Question 1099447
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The basic solutions are these 5 triples:

a       b       c
-------------------

0       0       15     (1)

0       9       12     (2)

2	5	14     (3)

2	10      11     (4)

5	10	10     (5)



1)  Triple (1) creates  3 permutations, that are the solutions, too.

                               Playing with the signs, each of these permutations provides  2 triples, that are the solutions, too.

    So, the triple (1) produces 2*3 = 6 triples that are the solutions.



2)  Triple (2) creates  6 permutations, that are the solutions, too.

                               Playing with the signs, we get {{{6*2^2}}} = 24 triples, that are the solutions, too.

    So, the triple (2) produces 24 triples that are the solutions.



3)  Triple (3) creates  6 permutations, that are the solutions, too.

                               Playing with the signs, each of these permutations provides 2 triples, that are the solutions, too.

    So, the triple (3) produces {{{6*2^3}}} = 48 triples that are the solutions.


4)  Triple (4) creates  6 permutations, that are the solutions, too.

                               Playing with the signs, each of these permutations provides 2 triples, that are the solutions, too.

    So, the triple (4) produces {{{6*2^3}}} = 48 triples that are the solutions.


5)  Triple (5) creates  3 permutations, that are the solutions, too.

                               Playing with the signs, each of these permutations provides 2 triples, that are the solutions, too.

    So, the triple (4) produces {{{3*2^3}}} = 24 triples that are the solutions.
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In all, &nbsp;there are &nbsp;&nbsp;6 + 24 + 48 + 48 + 24 = 150 &nbsp;&nbsp;different integer triples that are the solutions.