As the title says, I need help proving or disproving that there is only one Fibonacci number that's the sum of two (positive) cubes, $2$. I did a small brute force test with Fibonacci numbers below $10^{15}$ but I couldn't find anything.
Edit: A possibly fast way to factor sums of cubes is described in another question here.
I tested Hagen von Eitzen's $F_{3n} – F^3_{n+1}$ up to $n = 200000$ and brute force up to $10^{21}$ but no results. As the numbers grow bigger the chance of a cube gets smaller.
Best Answer
On the assumption there isn't an "easy" reason for the answer to be yes or no, here is a heuristic justification of the conjecture.
There are only $O(n^{2/3})$ ways to write a sum of two (positive) cubes that is a number less than $n$; a "random" number of size $\Theta(n)$ therefore has a probability $O(n^{-1/3})$ of being a sum of two cubes, with the hidden constant not being too small. (probably a little less than $1/3$)
The expected number of ways to write a Fibonacci number as a sum of two cubes is thus something like
$$ O\left( \sum_{n=0}^{\infty} F_n^{-1/3} \right) \approx O\left( \sum_{n=0}^{\infty} \varphi^{-n/3} \right) \approx O\left( \frac{1}{1 - \varphi^{-1/3}} \right)$$
so it's probably a small finite number. Your empirical evidence says the small finite number is probably $1$.
It is quite possible that the conjecture is true, but for no good reason at all, which would make it very hard to come up with a proof. It might even be independent of Peano's axioms!