Proving that every third Fibonacci number is divisible by F2=2 In our notation we have that $F_{n-1}$ is the $n$th Fibonacci number since we start with $F_{0}=1$.
We want to prove that every third Fibonacci number is divisible by $F_{2}=2$.  
The proof is the following using induction:
$F_{3n+2}=F_{3n+1}+F_{3n}$
$F_{3n+2}=F_{3n}+F_{3n-1}+F_{3n-1}+F_{3n-2}$  (A)
$F_{3n+2}=F_{3n-1}+F_{3n-2}+F_{3n-1}+F_{3n-2}$ (B)
$F_{3n+2}=2(F_{3n-1}+F_{3n-2})$
I don't understand how you go from step (A) to step (B) can anyone explain this to me?

There is a well-known property of Fibonacci numbers:
$$ \gcd(F_n,F_m)=F_{\gcd(n,m)} \tag{1} $$
Since $F_3=2$,
$$ \gcd(F_{3n},2)=\gcd(F_{3n},F_3)=F_{\gcd(3,3n)} = F_3 = 2 \tag{2} $$
hence every Fibonacci number of the form $F_{3n}$ is even.
