Determine which Fibonacci numbers are even (a) Determine which Fibonacci numbers are even. Use a form of mathematical induction to prove your conjecture.
(b) Determine which Fibonacci numbers are divisible by 3. Use a form of mathematical induction to prove your conjecture
I understand that for part a that all multiples of 3 of n are even. So F(0),F(3),F(6)... I just don't understand how to prove it.
For part B it is the same thing except multiples of 4
Please help, thank you!

Part A:
Base case: 
$F(0) = 0$, 0 is even.
$F(3) = 2$, 2 is even.
Inductive Hypothesis: 
Assume $F(k)$ is even for some arbitrary positive integer $k$ that is divisible by 3.
Want to prove: That $F(k+3)$ is even given the inductive hypothesis.
\begin{align*}
F(k+3) &= F(k+2) + F(k+1)\\
&= F(k+1) + F(k) + F(k+1)\\
&= F(k) + 2 F(k+1)
\end{align*}
We know by our hypothesis that $F(k)$ is even. We also know that $F(k+1)$ is an integer, so $2 F(k+1)$ is necessarily even by properties of integers. And because the sum of two even integers is also even, it follows that $F(k+3)$ is even. This holds for any arbitrary positive integer $k$ that is divisible by 3, hence we've proved that  $F(m)$ is even for all positive integers $m$ divisible by 3.
Part B:
Base case: 
$F(0) = 0$, 0 is a multiple of 3.
$F(4) = 3$, 3 is a multiple of 3.
Inductive Hypothesis: 
Assume $F(k)$ is a multiple of 3 for some arbitrary positive integer $k$ that is divisible by 4.
Want to prove: That $F(k+4)$ is a multiple of 3 given the inductive hypothesis. 
\begin{align*}
F(k+4) &= F(k+3) + F(k+2)\\
&= F(k+2) + F(k+1) + F(k+1) + F(k)\\
&= F(k+1) + F(k) + F(k+1) + F(k+1) + F(k)\\
&= 2 F(k) + 3 F(k + 1)
\end{align*}
By our hypothesis, $F(k)$ is a multiple of 3, so $2 F(k)$ must also be a multiple of 3. Furthermore, we know that $F(k+1)$ is some integer, so $3 F(k+1)$ must also be multiple of 3. Hence their sum, $F(k+4)$ must also be a multiple of 3.
