# Show the given property about divisibility

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?

• Can you prove $F_m\mid F_{mn}$, $F_n\mid F_{mn}$ and $\gcd(F_m,F_n)=1$? – Lord Shark the Unknown Oct 29 '18 at 19:43
• Could you give me a hint how we show these properties? @LordSharktheUnknown – Evinda Oct 29 '18 at 19:50
• @Evinda Follow the link in my answer for some proofs. – Bill Dubuque Oct 29 '18 at 19:56

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