2

I proved the claim that for every $m\in\mathbb{Z^+}$ the Fibonacci sequence repeats itself for modulo $m$. But how can I move on and make connections with it to prove that for every $m\in\mathbb{Z^+}$ there exists a Fibonacci number such that its last $m$ digits are $0$. Should I show it like using the modulo $10^m$ and show that by the Dirichlet's Box principle after $10^{2m}+1$ it should repeat itself? But does this have anything with my problem?

Aig
  • 4,178
Elfryionnn
  • 23
  • 5
  • 6
    You know the Fibonacci numbers are periodic modulo $10^m$. You also know that $F_0 \equiv 0 \bmod{10^m}$. – Haran Mar 18 '24 at 14:56
  • 1
    Just out of curiosity, what are the first Fibonacci numbers, ending with two or three zeroes? – Dominique Mar 18 '24 at 15:06
  • So can I conclude it like the next period that mirrors the first one has $m$ last digit 0's. Is it enough? – Elfryionnn Mar 18 '24 at 15:07
  • 1
    The smallest positive indeces $n$ , such that $F_n$ is divisible by $10,100,1\ 000,10\ 000,100\ 000$ are $15,150,750,7\ 500,75\ 000$ respectively. – Peter Mar 18 '24 at 15:24
  • 1
    @Elfryionnn You need to show that the sequence is "immediately periodic", and not just "eventually periodic". EG The sequence $0, 1, 1, 1, \ldots$ is eventually periodic, and never divisible by 10 thereafter. $\quad$. So while "yes we can conclude that the next period mirror the first one has ...", you just have to make sure that you showed the correct thing initially. – Calvin Lin Mar 18 '24 at 16:30
  • @CalvinLin What do you mean by immediately? Do you mean like I need to be sure that after some special $n\in\mathbb{N}$ $F_{n+1} \equiv 0 \pmod{10^m}$? Or did I get it wrong. – Elfryionnn Mar 18 '24 at 17:03
  • @Elfryionnn what Calvin means is that you need to show that the sequence is periodic from the start, i.e., not like the periodicity starts from the third or fourth number, and the previous ones do not appear again. – D S Mar 18 '24 at 18:24
  • 1
    @Elfryionnn You have to show that $ F_{n+k} \equiv F_n \pmod{10^m}$ holds true for all $n$ (including $n = 0$), and not just "$n$ large enough (like $ n \geq 1$)". $\quad$ This claim is true for $n=0$ (and even negative $n$). The typical approach is a forward-only induction proof, which will only show it for positive integers. – Calvin Lin Mar 18 '24 at 19:56

0 Answers0