A question regarding the divisibility of Fibonacci sequence

Question

Knowing that $F(n)$ is the Fibonacci sequence. The question is to prove that $F(n)$ is divisible by $F(m)$ if $n$ is divisible by $m$. I show my method in the picture below, however, I did not solve it. I hope to get some help.

Answer 1

Do you have to use Binet's formula, or can you use the recurrence $F(n)=F(n-1)+F(n-2)$ and the initial values $F(0)=0$, $F(1)=1$? Here is an easy proof by induction using the recurrence.

Consider a fixed integer $m$ and let $a=F(m+1)$. Then: $$F(m)\equiv aF(0)\pmod{F(m)}$$ $$F(m+1)\equiv aF(1)\pmod{F(m)}$$ $$F(m+2)=F(m+1)+F(m)\equiv aF(1)+aF(0)=aF(2)\pmod{F(m)}$$ $$F(m+3)=F(m+2)+F(m+1)\equiv aF(2)+aF(1)=aF(3)\pmod{F(m)}$$ etc. By induction, $$F(m+r)\equiv aF(r)\pmod{F(m)}$$ and in particular $$F(2m)=F(m+m)\equiv aF(m)\equiv0\pmod{F(m)}$$ $$F(3m)=F(m+2m)\equiv aF(2m)\equiv0\pmod{F(m)}$$ etc. By induction, $F(km)\equiv0\pmod{F(m)}$ for all $k$.