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prove that if $d$ divides $n$ then prove that
fibonacci of $d$ divides fibonacci of $n$.

i have tried to write $F(n)$ as a multiple of $F(d)$ using the fact that $n = ad$ for some natural $a$ but got nowhere..

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By the addition law $\rm\: f(x+y) = i\:f(x) + j\:f(y)\:$ for $\rm\:i,j\in \mathbb Z,\,$ so by $\,\rm\color{#c00}{induction}\,$ $\rm\: f(d)\mid f(nd)$

$$\rm f((n\!+\!1)d)\, =\, f(nd\!+\!d)\, =\, i\: \color{#c00}{f(nd)} + j\: f(d)\, =\, i\, \color{#c00}{k\, f(d)}\! + j\: f(d)\, =\, (i\,k+j)\: f(d) $$

Remark $\ $ More generally $\rm\, \gcd(f(k),f(n)) = f_{\,\gcd(k,n)}.\,$ For a proof see here.

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  • $\begingroup$ dont have enough rep to vote up $\endgroup$ – user2993422 May 9 '15 at 15:24
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You can prove it from Binet's formula: if $\varphi`$ is the golden ratio $\dfrac{1+\sqrt 5}2$, $\varphi'$ its conjugate, we have: $$F_n=\frac 1{\sqrt5}(\varphi^n-\varphi'^n)$$ so that $$\frac{F_n}{F_d}=\frac{\varphi^n-\varphi'^n}{\varphi^d-\varphi'^d}.$$ Now, if $n=md$, we get $$\frac{F_n}{F_d}=\frac{(\varphi d)^m-(\varphi'^{\mkern1mud})^m}{\varphi^d-\varphi'^{\mkern1mu d}}=\sum_{i=0}^{m-1}\varphi^{(m-1-i)d}\varphi'^{\mkern1mu id}$$ This sum is a symmetric polynomial of $\varphi$ and $\varphi'$, hence a polynomial in $\varphi+\varphi'=1$ and $\varphi\varphi'=-1$, so that it is an integer.

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