5
$\begingroup$

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 $p|S_N$ ?

$\endgroup$
3
$\begingroup$

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 $F_{p-1}$

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.