At most one representation as a sum of two fibonacci-numbers?

Question

I wanted to start a project to find primes of the form $F_m+F_n$ with integers $m,n$ satisfying $1<m<n$ , where $F_n$ denotes the $n$ th fibonacci-number.

I wondered whether duplicate numbers will appear.

Question : Does a positive integer $N$ always have at most one represenatition $$N=F_m+F_n$$ with positive integers $m,n$ , $1<m<n$ , $F_n$ the $n$ th fibonacci number ?

I created the sums for $2\le m<n\le 500$ and no duplicates occured.

Answer 1

Without any theorem, just with my fingers...

Consider $X=F_a+F_b$, with $a<b$

and $Y=F_c +F_d$ with $c<d$

And suppose $F_b < F_d$, so $b<d$, or in other words $b+1 \le d$ $X=F_a + F_b \le F_{b-1}+F_b = F_{b+1} \le F_d < F_d + F_c=Y$

Last step is strict : $F_d < F_d+F_c$, so we cannot have $X=Y$.

If we enlarge the conditions, and accept decompositions like $F_a+F_a$, it is more complex.

Answer 2

If $N$ is itself a Fibonacci number we have the case $N=F_n=F_{n-1}+F_{n-2}$. It's clear that no Fibonacci number has another representation of this form, as any other choices would be too small. So assume that $N$ is not itself a Fibonacci number.

But if $N$ is not a Fibonacci number then we know that $m,n$ are not consecutive, so we can use Zeckendorf's Theorem to conclude the uniqueness of the representation.