I found that the Fibonacci numbers $F_n$, taken$ \bmod n$, have the property that, for prime $n\ne 5$, $F_{n-1}$ and $F_{n+1}$ are $0$ and $1$, with the order depending on the value of $n\bmod 5$.

So if prime $n\equiv \pm 1\bmod 5$, $F_{n-1}\equiv 0$ and $F_{n+1}\equiv 1 \bmod n$
If prime $n\equiv \pm 2\bmod 5$, $F_{n-1}\equiv 1$ and $F_{n+1}\equiv 0 \bmod n$

It also turns out this is a fairly reliable test of whether $n$ is prime - that is, there are not many composites passing the test. The first composite number to masquerade as a prime is $4181$, then $5777, 6721, 10877, 13201, 15251$ are the remaining pseudoprimes under $20000$. This sequence is listed under OEIS A212424.

What I haven't been able to do is prove that this property does in fact hold for prime numbers. Can anyone produce a proof of the correctness of this relation?

Greg Martin gave some useful terminology:

The first positive integer $m$ for which $F_m\equiv 0 \bmod n$ is called the "rank of apparition" of $n$

Then ($\equiv \bmod n$) the "apparition multiplier" (my term) $k:\equiv F_{m-1}\equiv F_{m+1}\equiv F_{m+2}$ means that $F_{i+m}\equiv k F_i$ and $F_{i+jm}\equiv k^j F_i$. $k$ must be coprime to $n$ because the Fibonacci sequence can be generated in reverse so some $d\ne 1$ that divides both $k$ and $n$ would preclude $F_1=1$ (since $d$ would divide every element). This guarantees that $F_s\equiv 0 \Rightarrow m\mid s$ and that since at some point $k^j\equiv 1$, we will have the period $r, \bmod n$.

For the result above on primes we need $r\mid n{-}1$ in the first case and $m\mid n{+}1, r\mid 2(n{+}1)$ in the second case.

I was still thinking about what values the rank of apparition $m$ can take and working through http://www.fq.math.ca/Scanned/1-2/vinson.pdf .

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    $\begingroup$ Related : math.stackexchange.com/a/37957/78967 $\endgroup$ – mathlove Mar 19 '17 at 7:41
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    $\begingroup$ The first positive integer $m$ for which $F_m\equiv0\pmod n$ is called the "rank of apparition" of $n$; that will help you search for relevant mathematics. $\endgroup$ – Greg Martin Mar 19 '17 at 7:42

If $p\equiv \pm 1 \pmod 5$ then $x^2-x-1$ factors as $(x-\phi)(x-\overline{\phi})$ in $\mathbb{F}_p$. Then your identities follow from $$F_n\equiv \frac{\phi^n-\overline{\phi}^n}{\phi - \overline{\phi}} \pmod p$$ since in this case $\phi^{p-1}\equiv 1$ (by Fermat) and $$\phi^{p+1}\equiv \phi^2 \equiv \phi+1$$ and likewise for $\overline{\phi}$.

If $p\equiv \pm 2 \pmod 5$ then the same polynomial factors in $\mathbb{F}_{p^2}$ and the Frobenius automorphism exchanges the roots. That is, $\phi^p=\overline{\phi}$ and $\overline{\phi}^p=\phi$. Your identities now follow from $$\phi^{p-1}\equiv \overline{\phi} \phi^{-1} \equiv -\overline{\phi}^2 \equiv -\overline{\phi}-1$$ and $$\phi^{p+1}\equiv \overline{\phi} \phi \equiv -1$$ and likewise for $\overline{\phi}$.

  • $\begingroup$ Thanks, can you expand on how these work to produce the identities? $\endgroup$ – Joffan Mar 19 '17 at 18:10
  • $\begingroup$ Added some more details. $\endgroup$ – WimC Mar 19 '17 at 20:42

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