The sequence is 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, ...

The terms

f_{3}, f_{6}, f_{9}, f_{12}, f_{15} . . .

are even which leads to the conjecture that the Fibonnaci number

f_{n}

is even if and only if

n

is a multiple of 3.

The sequence of parities must be even, odd, odd, even, odd, odd, even, odd, odd, even, ...

The term

f_{3}

is even and the term before it is odd.

If

f_{n - 1}

is odd and

f_{n}

is even then

f_{n + 1}

is odd,

f_{n + 2}

, is odd and

f_{n + 3}

is even because it is the sum of two odd numbers so every third term, starting with

f_{3}

will be even.

It appears from the first few terms that as $f_{4}$ , $f_{8}$ , $f_{12}$ are divisible by 3 we might make the conjecture that $f_{s}$ is divisible by 3 if and only if $s$ is a multiple of 4. Using the defining relation for Fibonnaci numbers we get:

$\begin{matrix} f_{n + 4} & = f_{n + 3} + f_{n + 2} \\ = 2 f_{n + 2} + f_{n + 1} \\ = 3 f_{n + 1} + 2 f_{n} \\ = 5 f_{n} + 3 f_{n - 1} \end{matrix}$

This shows that if $f_{n}$ is a multiple of 3 then $f_{n + 4}$ is a multiple of 3. As we know $f_{4}$ is a multiple of 3 it follows, using the axiom of induction, that $f_{s}$ is divisible by 3 if $s$ is a multiple of 4.

We should still show that these are the only Fibonnaci numbers divisible by 3 to prove the 'only if' condition.

If any two consecutive Fibonnaci numbers have a common factor (say 3) then every Fibonnaci number must have that factor. This is clearly not the case so no two consecutive Fibonnaci numbers can have a common factor. If $f_{n}$ and $f_{n + 2}$ are both multiples of 3 then $f_{n + 1}$ must also be a multiple of 3 and hence all Fibonnaci numbers will be multiples of 3 which is not the case. This shows that if $f_{n}$ and $f_{n + 4}$ are multiples of 3 then no Fibonnaci numbers between them can be multiples of 3, that is $f_{s}$ is divisible by 3 only if $s$ is a multiple of 4.