Is 8 the largest cube in the Fibonacci sequence? Can you prove that 8 is the largest cube in the Fibonacci sequence?
 A: 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.
A: This paper, using some heavy-duty number-theoretic 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.
A: 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 $5a^6 = (b-2)(b+2)$, and using the fact that any common factor of $b-2$ 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:  $5a^6 = (b+2i)(b-2i)$.  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?
A: For a much more accessible treatment, and history of the result, see Andrejic (2006) "On Fibonacci powers"
