Let $\{F_n\}, n\in \mathbb{N}$ be the sequence of Fibonacci numbers such that $F_1=1$, $F_2=1$ and $F_n=F_{n-1}+F_{n-2}$ $\forall n\geq2$.

Define a new sequence $\{S_n\}$ such that $S_n=F_n+1$ $\forall n\in \mathbb{N}$.

Now the question is: For every prime $p$, does there exist an $N\in \mathbb{N}$, such that $p|S_N$ ?


Hint. The answer is yes. Show that for any prime $p\not=5$, $$p\;\mbox{divides}\;S_{p^2-3}=F_{p^2-3}+1.$$ See for example Jack D'Aurizio's answer here:Fibonacci Sequence problem. Prove that there are infinitely many prime numbers such that $p$ divides $F_{p-1}$


Well $S_{-2}=0$, but $-2\notin\Bbb N$. Never mind!

  • $\begingroup$ what do you mean by $S_-2$ ??? $-2$ is not a natural number actually, i believe i got what you wanted to say but explain further $\endgroup$ – Arpan1729 May 25 '17 at 17:02
  • $\begingroup$ I said that $-2$ wasn't a natural number, didn't I? The Fibonacci sequence does extend naturally to negative arguments, doesn't it? $\endgroup$ – Lord Shark the Unknown May 25 '17 at 17:03
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    $\begingroup$ Hmm, and the Fibonacci series is periodic (mod $n$) for any $n$ - also the naturally extended Fibonacci series. I think these two observations together solve the question. $\endgroup$ – Daniel Schepler May 25 '17 at 17:05
  • $\begingroup$ yes..........true $\endgroup$ – Arpan1729 May 25 '17 at 17:07

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