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=TA
40 for Co1=1 to 1 step 2
50 for Co2=1 to 1 step 2
60 C=3Co1*ACo2*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 20140322 13:17:21 