Selection tests for the Balkan Mathematical Olympiad 2013

Let $(a,b,c,n)$ be a quadruplet of positive integers with $a<b<c$ such that each of $a,b,c,a+n,b+n,c+2n$ is a term of the Fibonacci sequence, and let $$b+n=F_k$$ for some positive integer $k$. Because $b<b+n$ and $a+n<b+n$, we have $\max \{b,a+n\}\le F_{k-1}$. Assume $\min \{b,a+n\}\le F_{k-2}$. We have $a+n+b\le F_{k-1}+F_{k-2}=F_k=b+n$, which is impossible since $a>0$. Therefore, $$\begin{array}{c}a+n=b=F_{k-1},\\ n=(b+n)-b=F_k-F_{k-1}=F_{k-2},\end{array}$$ and $$a=(a+n)-n=F_{k-1}-F_{k-2}=F_{k-3}.$$ Let $c+2n=F_m$. We have $F_k\le c\le F_{m-1}$ and therefore $F_{m-2}\le 2n=2F_{k-2}\le F_k$. If $F_{m-2}=F_k$ then $F_{k-2}=F_{k-1}=1$ and $a=F_{k-3}=0$ which is impossible. Therefore $$c=b+n=F_{m-1}=F_k,$$ and $$c+2n=F_m=F_{k+1}$$ But $$2n=(c+2n)-c=F_{k+1}-F_k=F_{k-1}=a+n.$$ We deduce that $$F_{k-3}=a=n=F_{k-2}=1.$$ Hence, $k=4$, $(a,b,c,n)=(1,2,3,1)$ and we check easily that $$a=F_1,b=a+n=F_3,c=b+n=F_4\text{ and }c+2n=F_5.$$