Conjecture: Only one Fibonacci number is the sum of two cubes

Solution 1:

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!

Solution 2:

There are no further solutions for $n \leq 300$. Even dropping the positivity condition the only other solutions are $F_6 = 8 = 2^3 + 0^3 = 0^3 + 2^3$. This corroborates Hurkyl's heuristic that it is almost certain that there are no other Fibonacci numbers that are sums of two cubes. I doubt that this can be proved, though the search technique described below can surely be used well past $n=300$, because the main difficulty is factoring $F_n$, and $F_{300}$ has only about $60$ digits.

To try to express an integer $f$ as $a^3+b^3$, use the factorization $a^3 + b^3 = (a+b)(a^2-ab+b^2)$. For each pair $(x,y)$ such that $f=(x,y)$, check whether $3(4x-y^2)$ is a square, since that's the criterion for $(x, y) = (a^2-ab+b^2, a+b)$ [the point is that $4x-y^2 = 3(a-b)^2$].

I did this for $f=F_n$ and all $n \leq 300$ using this table of Fibonacci factorizations (discarding the 100 values of $n$ for which $F_n\bmod 9 \in \{3,4,5,6\}$, since such a number can never be the sum of two cubes). This took only ~30 seconds in gp, even though some candidates $F_n$ in this range have more than a million factors. Only the known solutions for $F_3=2$ and $F_6=8$ were found.

(also reported at MSE question 822435.)