Prove for Fibonacci numbers: $3\mid f(n) \iff 4\mid n$

Question

Let $f(n)$ be the $n$th Fibonacci number.

Prove that $$3\mid f(n) \iff 4\mid n$$

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I tried to use induction to prove it but I couldn't continue when I reached $n+1$ case.

Answer 1

Let $g(n)=f(n)\bmod 3$. Then $g(0)=0$, $g(1)=1$, and $g(n+1)=\big(g(n)+g(n-1)\big)\bmod 3$ for $n>0$. The next few values are $g(2)=1$, $g(3)=2$, $g(4)=(1+2)\bmod 3=0$, and $g(5)=2$. Continuing in that vein, we can compute the following values, where I’ve starred the rows in which $3\mid f(n)$:

$$\begin{array}{c|c|cc} n&f(n)&g(n)&\\ \hline 0&0&0&*\\ 1&1&1\\ 2&1&1\\ 3&2&2\\ 4&3&0&*\\ 5&5&2\\ 6&8&2\\ 7&13&1\\ \hline 8&21&0&*\\ 9&34&1\\ \end{array}$$

Can you see why I put a bar between $n=7$ and $n=8$? Remember, each value of $f$ or $g$ depends only on the immediately preceding two values. The induction with $g(n)$ is a little easier to see than the induction with $f(n)$.

Answer 2

Hint: use the recurrence three times to write $f_{n+1}=3f_{n-2}+2f_{n-3}$. Then use the induction hypothesis, and the fact that $3\mid 2a \Longleftrightarrow 3 \mid a$ for any integer $a$.

:)

Answer 3

Hint $ $ Show $\rm\ \color{#0a0}{f_{\,n}} \equiv\: -f_{\,n-4}\pmod{\!\color{#c00} 3},\:$ so for $\,\rm n\ge 4\,$ this yields our inductive step,
i.e. $\rm\ \,\color{#c00}3\mid \color{#0a0}{f_{\,n}}\!\iff \color{#c00}3\mid f_{\,n-4}\!\!\!\overset{\rm induct\!\!\!}\iff\! 4\mid n\!-\!4\!\iff\! 4\mid n,\,$ so it suffices to verify base cases $\rm\,n<4$.

Remark $ $ The above congruence is a special case $\rm\:k = 4\:$ of $\rm\:f_{\:n}\equiv f_{\:k-1}\ f_{\:n-k} \pmod{f_{\:k}}$.
This leads to a simple inductive proof that $\rm\:gcd(f_n,f_k) = f_{\:gcd(n,k)},\ $ so $\rm\ f_{\:k}\ |\ f_{\:n}\! \iff k\ |\ n$.
Specializing $\rm\:k = 4\:$ yields $\rm\ 3\ |\ f_{\:n}\!\iff 4\ |\ n,\,$ so the OP is a special case of this.