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.

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