# Thread: Fibonacci sequence divisibility proof

1. ## Fibonacci sequence divisibility proof

Hello All,

I am thoroughly lost with how to construct this proof:

Let F(sub)n be the nth Fibonacci number.

Prove that, if 3|n then 2|F(sub)n.

Would induction be a recommendation, and if so, could anyone please recommend a base assumption to begin with?

Thank you very much for your time,

Panglot

2. ## Re: Fibonacci sequence divisibility proof

Originally Posted by panglot
Hello All,

I am thoroughly lost with how to construct this proof:

Let F(sub)n be the nth Fibonacci number.

Prove that, if 3|n then 2|F(sub)n.

Would induction be a recommendation, and if so, could anyone please recommend a base assumption to begin with?

Thank you very much for your time,

Panglot

If and only if can be placed.

Are you familiar with the theorem that says:

$\displaystyle F_m|F_n$ if and only if $\displaystyle m|n$.

3. ## Re: Fibonacci sequence divisibility proof

What you want to prove is that
$\displaystyle F_{3n}$
is divisible by 2 for any positive integer n. It is easy to do that by induction on n.

If n= 1,
$\displaystyle F_{3(1)}= F_3= 2$
which is divisible by 2.

Assume that, for some k,
$\displaystyle F_{3k}$
is divisible by 2. That is, that
$\displaystyle F_{3k}= 2m$
for some integer m.

Then
$\displaystyle F_{3(k+1)}= F_{3k+ 3}= F_{3k+1}+ F_{3k+2}= F_{3k+1}+ (F_{3k}+ F_{3k+1})= 2m+ 2F_{3k+1}$.

4. ## Re: Fibonacci sequence divisibility proof

Originally Posted by HallsofIvy
What you want to prove is that
$\displaystyle F_{3n}$
is divisible by 2 for any positive integer n. It is easy to do that by induction on n.

If n= 1,
$\displaystyle F_{3(1)}= F_3= 2$
which is divisible by 2.

Assume that, for some k,
$\displaystyle F_{3k}$
is divisible by 2. That is, that
$\displaystyle F_{3k}= 2m$
for some integer m.

Then
$\displaystyle F_{3(k+1)}= F_{3k+ 3}= F_{3k+1}+ F_{3k+2}= F_{3k+1}+ (F_{3k}+ F_{3k+1})= 2m+ 2F_{3k+1}$.
I'm sorry, but I don't quite follow how you arrived at the last two in the series of equivalences:

= F_{3k+1}+ (F_{3k}+ F_{3k+1})= 2m+ 2F_{3k+1}

5. ## Re: Fibonacci sequence divisibility proof

Originally Posted by panglot
I'm sorry, but I don't quite follow how you arrived at the last two in the series of equivalences:

= F_{3k+1}+ (F_{3k}+ F_{3k+1})= 2m+ 2F_{3k+1}
By definition of Fibonacci sequence.

F_{k+2}=F_{k+1}+F_{k}.