Find all quadruplets of integers $(a, b, c, d) \in \mathbb{N}_{\geqslant 2}$ such that $F_{a}+F_{b}=F_{c}+F_{d}$.
We can assume without loss of generality that $a \geqslant b, c, d \geqslant 2$ as well as $c \geqslant d$, then either $a=c$ which gives $b=d$.

Or $a \geqslant c+2$, then $F_{b}+F_{a}>F_{a}=F_{a-1}+F_{a-2}>F_{c}+F_{d}$.

The remaining case is where $c=a-1$, then if $d \geqslant a-2$ we have $F_{a}+F_{b}>F_{a}=F_{a-1}+F_{a-2}=F_{c}+F_{a-2} \geqslant$ $F_{c}+F_{d}$ which is a contradiction with the relation we want. Therefore, $d=a-1$, then $F_{b}=-F_{a}+2 F_{a-1}=F_{a-1}-F_{a-2}=F_{a-3}$.

Thus, we obtain the following solutions: $(a, b, a, b),(a, b, b, a),(a, a-3, a-1, a-1),(a-3, a, a-1, a-1),(a-1, a-1, a, a-3)$ and $(a-1, a-1, a-3, a)$.