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From AP to cubes (Posted on 2024-12-07) Difficulty: 3 of 5
Let a, b, and c be positive integers such that a1/3, b1/3, and c1/3 are terms of an arithmetic sequence. Prove that abc is a perfect cube.

No Solution Yet Submitted by Danish Ahmed Khan    
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First thoughts: a conjecture | Comment 2 of 3 |
Observation:  for any 3 integers, x,y,z, the product of their cubes is a perfect cube; (x^3)(y^3)(z^3) = (xyz)^3

Conjecture: that for integers a,b,c, their cube roots can only be in arithmetic sequence if all 3 cube roots are integers.

I'm not certain that the conjecture is true, but if it is then it would be sufficient for a proof.

But proving the conjecture might be more difficult than proving the original problem, particularly since the conjecture may be wrong.

  Posted by Larry on 2024-12-07 15:13:02
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