When I compute the following expression to find integer solutions of the equation (x^2 + y^2 + z^2 == 14^2)
Solve[x^2 + y^2 + z^2 == 14^2 && x > 0 && y > 0 && z > 0, {x, y, z}, Integers]
Mathematica returns
{{x -> 4, y -> 6, z -> 12}, {x -> 4, y -> 12, z -> 6}, {x -> 6, y -> 4, z -> 12},
{x -> 6, y -> 12, z -> 4}, {x -> 12, y -> 4, z -> 6}, {x -> 12, y -> 6, z -> 4}}
which are permutations of the only solution: x=4, y=6, z=12.
How can I remove other "solutions" by somehow merging the permutations? For me, x, y, and z are equivalent.
I don't want to do the following:
First[Solve[x^2 + y^2 + z^2 == 14^2 && x > 0 && y > 0 && z > 0, {x, y, z}, Integers]]
Because I will also solve:
Solve[x^2 + y^2 + z^2 == 14^3 && x > 0 && y > 0 && z > 0, {x, y, z}, Integers]
which has more than one solution.
Edit (by belisarius)
Is there a way to specify the equivalency to Solve[] or Reduce[] so to spare the results post-processing stage?
