Every integer can be written as the linear combination of two Fibonacci numbers.
For example, if we want to write $N = aF_m + bF_n$, we can choose two Fibonacci numbers, $F_m, F_n$ that are coprime and then solve for $a, b$ using the Extended Euclidean algorithm. Is the converse also true? i.e., Given an integer $N$, does there exist a pair of integers $a,b$ such that all divisors $d$ of $N$ may be written as $d = aF_m + bF_n$?
i.e., Given $N \in \mathbb{Z}$, is there an identity similar to the Brahmagupta-Fibonacci identity where
$$N = aF_m + bF_n = (aF_r + bF_s)(aF_t + bF_u)$$
for some integers $m, n, r, s, t, u$.
Clearly $a,b$ can be chosen such that $1$ and $d|N$ are represented, at least for the following small examples with $a = 1, b = -2$:
$$ 1 = F_4 - 2F_1, \\ 2 = F_6 - 2F_4, \\ 3 = F_5 - 2F_1, \\ 4 = F_6 - 2F_3, \\ 5 = F_8 - 2F_6, \\ 6 = F_6 - 2F_2, \\ 7 = F_7 - 2F_4, \\ 8 = F_9 - 2F_7, \\ 9 = F_7 - 2F_3, \\ 10 = F_6 - 2F_{-2}, \\ 11 = F_7 - 2F_1, \\ 12 = ? \\ 13 = F_{10} - 2F_8, \\ 14 = F_6 - 2F_{-4}, \\ 15 = F_8 - 2F_4, \\ 16 = F_0 - 2F_{-6}, \\ 17 = F_8 - 2F_3, \\ 18 = F_9 - 2F_6, \\ 19 = F_8 - 2F_2, \\ 20 = ? \\ 21 = F_9 - 2 F_0, \\ 22 = ? \\ 23 = F_9 - 2F_{-2}, \\ 24 = F_9 - 2F_5 $$
Note that we are extending the Fibonacci series in the negative numbers as well (as can be seen with the example for $10, 14, 16, 23$). At least with $N$ up to $11$, we see that all divisors of $N$ can be represented with $a = 1, b = -2$. For the divisors of $12$, the pair $a=1, b = -3$ generates $1, 2, 3, 4, 6$ and $12$. It is necessary that $a,b$ are coprime (so that a prime factor $p$ of $N$ can be represented by $aX + bY = p$).
Question: I am asking if we can always find $a,b$ such that all divisors of a given $N$ can be represented as the linear combination $aX + bY$ where $X,Y$ are Fibonacci numbers.
Note: For the representations of $12, 20, 22$ and their respective divisors see this related question:
They use different $a, b$ pairs for the representation of the divisors.