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I wanted to find the $gcd(f_k, f_{k+3})$, where $f_k$ is the k'th Fibonacci number (i.e. $f_0=0, f_1=1, f_k=f_{k-1} + f_{k-2}$ for $k \geq 2$. So far I've tried to exprss $f_{k+3}$ as $2f_{k+1}+f_k$. Using this it follows that $gcd(f_k, f_{k+3}) = gcd(f_k, 2f_{k+1}$), but I don't see how I can continue from there one. Can someone help me out?

Thanks!

Gorid
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  • Hint: $\gcd(f_k, f_{k+1})=1$. – lhf Dec 12 '16 at 15:38
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    Also $\gcd(F_m, F_n) = F_{\gcd(m,n)}$. The above property holds because Fibonacci Numbers follow Divisibility Sequence, i.e., if $M$ divides $N$, then $F_M$ also divides $N$. For example, $F_{3} = 2$ and every third third Fibonacci Number is even. –  Dec 12 '16 at 15:50
  • Thank you for your help. @lhf: I’ve been able to show that $gcd(f_k,f_{k+1})=gcd(f_k,f_{k+2})=1$, and I think that I should somehow relate $gcd(f_k,2f_{k+1})$ to $gcd(f_k,f_{k+1})$, but I don’t quite see how. – Gorid Dec 12 '16 at 15:54

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Since $\gcd(f_k, f_{k+1})=1$, we have $\gcd(f_k, 2f_{k+1})=\gcd(f_k, 2)$.

Since $f_3=2$, we have $\gcd(f_k, 2)=2$ iff $n$ is a multiple of $3$. Otherwise, $\gcd(f_k, 2)=1$.

lhf
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  • See also http://math.stackexchange.com/questions/815004/determine-which-fibonacci-numbers-are-even and https://en.wikipedia.org/wiki/Fibonacci_number#Divisibility_properties. – lhf Dec 12 '16 at 15:50