Conjecture: Suppose $N = \sum_{i=0}^{k} F_{c_i}, (c_i \ge 2)$ is a Zeckendorf representation of $N \in \mathbb{Z}$ and $\zeta_N$ be the set of Fibonacci numbers in the Zeckendorf representation of $N$.
$\exists \langle F_A, F_B \rangle$, a pair of Fibonacci integers in $\zeta_N$ such that $N = aF_A + bF_B$ with $a,b \in \mathbb{Z}$ and the divisors $d|N$ can be represented as a linear combination $d = aF_m + bF_n$ for some integers $m, n$ (with $F_m, F_n$ not necessarily members of $\zeta_N$).
Question: Is this conjecture true?
Obviously, the first part of representing $N = aF_A + bF_B$ with $F_A, F_B \in \zeta_N$ is possible. $\forall \langle F_A, F_B \rangle$ with $F_A, F_B \in \zeta_N$ we can write the other elements $F_C$ as a linear combination of $F_A, F_B$ as long as $\gcd(F_A, F_B) | F_C$. If $\gcd(F_A, F_B) = 1$ then this linear combination is possible for any $F_C$. Therefore, after reducing all elements into linear combinations of $F_A, F_B$, we have $N$ itself as a linear combination of $F_A, F_B$.
What remains to be proven is that the divisors of $N$ is represented by a similar linear combination of Fibonacci numbers with the coefficients $a,b$ determined above. Suppose $(a,b) = h$. If $h | N$ and if $h \gt 1$ then we have a non-trivial divisor of $N$. We can solve $\frac{a}{h}X^{\prime} + \frac{b}{h}Y^{\prime} = \frac{N}{h}$ and solve for $X^{\prime}, Y^{\prime}$. Then, $X = hX^{\prime}, Y = hY^{\prime}$.
If $h = 1$ then $d|N = aX + bY$ has integer solutions.
What remains to be proven is $X, Y$ are Fibonacci numbers.
This conjecture came up while attempting to answer this question and this question.