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Simultaneous Settlement IV (Posted on 2014-03-22) Difficulty: 3 of 5
Determine all possible ordered triplets (A,B,C) of integers satisfying this simultaneous equation:

A3 + B3 + C3 = 3 and:
A + B + C = 3

Prove that there are no others.

No Solution Yet Submitted by K Sengupta    
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Solution Solution. Comment 2 of 2 |
Rewrite the second equation as A+B=3-C
Then the first as A3 + B3 = 3 - C3
factor and substitute
(A+B)(A-AB+B) =3 - C3
(3-C)(A-AB+B) =3 - C3
(A-AB+B) =(3-C3)/(3-C)
the RHS must be an integer, the division leads to
=C+3C+9+ 24/(C-3)
which means (C-3) 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 (3-C3)/(3-C)
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.

(A-AB+B)= Q
By the quadratic formula
A=.5B.5√(4Q-3B)

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,or4
leaving A to be one of 9,5,or4
The only triple of these that actually works is {4,4,-5}
and its permutations

  Posted by Jer on 2014-03-22 23:48:51
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