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Prove that $\gcd(F_n,F_{n+3})=1$ or $2$ for $n\geq 2$.

The excercise has a slightly confusing hint. It says "Let $d|\gcd(F_n,F_{n+3})$" and also asks to show $d|2$ which I understand.

I don't know what "$d$" is supposed to be; perhaps the gcd of $F_n$ and $F_{n+3}$, or is it related to induction somehow?

Here is my attempt if you want to have a look:

I just took it for granted and came up with this.

Suppose that $d|(F_n,F_{n+3})$, and make $e=(F_n,F_{n+3})$ then $d|e$ and $e|F_n$ and $e|F_{n+3}$ therefore $d|F_n$ and $d|F_{n+3}$, therefore $d|xF_n+yF_{n+3}$ for all $x,y$.

Then I apply the definition of the fibonacci sequence. I got very close anwsers one being $d|xF_n+yF_{n+1}+yF_n+yF_{n+1}$. Almost there, since $(F_n,F_{n+1})=1$. Then I tried again and got $d|xF_{n+2}+yF_{n+1}+(-x)F_{n+1}+yF_{n+2}$ (trying to use the fact that $(F_{n+1},F_{n+2})=1$).

Arthur
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cheesewiz
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    Fun fact: $\gcd(F_m,F_n)=F_{\gcd(m,n)}$. – anon Nov 29 '22 at 06:47
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    Hint $F_{n+3}=2F_{n+1}+F_n$, and you know that $gcd(F_{n+1},F_n)=1$, then by Euclid algorithm... – Hypernova Nov 29 '22 at 06:51
  • @Hypernova I attempted this and applied Euclids algorithm but it ended up in a cyclical manner – cheesewiz Nov 29 '22 at 07:20
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    @cheesewiz $gcd(F_{n+3},F_n)=gcd(2F_{n+1},F_n)=gcd(2, F_n)$ so we are done. – Hypernova Nov 29 '22 at 07:46
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    By the linked dupe $,(f_{\large n},f_{\large n+3}) = f_{\large (n,n+3)}! = f_{\large\color{#c00}{(n,3)}} = f_{\large \color{#c00}1}(=1)$ or $f_{\large \color{#c00}{3}}(=2)\ \ $ – Bill Dubuque Nov 29 '22 at 08:06
  • You may find it instructive to study how the arguments above are special cases of the linked simple inductive proof for the lgeneral case. – Bill Dubuque Nov 29 '22 at 08:10
  • Consider this solution : https://imgur.com/a/OZ4OvlG – D S Nov 29 '22 at 09:13
  • @Moderator Can I request you reopen this post, I would like to put an anwser for future students who come across this as the duplicate did anwser it however it already assumes some knowledge a student might not know, for example this excercise can be proven in different manner with only theorems already discussed in my book. – cheesewiz Mar 09 '23 at 09:22

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