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$f_n$ is divisible by 3 if and only if n is divisible by 4.

This seem to be true as $f_4=3$ $f_8=21$

So would I do $F_{4n}=F_{4n-1}+f_{4n-2}$

The turn n into n+1 using the induction.

$F_{4n+1}=F_{4n}+F_{4(n-1)}$ ??

Another question I have is similiar if $f_n$ divisible by 4 if and oly if n is divisible by 6.

It also seem true $f_6=8$

Would I use same method to show this.

Fernando Martinez
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  • If you wanted to do induction it would be more natural try to show that $F_{4(n+1)}$ is disvisible by three if $F_{4n}$ is divisible by three. – Tim Seguine Mar 30 '14 at 19:18
  • One or two examples tells you very little. You should be very weary of making conjectures like that. 2 is prime, so even even number is prime. 3,5,7 are all prime so every odd >1 is prime. For a less obvious example, consider $T_n$ to be the maximum number of regions formed by intersecting a circle with n lines. This is a power of 2 up until you hit 31. – Stella Biderman Mar 30 '14 at 19:18
  • You're inductive statement should be about $F_{4(n+1)}=F_{4n+4}$ by the way. – Stella Biderman Mar 30 '14 at 19:19
  • Note that if you count the Fibonacci Numbers correctly you have $f_r|f_{nr}$. – Mark Bennet Mar 30 '14 at 19:24

2 Answers2

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Let $\overline f_n$ be the residue of the $n$th Fibonacci number modulo $3$. We have $$ \overline f_n = \overline f_{n-1} + \overline f_{n-2} \pmod{3} $$ That is, the residues form a "Fibonacci sequence" modulo $3$. We compute: $$ \overline f_0 = 0\\ \overline f_1 = 1\\ \overline f_2 = 1+0 \equiv 1\\ \overline f_3 = 1+1 \equiv 2\\ \overline f_4 = 2+1 \equiv 0\\ \overline f_5 = 0+2 \equiv 2\\ \overline f_6 = 2+0 \equiv 2\\ \overline f_7 = 2+2 \equiv 1\\ \overline f_8 = 1+2 \equiv 0\\ \overline f_9 = 0+1 \equiv 1\\ $$ From there, the cycle repeats. The same idea works for any divisor. The length of this cycle for a divisor $n$ is known as the $n$th Pisano period.

Ben Grossmann
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From the Binet formula it can be shown that: $F_{n+k} = F_kF_{n+1} + F_nF_{k-1}$. Using this and induction on $m$ we can prove that $F_k$ divides $F_{mk}$ for all natural number $m$. Now letting $k=4$ we have that $3 = F_4$ divides $F_{4m}$ for all $m$.

DeepSea
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