Sum of Fibonacci Number and Product of Fibonacci Number are equal? Let $a,b,c,d$ be nonnegative integers. Consider the equation
$$F_a+F_b=F_cF_d$$
WLOG, I can assume that $a \geq b$ and $c \geq d$. I try to solve for all subscripts $a,b,c,d$ in the above equation.
It is definitely solvable. For example, we have $F_d=1$ for $d=1,2$, the equation turns into
$$F_a+F_b=F_c$$
By definition of Fibonacci number, $$(a,b,c)=(n-1,n-2,n)$$ is one of the solutions and other solutions can be found in lemma 2.3 in this link. https://www.researchgate.net/publication/337918351_On_the_sum_of_three_arbitrary_Fibonacci_and_Lucas_numbers.
For $d=3$, the equation becomes
$$F_a+F_b=2F_c$$
Clearly, if $a=b=c=n$, we have another set of the solutions.
However, my hunch tells me there are other solutions for $d \geq 3$, I even try writing program on MATHEMATICA to look out other forms of solution.
Anyone has a nice idea to attack this problem? Actually, I try to do for Lucas numbers, too.
$$L_a+L_b=L_cL_d$$
$$L_a+L_b=F_cF_d$$
$$F_a+F_b=L_cL_d$$
 A: All solutions to $F_a+F_b=F_cF_d$ with $a \geq b$, $c \geq d$ are given by ($n \geq 1$):
\begin{align}
(a,b,c,d) \in \{(n+5,n+1,n+3,4),(n+3,n,n+2,3),(n+1,n,n+2,2),(n+1,n,n+2,1),(n+2,n+2,n+2,3),(1,1,3,1),(1,1,3,2),(2,2,3,1),(2,2,3,2),(3,1,4,1),(3,1,4,2),(4,2,3,3),(5,1,4,3),(6,1,4,4)\}.
\end{align}
To prove this we use Zeckendorf's theorem: every positive integer can be represented uniquely(!) as the sum of one or more distinct non-consecutive Fibonacci numbers. Together with known formula for Zeckendorf's representation of a product $F_cF_d$, this gives the complete solution.
Proof.
The left hand side $F_a+F_b$ is already in Zeckendorf's form unless $a=b$ or $a=b+1$. We omit these cases for now and assume $a \geq b+2$. The right hand side $F_cF_d$ has well-known Zeckendorf representation for $c \geq d \geq 2$ (see for example here):
$$
F_cF_d=\sum_{r=1}^{\lfloor d/2 \rfloor } F_{c+d+2-4r}
$$
when $d$ is even. When $d$ is odd, we have to add $F_{c-d+1}$ if $c>d$ and $F_2$ if $c=d$. Going case by case, we can infer that by Zeckendorf's theorem the sum must contain exactly two terms and they need to be $F_a$ and $F_b$.
If $d$ is even, then the sum has exactly $d/2=2$ terms, so we must have $d=4$. The terms are then $F_{c+4+2-4}=F_{c+2}$ and $F_{c+4+2-8}=F_{c-2}$ and we conclude $a=c+2$, $b=c-2$ (recall $a \geq b$). This corresponds to
$$
F_{c+2}+F_{c-2}=F_cF_4=3F_c.
$$
Here we must be careful and notice that the identity we used works for $d\geq 2$ only, but since $F_1=F_2$, we get additional solution from above when $c=4$ and replacing $F_2$ with $F_1$. Namely $F_6+F_1=F_4F_4$.
If $d$ is odd and $c>d$, then we have $\lfloor d/2 \rfloor+1=2$ terms, hence $d=1$ or $d=3$. However by assumption $d \geq 2$, hence we consider only $d=3$. The two terms are then $F_{c+d+2-4}=F_{c+1}$ and $F_{c-d+1}=F_{c-2}$. This corresponds to
$$
F_{c+1}+F_{c-2}=F_{c}F_3=2F_c.
$$
As before, for $c=4$ and replacing $F_2$ with $F_1$ we get additional solution $F_5+F_1=F_4F_3$.
If $d$ is odd and $c=d$, then again we have $d=3$. The two terms are now $F_{c+d+2-4}=F_{4}$ and $F_2$. This corresponds to
$$
F_4+F_2=F_3F_3.
$$
This also gives $F_4+F_1=F_3F_3$, but this is already covered by another parametrization so we do not mention this case explicitly in the summary.
If $d=1$, then you have already found all solutions to $F_a+F_b=F_cF_d=F_c$, so we are done there.
So the only remaining cases are then $a=b$ and $a=b+1$.
The case $a=b+1$ is simple as then $F_a+F_b=F_{b+2}$ is its own Zeckendorf representation, so there must be exactly one term in representation $F_cF_d$ and so $d$ is even, $d/2=1$, thus $d=2$. The only term is $F_{c+d-2}=F_{c}=F_{b+2}$ and so $c=b+2$. This retrieves
$$
F_{b+1}+F_b=F_{b+2}F_{2}=F_{b+2}.
$$
So the recurrent relation is the only solution in this case.
If $a=b$, then we want to find all solutions of $2F_a=F_cF_d$. We can view the left side as $F_3F_a$. First for $a > 3$, we have
$$
F_aF_3=F_{a+3+2-4}+F_{a-3+1}=F_{a+1}+F_{a-2}.
$$
So similar as before, looking at representations of $F_cF_d$ with $c \geq d \geq 2$ and have exactly two terms, we find either $d=4$ and $F_cF_d=F_{c+2}+F_{c-2}$ (incompatible $F_{a+1}+F_{a-2}$), $d=3$ and $F_{c+1}+F_{c-2}$ (which corresponds to $a=c$ trivially) or $d=3$ and $F_4+F_2$ (incompatible with $F_{a+1}+F_{a-2}$).
For $a \leq 3$, we can consider just case by case $a=b=1$, $a=b=2$ and $a=b=3$ and we find few cases $(a,b,c,d) \in \{(1,1,3,1),(1,1,3,2),(2,2,3,1),(3,3,3,3)\}$.
$\square$
