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What's special about this number states $8$ to be the largest fibonacci number who's also a cube (the third power of some integer). So it's basically the only one beacuse forget about $1$.

I Personally think $8$ can do better than that (e.g, it's the smallest order of a hamiltonian group - the quaternions) but we'll leave that for now.

I've searched for a proof here and suprisingly couldn't find one (maybe I haven't searched good enough?) so I would be curious to here why is that so. Thanks!

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  • $\begingroup$ Doesn't appear to be trivial. here is a discussion. $\endgroup$ – lulu Apr 15 '17 at 18:42
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    $\begingroup$ Nitpick: also $0$, $-1$, and $-8$ are cubes, but people usually forget about the zeroeth and negative places. (which is disappointing; I find $0,1$ a nicer place to start than $1,1$) $\endgroup$ – Hurkyl Apr 15 '17 at 18:47
  • $\begingroup$ @lulu Huh, and I innocently thought that running possibilities for cubic residues will do... Thanks for the reference $\endgroup$ – 35T41 Apr 15 '17 at 18:56
  • $\begingroup$ If you are interested in one cube for Fibonacci, you might also like two cubes. $\endgroup$ – Dietrich Burde Apr 15 '17 at 20:54
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An answer, together with references has been given on MO. In particular, this paper by Burgeaud et al. shows that 1, 8, and 144 are the only perfect powers in the Fibonacci sequence, which in particular implies that 8 is the largest cube.

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Wikipedia references a proof that $8$ and $144$ are the only perfect powers in the sequence of Fibonacci numbers. (presumably, restricting to $a^b$ with $a,b > 1$)

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