Find six distinct positive integers A, B, C, D, E, F, G satisfying:
A3 + B3 = C3 + D3 = E3 + F3 = 19G3.
Please submit primitive solutions only, that is, A, B, C, D, E, F, G should not have a common factor.
clearvars,clc
values=double.empty(0,3);
for a=1:4000
for b=1:4000
if b>a
g3=(a^3+b^3)/19;
if g3==floor(g3)
g=round(g3^(1/3));
if g^3==g3
values(end+1,:)=[a,b,g];
end
end
end
end
end
v=sortrows(values,3);
for i=1:length(v)-2
if v(i,3)==v(i+1,3) && v(i,3)==v(i+2,3)
disp(v(i,:))
disp(v(i+1,:))
disp(v(i+2,:))
disp(' ')
end
end
finds answers that are all multiples of the sought solution shown here:
G
A, B: 70 560 210
C, D: 198 552 210
E, F: 315 525 210
Of course A and B can be interchanged, etc. and the pairs themselves can be permuted, so long as 70 is still paired with 560, etc.
The cubes of the members of the pairs are
343000 175616000
7762392 168196608
31255875 144703125
and
19*G^3 = 175959000
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Posted by Charlie
on 2024-04-18 08:57:42 |
computer solution
