General Question

6rant6's avatar

Can you find the smallest radius sphere that has points with integer values?

Asked by 6rant6 (13705points) May 15th, 2012

Stepping up from this.

How about if we take a sphere instead of a circle. For example, a sphere of radius sqrt(38) will have integer coordinates here:

2,3,5 (and -2,3,5 2,-3,5, 2,3,-5, -2,-3,5 -2,3,-5 -,2,-3,5 -2,3,5)
2,5,3 (and 7 more)
3,2,5 (and 7 more)
3,5,2 (and 7 more)
5,2,3 (and 7 more)
5,3,2 (and 7 more)

6,1,1 (3 permutations * 2^3 arrangements of negative signs = 24 total)

for a total of 72.

Triples of the form x^2 +y^2 + z^2 = r^2 will generate
42 points if x,y,and z are unique
or 24 points if there are two unique numbers
or 8 it they are all the same.

Is 7 the lowest radius for 54 points? Is it the smallest radius for an integer sphere with more than 15 contact points?

Is sqrt(38) the lowest radius for 72 points?

Is the sphere with the most integer intersects always centered on the origin?

What are some other minimum radii for integer coordinate points on the sphere?

sqrt(86) has 108 points. Is that the minimum?
sqrt(101) has 144 points. Is that the minimum?

What’s the minimum integer radius for a sphere with more than 95 points?

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7 Answers

PhiNotPi's avatar

When you say that there will be 42 points if the x y z are unique, did you mean 48? There are six ways to arrange the numbers and eight combinations of the signs.

Besides that, one other observation might be determining not the possible radii, but the different possible values for the number of points intersected. Will it always be a multiple of two (I think so)? What conditions cause it to be a multiple of four or eight? With this info, we can reverse- engineer the answer.

For the question of >95, it becomes the sphere with at least 96, which is a multiple of eight. If we could find a pattern as to which radii cause multiples of eight, we could limit the number of solutions.

6rant6's avatar

Radius 15 has 102 intersect, I think.

(11,10,2) – 48
(14,5,2) – 48
(15,0,0) -6

6rant6's avatar

@PhiNotPi Yeah, typo. 48, not 42.

The idea of patterns is complicated by the fact that some triples create 48 points (x,y,z), some 24 (x,x,y) and (x,y,0), some 8 (x,x,x) some 12 (x,x,0) , some 6 (x,0,0).

6rant6's avatar

161 has 48 *4 = 192 points
(triples 9,8,4 10,6,5 11,6,2 12,4,1)

593 has 48*5 = 240 points
(triples 18,13,10 19,14,6 20,12,7 22,10,3 24,4,1)

5555 has 48*12 =576 points
(triples 45,43,41 53,39,35 57,41,25 59,43,15 59,45,7 63,31,25 63,35,19 67,25,21 67,29,15 69,25,13 71,17,15 73,15,1) [It’s definitely not the smallest.]

6rant6's avatar

points, radius
1, 0
6, 1
12, 2
24, 5
30, 9
48, 14
72, 26
96, 41
120, 74
144, 89
168, 101
192, 146
240, 194

BonusQuestion's avatar

Do you only consider spheres centered at zero? If so, I don’t think a closed formula can be achieved in general. It is closely related to the function r_3(n): http://mathworld.wolfram.com/SumofSquaresFunction.html

Guass has a formula for r_3(n) when n is square-free but even that involves class numbers.

6rant6's avatar

@BonusQuestion In my OP, one of the subsidiary questions was “Is the sphere with the most integer intersects always centered on the origin?” which is equivalent to centering them on zero. I would be very interested to learn there was a least radius sphere that wasn’t on the origin.

A related concept which may shed some light (when I’m less busy) is Pythagorean quadruplets.

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