# $8$ is the only fibonacci number who's a cube (except $1$)

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!

• Doesn't appear to be trivial. here is a discussion. – lulu Apr 15 '17 at 18:42
• 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 $1,1$) – Hurkyl Apr 15 '17 at 18:47
• @lulu Huh, and I innocently thought that running possibilities for cubic residues will do... Thanks for the reference – 35T41 Apr 15 '17 at 18:56
• If you are interested in one cube for Fibonacci, you might also like two cubes. – Dietrich Burde Apr 15 '17 at 20:54

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 $a,b > 1$)