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

Question

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..

Answer 1

Addition law $\Rightarrow \rm f(x\!+\!y) = i\:f(x) + j\:f(y),\,$ $\rm i,j\in \mathbb Z,\,$ so if $\color{#c00}{\rm\: f(d)\mid f(nd)}$ our inductive step is

$$\rm \color{#0a0}{f((n\!+\!1)d)}\, =\, f(nd\!+\!d)\, =\, i\: \color{#c00}{f(nd)} + j\: f(d)\, =\, i\, \color{#c00}{k\, f(d)}\! + j\: f(d)\, =\, (i\:\!k\!+\!j)\, \color{#0a0}{f(d)} $$

so $\rm \,\color{#c00}{P(n)}\Rightarrow \color{#0a0}{P(n\!+\!1)},\,$ i.e. $\rm\,\color{#c00}{f(d)\mid f(nd)}\Rightarrow \color{#0a0}{f(d)\mid f((n\!+\!1)d}\,$ demonstrates our inductive step, and the base case $\,\rm n=0\,$ is trivially true since $\rm\:\!\ f(d)\mid f((0\!+\!1)d)$

Remark $ $ More generally $\rm\, \gcd(f(d),f(n)) = f_{\,\gcd(d,n)}\,$ so $\rm\,f(d)\mid f(n)\!\iff\! d\mid n,\,$ proved here.

Answer 2

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.