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I want to show that if $m \geq 1, n \geq 1$ and $gcd(m,n)=1$, then $F_m F_n \mid F_{mn}$.

$F_n$ is the $n$-th Fibonacci number.

I have tried the following so far:

$$F_{mn}=F_{mn-1}+F_{mn-2}=2F_{mn-2}+F_{mn-3}=2(F_{mn-3}+F_{mn-4})+F_{mn-3}\\=3F_{mn-3}+2F_{mn-4}=\dots=\lambda F_{mn-\lambda}+(\lambda-1)F_{mn-(\lambda+1)}$$

Is it right so far? How can we continue in order to get the desired result?

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  • $\begingroup$ Can you prove $F_m\mid F_{mn}$, $F_n\mid F_{mn}$ and $\gcd(F_m,F_n)=1$? $\endgroup$ – Lord Shark the Unknown Oct 29 '18 at 19:43
  • $\begingroup$ Could you give me a hint how we show these properties? @LordSharktheUnknown $\endgroup$ – Evinda Oct 29 '18 at 19:50
  • $\begingroup$ @Evinda Follow the link in my answer for some proofs. $\endgroup$ – Bill Dubuque Oct 29 '18 at 19:56
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By here the $f_k$ are a strong divisibility sequence $\,\gcd(f_j,f_k) = f_{\large \gcd(j,k)}.\,$ In particular $\,f_j\mid f_{jk}$. Thus $\,\gcd(f_m,f_n) = f_{\gcd(m,n)}\! = f_1\! = 1\,$ and $\,f_m,f_n\mid f_{mn}\Rightarrow\, {\rm lcm}(f_m,f_n) = f_m f_n\mid f_{mn}$

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  • $\begingroup$ Above we used $\,\gcd(a,b)=1\,\Rightarrow\, {\rm lcm}(a,b) = ab.\ $ A proof is here. $\endgroup$ – Bill Dubuque Oct 29 '18 at 20:04

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