Rewrite the second equation as A+B=3C
Then the first as A
^{3} + B
^{3} = 3  C
^{3} factor and substitute
(A+B)(A²AB+B²) =3  C
^{3} (3C)(A²AB+B²) =3  C
^{3} (A²AB+B²) =(3C
^{3})/(3C)
the RHS must be an integer, the division leads to
=C²+3C+9+ 24/(C3)
which means (C3) is a factor of 24 so C must be one of
{21,9,5,3,1,0,1,2,4,5,6,7,9,11,15,26}
[And by the symmetry of A, B and C, they all must be one of these numbers.]
Put each of the above into (3C
^{3})/(3C)
To create a new list of quotients Q={386,61,5,1,1,1,5,61,61,71,89,121,166,281}
The two values that occur thrice: 1 and 61 are the only ones we need, they arise from permutations of A,B,C by the symmetry of these variables.
(A²AB+B²)= Q
By the quadratic formula
A=.5B±.5√(4Q3B²)
When Q=1, (C could be 1,0,or1)
the discriminant is only a perfect square if
B is 1,0,or1
leaving A to be one of 1,0,1 as well.
The only triple of these that actually works is
{1,1,1}When Q=61, (C could be 9,4 or 5)
B is ±9,±5,or±4
leaving A to be one of ±9,±5,or±4
The only triple of these that actually works is
{4,4,5}and its permutations

Posted by Jer
on 20140322 23:48:51 