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:
$F_{3n+2}=F_{3n}+F_{3n-1}+F_{3n-1}+F_{3n-2}$ (A)
$F_{3n+2}=F_{3n-1}+F_{3n-2}+F_{3n-1}+F_{3n-2}$ (B)
I don't understand how you go from step (A) to step (B) can anyone explain this to me?


4 Answers 4


There's a missing term in (B) as you noticed. A correct proof is by induction. Show that $F_2$ is even (immediate). Then write

$$ \begin{align} F_{3n+2} &= F_{3n+1} + F_{3n} \\ &= 2F_{3n} + F_{3n-1} \\ &= 2F_{3n} + F_{3(n-1)+2} \enspace, \end{align} $$

which is even because it's the sum of two even numbers. ($F_{3(n-1)+2}$ is even by the induction hypothesis.)


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.


Base case:

$F_1=1\\ F_2=1\\ F_3=2$

Inductive hypothesis: Suppose $F_{3k-2}$ is odd, and $F_{3k-1}$ is odd, and $F_{3k}$ even.

We must show that $F_{3(k+1)-2}$ and $F_{3(k+1)-1}$ are odd, and $F_{3(k+1)}$ is even

$F_{3(k+1)-2} = F_{3k-1}+F_{3k}$

Based on the inductive hypothesis, $F_{3(k+1)-2}$ must be odd as the sum of an odd number and an even number are odd.

I will leave it to you to finish from here.


The rank of apparition of $2$, denoted $\omega(2)$, in the Fibonacci sequence is $3$; that is, the first term in the Fibonacci sequence that contains $2$ as a divisor is $F_{3} = 2$.

Since the Fibonnacci sequence is a member of the family of Lucas sequences, divisibility properties of the Lucas sequences are inherited by the Fibonacci sequence. Hence, $F_{3} | F_{k}$ if and only if $k = r\omega(2) = 3r$, for any positive integer $r$.

Therefore $2$ divides $F_{3}$ and every third term thereafter (and no others) in the Fibonacci sequence.


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