I use Mathematica. This code finds integer points on the sphere

(x-2)^2 + (y-4)^2 + (c-6)^2 =15

and select two of them so that distance of two this points equal to 4.

ClearAll[a, b, r, c];
a = 2;
b = 4;
c = 6;
r = 15; ss =
Subsets[{x, y, z} /.
Solve[{(x \[Minus] a)^2 + (y \[Minus] b)^2 + (z \[Minus] c)^2 ==
r^2, x != a, y != b, z != c, x y z != 0}, {x, y, z},
Integers], {2}];
t = Select[ss, And @@ Unequal @@@ Subsets[Flatten[#], {2}] &];
Length[t]
Select[ss, Apply[EuclideanDistance, #] === 4 &]

and this code select four points on the shere so that none of three points make a right triangle

ClearAll[a, b, r, c];
a = 2;
b = 4;
c = 6;
r = 15;
ss = Subsets[{x, y, z} /.
Solve[{(x - a)^2 + (y - b)^2 + (z - c)^2 == r^2, x != a, y != b,
z != c, x y z != 0, x > y}, {x, y, z}, Integers], {4}];
nonright =
Pick[ss, (FreeQ[#, \[Pi]/2] &) /@ ({VectorAngle[#2 - #1, #3 - #1],
VectorAngle[#1 - #2, #3 - #2],
VectorAngle[#1 - #3, #2 - #3]} & @@@ ss)];
Select[nonright, (12 == Length[Union @@ #] &)]

 I am looking for a  procedure in Maple.  I have some problems with this sphere. For example:

Choose four points so that 12 coordinates difference and it makes a square.

Can your code improve with sphere?

I want to reduce all solution of the equation sin(x)^2=1/4

restart:
sol:=solve(sin(x)^2=1/4, x, AllSolutions);

and

restart:
k:=combine((sin(x))^2);
sol:=solve(k=1/4, x, AllSolutions = true, explicit);
simplify(sol);

How can I reduce solution sol := -1/3*Pi*_B3+1/6*Pi+Pi*_Z3 ?

How can I get x= pi/6+k*pi and x= -pi/6+k*pi?

How can I solve this equation 18 *9^(x^2 + 2* x) + 768* 4^((x + 3)* (x - 1)) - 5 *6 ^((x + 1)^2)?

I tried

restart:

A:=18 *9^(x^2 + 2* x) + 768* 4^((x + 3)* (x - 1)) - 5 *6 ^((x + 1)^2);
solve(A=0);

I see that, the equation has three solutions: x = -2, x = -1 and x = 0. I check

f:=x->18 *9^(x^2 + 2* x) + 768* 4^((x + 3)* (x - 1)) - 5 *6 ^((x + 1)^2);

f(-2);

f(-1);

f(0);

Another question, Maple can not solve inequality 

18 *9^(x^2 + 2* x) + 768* 4^((x + 3)* (x - 1)) - 5 *6 ^((x + 1)^2) > = 0.

PS. We can easy to solve the above inequality with Mathematica

Reduce[18 9^(x^2 + 2 x ) + 768 4^((x + 3) (x - 1)) - 5 6 ^((x + 1)^2) >= 0 , x, Reals]

I got x <= -2 || x == -1 || x >= 0

When I use fsolve with equation 

-x^2 + 2*x + 5 + (x^2 + 2*x - 1)* sqrt(2 - x^2)=0

I got only one solution.

fsolve(-x^2 + 2*x + 5 + (x^2 + 2*x - 1)* sqrt(2 - x^2)=0,x);

In fact, it have two reals solutions.  

I posted at here

http://mathematica.stackexchange.com/questions/83985/does-the-equation-have-two-roots/83991#83991

I have a list L:={1,2,3,4,5,6,7,8}. I choose three elements from list L. How many different combinations of three numbers can be selected from L so that the numbers could represent the side lengths of a triangle?