(In reply to
First thoughts: a conjecture by Larry)
Conjecture: that for integers a,b,c, their cube roots can only be in arithmetic sequence if all 3 cube roots are integers.
Let a^(1/3) = 2*3^(1/3), let b^(1/3)=5*3^(1/3), let c^(1/3)=7*3^(1/3) (the puzzle does not specify necessarily consecutive terms)
then abc = 210^3
It is also possible to construct common differences that are not rational, e.g. 2(243)^(1/3)/3*3^(1/3), 3(243)^(1/3)/3*3^(1/3), 5(243)^(1/3)/3*3^(1/3), but these depend on cancellation between the common difference and the cube root.
Absent such tricks, the common difference must be integral, say np, where the terms again need not be consecutive, say np, xnp, ynp, such that xnp, ynp continue to be integral.
But then ((np(a^(1/3))(xnp(b^(1/3))(ynp(c^(1/3)))^3 is obviously a cube.
Edited on December 8, 2024, 12:44 am
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Posted by broll
on 2024-12-08 00:39:27 |
re: First thoughts: a conjecture
