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I'm having troubles proving that in a fibonacci sequence if n is divisible by four, then Fn is divisible by three

So when Fn is 6, n is 8 and so on. I was thinking maybe I could use mod 3 or mod 4 but don't really know what to do with it.

user2980566
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    Try reducing the whole Fibonacci sequence modulo 3. You will see the pattern. – Nocturne Nov 05 '14 at 23:03
  • @Nocturne But how would I write that as a formal proof? – user2980566 Nov 05 '14 at 23:04
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    Try uisng induction on Fibonacci [mod 3]. $F_4=0$ will be the base case. Now, suppose $F_{4n}=0$ for $n \geq 1$ and prove that $F_{4(n+1)}=F_{4n+4}=0$. Just write $F_{4n+4}$ as a combination of $F_{4n+1}$ and $F_{4n}$ and use the fact that you count modulo 3 to prove that $F_{4n+4}=0$, i.e $F_{4n+4}=0$ is divisible by 3. – Nocturne Nov 05 '14 at 23:11
  • @Nocturne Thanks man! I'm just confused on how you write F4n+1 and F4n as a combination? – user2980566 Nov 06 '14 at 02:28
  • Just use the definition of the Fibonacci sequence down to $F_{4n}$ and $F_{4n+1}$. $F_{4n+4}=F_{4n+3}+F_{4n+2}=(F_{4n+2} +F_{4n+1}) + (F_{4n+1}+F_{4n})=((F_{4n+1}+F_{4n}) +F_{4n+1}) + (F_{4n+1}+F_{4n})=3F_{4n+1}+2F_{4n}$. Now, if you look the sequence modulo 3, the first term disappears because is multiplied by 3 and the second is zero by hypothesis, so $F_{4n+4}=F_{4(n+1)}=0$. – Nocturne Nov 07 '14 at 17:29

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This follows from the matrix formulation, which is well worth knowing and easily proved: $$ \begin{pmatrix}1&1\\1&0\end{pmatrix}^n= \begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix} $$

We will prove by induction that $3$ divides $F_{4k}$. Let $ A=\begin{pmatrix}1&1\\1&0\end{pmatrix} $. Then $A^{4(k+1)}=A^{4k}A^4$ : $$ \begin{pmatrix}F_{4(k+1)+1}&F_{4(k+1)}\\F_{4(k+1)}&F_{4(k+1)-1}\end{pmatrix} = \begin{pmatrix}F_{4k+1}&F_{4k}\\F_{4k}&F_{4k-1}\end{pmatrix} \begin{pmatrix}F_{5}&F_4\\F_4&F_{3}\end{pmatrix} $$ and so, by looking at position $2,1$, $$ F_{4(k+1)}=F_{4k}F_{5}+F_{4k-1}F_4 $$ By induction, $F_4=3$ divides $F_{4k}$ and so $F_4=3$ divides $F_{4(k+1)}$.

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lhf
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    More generally, if $m$ divides $n$ then $F_m$ divides $F_n$. Still more generally, $F_{\gcd(m,n)}=\gcd(F_m,F_n)$. Everything follows from the matrix formulation. – lhf Nov 05 '14 at 23:17
  • Thanks! I'm just confused on this part "F4 = 3 divides F4k and so F4 = 3 divides F4(k+1)" From that equation, how do we know that 3 divids F4k and 3 divides F4(k+1)? – user2980566 Nov 06 '14 at 03:15