$(x-\frac 1 3 )^2 + y^2 = \frac 1 9 5^{2k}$ for n=2k+1

The answer is yes, and there is an explicit formula to construct circles for every n.

 

[anuttarasammyak]

Interesting problem. Without losing generosity we can say (0, 0) is on the circle so its equation is
$$(x-a)^2+(y-b)^2=a^2+b^2$$
for the circle at least one lattice point is on it. Center of the circle in the 1st quadrant is enough so
$$a,b \geq 0$$
Furhter its half
$$0 \leq b \leq a$$
is enough for consideration.
$$x^2+y^2=2ax+2by$$
We know a and b should be rational for more lattice points on the circle.
So the problem is restated: Tune rational numbers $0 \leq b \leq a$ so that N integer (m,n) pairs satisfy the above equation.

Obviously say $a=\frac{m}{2}$, lattice point (m,0) satisfies the condition. Say $a=\frac{m}{2},b=\frac{n}{2}$ (m,n) satisfies the condition so there are at least four points including (0,0) satisfy the condition.