Can you prove that 8 is the largest cube in the Fibonacci sequence?
For a much more accessible treatment, and history of the result, see Andrejic (2006) "On Fibonacci powers"

1$\begingroup$ In case the link doesn't work, try pefmath.etf.bg.ac.rs/to appear/rad520.pdf (.yu domain changed to .rs) $\endgroup$ – Mio Apr 7 '10 at 17:59

1$\begingroup$ I made a URL that actually works: tinyurl.com/y8afbvb $\endgroup$ – Dan Piponi Apr 7 '10 at 18:25

3$\begingroup$ Let's try again: doiserbia.nb.rs/img/doi/03538893/2006/035388930617038A.pdf $\endgroup$ – Andrés E. Caicedo Jun 29 '12 at 19:26

3$\begingroup$ The link works but most of the definite articles are missing. $\endgroup$ – Emilio Pisanty Mar 19 '13 at 1:39
This paper, using some heavyduty numbertheoretic machinery, 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. There may be an easier way to prove it for cubes, however.

1$\begingroup$ The Fibonacci Series, slightly wonky powers of 2. $\endgroup$ – alan2here Nov 3 '15 at 19:00

$\begingroup$ It seems, at least at first, that the mode average (of the simplest form of) every fraction of pairs of values from the sequence, draws for first place, at 2, and 3, and 5, and 8, and 13, and 21, etc... It's strangely meta :) $\endgroup$ – alan2here Nov 3 '15 at 19:05

$\begingroup$ Name of this paper: Bugeaud, Mignotte, and Siksek  Classical and modular approaches to exponential Diophantine equations. I. Fibonacci and Lucas perfect powers. $\endgroup$ – LSpice Feb 10 at 19:38
This is only a partial answer, but: one characterization of Fibonacci numbers is that an integer $a$ is Fibonacci if and only if $5a^2 \pm 4$ is a perfect square.
Because of this: finding Fibonacci cubes is equivalent to solving the Diophantine equation(s) $5a^6 \pm 4 = b^2$. There are standard techniques for doing this: for example, in the case of $5a^6 + 4 = b^2$, the equation can be rewritten and factored as $5a^6 = (b2)(b+2)$, and using the fact that any common factor of $b2$ and $b+2$ must divide 4, you can get a lot of information out of this situation by looking at the prime factorizations of both sides.
In the case of $5a^6  4 = b^2$, you have to pass to the Gaussian integers (numbers of the form $a+bi$ with $a$ and $b$ both integers) to obtain a factorization: $5a^6 = (b+2i)(b2i)$. Again, you can apply the same approach (Gaussian integers also have unique factorization): it will be a little bit trickier, but I've seen problems like this solved by the same method.
Anyone want to try to fill in this sketch of a proof?
Edit: Having thought about it a bit more, I'm less optimistic. The problem is dealing with the factors of 5: the methods I describe are excellent for dealing with questions like $a^6 \pm 4 = b^2$ or even $a^3 \pm 4 = b^2$, but those arguments break down when you take the 5 into account. Also, any method that uses either of the factorizations I mentioned will also have to say something about solutions to the equation $5a^2 \pm 4 = b^2$: but we know that this equation has many solutions (since a can be any Fibonacci number). There are other potential ways we could rearrange this equation to factor it, but analyzing them won't be easy.
However, rephrasing this question as a Diophantine equation is still helpful. There's a general and rather hard theorem (Siegel's Theorem) that any Diophantine equation of genus $\ge1$ has only finitely many solutions. What "genus" is is a bit technical, but most equations of degree $\ge3$ have this property, and in particular Siegel's equation applies to equations of the form $y^2 = P(x)$ where $P(x)$ is a polynomial in $x$ of degree $\ge3$ (with no repeated roots), which includes our two equations.
So we can deduce from this that there are only finitely many Fibonacci numbers that are cubes. Unfortunately, Siegel's result isn't very helpful for the original problem because it doesn't give you any way of telling when you've found all of them.
Additionally, we know that if we have an solution to $5a^6 \pm 4 = b^2$, setting $c = a^2$ gives us a solution to the equation $5c^3 \pm 4 = b^2$. This is the equation of an elliptic curve, and there is a fair amount known about how to find the integer points on an elliptic curves (and more generally, rational points on elliptic curves), although the computations involved are usually not fun to do by hand. Anyone know how hard it would be to get a computer to solve this problem?

8$\begingroup$ It's really easy to get a computer to find all the integer points on an elliptic curve, as I was just saying the other day at mathoverflow.net/questions/20286/… . Multiply up by 25 and we're looking for integer points to X^3+100=Y^2 (with X,Y both multiples of 5). Both magma and sage have got this implemented now: you just type "IntegralPoints(EllipticCurve([0,100]));" into magma for one, and change 100 to 100 for the other. Both commands terminate in about 0.2 seconds. $\endgroup$ – Kevin Buzzard Apr 7 '10 at 22:35

1$\begingroup$ [remark: the biggest integer point on either curve is 5*532^3+4=27438^2, but, unfortunately, 532 isn't a square] $\endgroup$ – Kevin Buzzard Apr 7 '10 at 22:40

1$\begingroup$ If you look through all the results for ones that are 5 times a perfect square, it's only 0, 5 and 20. So that seems to answer the question completely. $\endgroup$ – Owen Dec 12 '15 at 23:32
In Siegel's article Zum Beweise des Starkschen Satzes. Inventiones mathematicae (1968), Siegel reduces the class number one problem for imaginary quadratic fields to the determination of Fibonacci cubes. The latter problem he solves in an elementary way. However, he does not mention Fibonacci numbers.