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Show that the sequence $\Big<\Big\{\frac{f_k}{n}\Big\}\Big>_{k=1}^\infty$ is periodic, where $\{\mathrm{F}\}$ is the fractional part of $\mathrm F$ and $f_k$ is the $k^\text{th}$ Fibonacci number starting with $f_1=0, f_2=1.$

This is how I tried to tackle the problem: Consider the numbers modulo $n$. If we show the existence of $0$ followed by $1$ or $1$ followed by $0$, we will be done. The reason is short and simple- the point of periodicity of the sequence is $(\cdots, 1,0,1,\cdots)$ modulo $n$. Now, we know that $\{F_{i-1}, F_i\}$ has $n^2$ possible pairs, since $n$ can have at most $n$ different remainders upon division of other numbers. Therefore, after the $n^2+1$th pair, there will be the recurrence of one of the pairs. But how do we know that $(0,1)$ pair will exist? How to proceed from here?

amWhy
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Mathejunior
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    One way of thinking about it is to observe that you can also run the Fibonacci recurrence backwards: $F_n=F_{n+2}-F_{n+1}$. If you reached the pair $(a,b)$ (modulo $n$) twice, you know for a fact that running it backwards from the 1st occurrence of $(a,b)$ will bring it back to $(0,1)$ in a certain number of steps. Guess what happens when you run it backwards from the second occurrence of $(a,b)$ the same number of steps :-) – Jyrki Lahtonen Jun 30 '18 at 12:31
  • +1 for including the good start, and explaining your problem. – Jyrki Lahtonen Jun 30 '18 at 12:33
  • Oh yes, that's a nice idea. Thank you! – Mathejunior Jun 30 '18 at 12:42
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    The "running backwards" idea has appeared at least here. – Jyrki Lahtonen Jun 30 '18 at 12:47

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