Find positive integers a, b, and c, all different, such that
a^3 + b^3 = c^4.

Please obey the title !

(In reply to re: New solutions not relatively prime by Brian Smith)

It's fortunate this came up, because I was motivated to find my original workings for Taking the Fifth.

That problem was based on a conjecture, in turn based on an observation.

Start with Sloane A003325, Numbers that are the sum of 2 positive cubes {2, 9, 16, 28, 35,...}. These numbers can be square: A050801, the squares at A050802. A051386 contains Numbers whose 4th power is the sum of two positive cubes. It starts the same as A003325, but has new entries, e.g. 134, 182:

134 = (2*67)^4 = 469^3+603^3 = (7*67)^3+(9*67)^3, because 7^3+9^3=1072 = 2^4*67,and 7+9=16.
182,455,1001; 2*(7*13), 5*(7*13), 11*(7*13), because 5^3+11^3 = 1456 = 2^4*(7*13), and 5+11=16.

Generally, 16(12n^2 - 84n + 211)^4 = (2n + 1)^3(12n^2 - 84n + 211)^3 + (15 - 2n)^3 (12n^2 - 84n + 211)^3, always true. Some solutions may be 'trivial'.

Similarly,
183,793,854; 3*61,13*61,14*61, because 13^3+14^3 = 4941 = 3^4*61, and 13+14=27
201,737,1072; 3*67,11*67,2^4*67, because 11^3+(2^4)^3 = 5427 = 3^4*67, and 11+2^4=27,...

Generally, 81(n^2 - 27n + 243)^4 = (n^3 - 27n^2 + 243n)^3 + (27 - n)^3 (n^2 - 27n + 243)^3, always true. Some solutions may be 'trivial'.

But also:
497,1136,3905; 7*71,2^4*71, 5*11*71, because 16^3+55^3 = 170471 = 7^4×71, and 55+16=71,...
and
658,3666,5170; (2*7)*47, (2*3*13)*47,(2*5*11)*47, because (2*3*13)^3+(2*5*11)^3 = (2*7)^4*47, and (2*3*13)+(2*5*11) = 2^2*47, a multiple of 47

Generally, if a solution to A^3 + B^3 = C^4 is 'non-trivial' in these cases, then C^4 can be expressed as (c*(a+b))^4 and we have something like (c*(a+b))^4 = (a(a+b))^3+(b(a+b))^3, [e.g.(7*(16+55))^4 = (16(16+55))^3+(55(16+55))^3]
which simplifies to [k]c^4=a^2-ab+b^2, with [k] the common factor multiple, if any.

But also, (a(a+b))^3+(b(a+b))^3 = (a + b)^3 (a^3 + b^3); the first term is a cube, while the second is another sum of cubes.

This led to the conjecture; given that a number (in this case a 4th power, it can be extended, e.g. to 5th powers) is a 'non-trivial' sum of cubes, it must be possible to sever it into a cube and a sum of smaller cubes.

For example, 3549^5, the smallest 'non-trivial' 5th power sum of cubes. 3549 = 3*7*13^2;  3549^5 = c^3 (a^3 + b^3) and it is now straightforward to compute that  a=403, b=650 (or vice versa) and c=1183;(7*13^2)^3((13*31)^3+(2×5^2×13)^3) = 3549^5.

Edited on May 8, 2017, 6:45 am

Posted by broll on 2017-05-08 06:39:33