# How to merge permutations obtained from Solve on multiple variables?

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?

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• Related: (3554) – Mr.Wizard Feb 3 '15 at 7:25

You can use DeleteDuplicates, DeleteDuplicatesBy (Version 10) or GatherBy as follows:

ddF = DeleteDuplicates[#, Sort[Last /@ #] == Sort[Last /@ #2] &] &;
ddbF = DeleteDuplicatesBy[#,Sort[Last/@#]&]&;
fgbF = First /@ GatherBy[#, Sort[Last /@ #] &] &;


Examples:

sol1 = Solve[ x^2 + y^2 + z^2 == 14^2 && x > 0 && y > 0 && z > 0, {x, y, z}, Integers];
sol2 = Solve[x^2 + y^2 + z^2 == 14^3 && x > 0 && y > 0 && z > 0, {x, y, z}, Integers];

ddF[sol1]
(* {{x -> 4, y -> 6, z -> 12}} *)

ddF[sol2]
(* {{x -> 2, y -> 6, z -> 52}, {x -> 2, y -> 36, z -> 38},
{x -> 10,  y -> 12, z -> 50}, {x -> 12, y -> 22, z -> 46},
{x -> 12, y -> 34,  z -> 38}, {x -> 14, y -> 28, z -> 42},
{x -> 18, y -> 22,  z -> 44}, {x -> 20, y -> 30, z -> 38}} *)


ddbF and fgbF produce the same output.

For the given problem it is better to use PowersRepresentations or IntegerPartitions which not only avoid the problem but are far faster as well:

IntegerPartitions[14^2, {3}, Range^2] // Sqrt

{{12, 6, 4}}

PowersRepresentations[14^2, 3, 2] ~DeleteCases~ {___, 0, ___}

{{4, 6, 12}}


IntegerPartitions[14^3, {3}, Range^2] // Sqrt

{{52, 6, 2}, {50, 12, 10}, {46, 22, 12}, {44, 22, 18},
{42, 28, 14}, {38, 36, 2}, {38, 34, 12}, {38, 30, 20}}

PowersRepresentations[14^3, 3, 2] ~DeleteCases~ {___, 0, ___}

{{2, 6, 52}, {2, 36, 38}, {10, 12, 50}, {12, 22, 46},
{12, 34, 38}, {14, 28, 42}, {18, 22, 44}, {20, 30, 38}}


Your two example equations are symmetric with respect to $x$, $y$, and $z$. The symmetry leads to the redundancy you want to avoid. Rather than calculate all possible redundant solutions and delete duplicates, you could impose an extra constraint, $x\le y\le z$, to eliminate the duplicates. Hence,

Solve[x^2 + y^2 + z^2 == 14^2 && x > 0 && y > 0 && z > 0 && x <= y <= z,
{x, y, z}, Integers][[All, All, 2]]

{{4, 6, 12}}


and

Solve[x^2 + y^2 + z^2 == 14^3 && x > 0 && y > 0 && z > 0 && x <= y <= z,
{x, y, z}, Integers][[All, All, 2]]

{{2, 6, 52}, {2, 36, 38}, {10, 12, 50}, {12, 22, 46}, {12, 34, 38},
{14, 28, 42}, {18, 22, 44}, {20, 30, 38}}


Mr.Wizard's answer, which uses IntegerPartitions and PowersRepresentations, implicitly avoids the redundancy.

• This is very natural and doesn't involve another function. Thanks! – Richard Cox Jun 14 '15 at 1:31

One way to approach this is to find all the answers, sort the numbers, and then remove the duplicates:

sol = Solve[x^2 + y^2 + z^2 == 14^2 && x > 0 && y > 0 && z > 0, {x, y, z}, Integers];
Union@Map[Sort, sol[[All, All, 2]], 1]


which returns the unique solution for the three variables. For the 14^3 problem, the same procedure returns a collection of distinct answers

sol = Solve[x^2 + y^2 + z^2 == 14^3 && x > 0 && y > 0 && z > 0, {x, y, z}, Integers];
Union@Map[Sort, sol[[All, All, 2]], 1]

{{2, 6, 52}, {2, 36, 38}, {10, 12, 50}, {12, 22, 46}, {12, 34, 38},
{14, 28, 42}, {18, 22, 44}, {20, 30, 38}}