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Let $F_n$ denote the $n$th Fibonacci number, and $p_n$ the $n$th prime.

Let $a(n)$ be the smallest positive integer such that $p_n$ is a factor of $F_{a(n)}$.

How can I see that it follows that exactly one of $\frac{p_n+1}{a(n)},\frac{p_n}{a(n)},\frac{p_n-1}{a(n)}$ an integer?

Context required by the admins: I observed this fact in numerics up to $n=80$, beyond which it becomes infeasible to continue to evaluate. I looked around and noticed it was stated as a property on OEIS A001177 but no links or citations were provided.

I am interested in knowing how this problem is approached, but I have no formal education in pure maths. This is not homework.

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    $\begingroup$ Related information for example here. Searching for Pisano periods gives a lot of good stuff. A lot hinges on whether $5$ is a quadratic residue or not. Hardly a surprise in view of Binet's formula. $\endgroup$
    – Jyrki Lahtonen
    Commented Apr 15, 2017 at 21:05
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    $\begingroup$ Apply the binomial theorem expansions to $F(n)=[(1+\sqrt 5)^n-(1-\sqrt 5)^n]/ [2\sqrt 5]^n$ for $n=p$ and for $n=p+1$ when $p$ is prime and $2\ne p\ne 5.$ Use these facts: $p$ divides $\binom {p}{j}$ when $0<j<p,$ and $p$ divides $\binom {p+1}{j}$ when $1<j<p,$ and $2^p\equiv 2 \pmod p$ and $ 5^{(p-1)/2}\equiv \pm 1 \pmod p,$.... to show that $p$ divides $F(p+1)$ or that $p$ divides $ F(p+1)-F(p)=F(p-1).$ $\endgroup$
    – DanielWainfleet
    Commented Apr 16, 2017 at 7:38
  • $\begingroup$ Thanks for the useful comments. $\endgroup$
    – ComptonScattering
    Commented Apr 16, 2017 at 17:03
  • $\begingroup$ @user254665, I may have misunderstood your comment but the question does not regard $F_p$, but $F_a$ where $a$ is defined as above. $\endgroup$
    – ComptonScattering
    Commented Apr 16, 2017 at 17:08
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    $\begingroup$ $F_a$ divides $F_{ab}.$ So if $p|F_{p+1}$ ($2\ne p\ne 5$) then the smallest $n$ such that $p|F_n$ is a divisor of $p+1.$ $\endgroup$
    – DanielWainfleet
    Commented Apr 16, 2017 at 18:04

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