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 Simultaneous Settlement IV (Posted on 2014-03-22)
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 No Rating

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 computer exploration | Comment 1 of 2

The following program was manually stopped at a point where the total of A's and B's absolute values happened to reach 53,974:

10   for T=2 to 99999999
20     for A=1 to T\2
30        B=T-A
40        for Co1=-1 to 1 step 2
50          for Co2=-1 to 1 step 2
60            C=3-Co1*A-Co2*B
70            if Co1*A*A*A+Co2*B*B*B+C*C*C=3 then print Co1*A;Co2*B;C
80          next
90        next
100     next
110   next
OK
run
1  1  1
4  4 -5
4 -5  4
Break in 80
?t
53974
OK

so, together with (-5, 4, 4), the ordered triplets are (1,1,1), (4,4,-5), and (4,-5,4).

Any further solutions would need |A| + |B| to exceed 53973, or actually any such pair from the three, due to the symmetry of the situation.

 Posted by Charlie on 2014-03-22 13:17:21

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