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$M>0$ is an integer.

For every $n>-0$ the remainder for the n Fibonacci number divided by m is: $r_n = f_n mod n$.

I need to prove that in :

$((r_n,r_n+1)) = (r_0,r_1),(r_1,r_2),(r_2,r_3)...$ must be repeats of pairs

Will appreciate some guidance because I don't know where to start...

baaa12
  • 754

2 Answers2

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As anon noted, there are only finitely many (exactly $M^2$) pairs, so in the infinite sequence there must be a repetition, say $(r_n,r_{n+1})=(r_k,r_{k+1})$, and $n>k$. Then, you should conlcude that $n-k$ is a period of this whole sequence.

Berci
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Hint $\rm\ (f_{n+1},f_{n+2}) = (f_n,f_{n+1}) A\:$ for an invertible matrix $\rm\:A,\:$ so $\rm\:(f_{n+k},f_{n+k+1}) = (f_n,f_{n+1}) A^k,\:$ and, by the pigeonhole/box principle, $\rm\:A^k\:$ has finite order mod $\rm\:n,\:$ yielding the sought cyclicity.

Alternatively $\rm\:A\:$ is an invertible map on a finite set $\rm\,\Bbb Z_n^2,\:$ i.e. a permutation, thus it has finite order. See this answer for further discussion of this viewpoint, where it is used to tackle the following problem: a sequence $\rm\:f(n)\:$ satisfies the relation $\rm\:f(n+2) = f(n+1)^2 - f(n),\,$ $\rm\,f(1) = 39,\ f(2) = 45.\:$ Prove that $1986$ divides infinitely many terms of the sequence.

Bill Dubuque
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