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

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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?

Gold Member
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...)

 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.

Mentor
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.

Gold Member
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.

Gold Member
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.

Mentor
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.

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Mentor
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.

Gold Member
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.

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