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This came up in a series of notes on number theory I'm reading. The question is:

Prove that the $(p-1)$th or the $(p+1)$th Fibonacci number is divisible by $p$ for some prime $p$

I'm very much lost on this question, so I thought I'd see if I could find a proof here.

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    If it were true that $F_{p-1}$ and $F_{p+1}$ were both divisible by $p$, then so too must $F_p$ since $F_{p+1}-F_{p-1}=F_p$ is the difference of two numbers, both of which divisible by $p$. But then... that implies that $F_{p-2}$ is divisible by $p$ for the same reason as $F_{p-2}=F_p-F_{p-1}$ and so on so forth, – 2017-01-16
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    It does seem like one or the other of those two numbers is always divisible by prime $p$, and the other one is $\equiv 1 \bmod p$. (with the exception of $p=5$). Interesting. – 2017-01-16
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    @Dave that is for a specific prime $p$, it wasn't supposed to be true for all prime $p$'s – 2017-01-16
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    @Paolo if that was in reference to my comment, in my comment I give the important realization that leads to a proof that there is never any prime $p$ for which this is true, and in general can be used to prove that $F_{n+1}$ is relatively prime to $F_{n-1}$ in every case – 2017-01-16
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    I realise I've made a mistake, need to replace 'and' with 'or' in the question – 2017-01-16
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    Geom, is the thing you want true for primes 2, 3, 5? – 2017-01-16
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    @WillJagy as far as I can see (now it's revised) it may well be true for *every* prime except $p=5$. – 2017-01-16
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    @Joffan, fine. Mostly I want the OP to check the first few cases, indeed, before asking about general primes. – 2017-01-16
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    In conclusion, $p\neq 5$ divides *exactly one* between $F_{p+1}$ and $F_{p-1}$ :) – 2017-01-16

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Fix a prime $p\ge 7$. We have by Cassini's identity that $$ F_{p-1}F_{p+1} = F_p^2-(-1)^{p-1} \,\,\,\,(=\, F_p^2-1). $$ Hence, it is enough to prove that $$ F_p^2 \equiv 1\pmod{p}. $$ But we have \begin{align*} F_p^2&=\left(\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^p-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^p\right)^2 \\\ &=\frac{1}{5}\left(\frac{1+\sqrt{5}}{2}\right)^{2p}+\frac{1}{5}\left(\frac{1-\sqrt{5}}{2}\right)^{2p}-\frac{2}{5}(-1)^p \\\ &=\frac{1}{5\cdot 2^{2p}}\left(1+\sqrt{5}\right)^{2p}+\frac{1}{5\cdot 2^{2p}}\left(1-\sqrt{5}\right)^{2p}+\frac{2}{5} \\\ &=\frac{1}{5\cdot 2^{p}}\left(3+\sqrt{5}\right)^{p}+\frac{1}{5\cdot 2^{2p}}\left(3-\sqrt{5}\right)^{2p}+\frac{2}{5} \\\ &=\frac{1}{5\cdot 2^{p}}\left(2\sum_{i=0, i\text{ even}}^p\binom{p}{i}3^{p-i}\sqrt{5}^i\right)+\frac{2}{5} \\\ &=\frac{1}{5\cdot 2^{p-1}}\left(\sum_{j=0}^{\frac{p-1}{2}}\binom{p}{2j}3^{p-2j}5^j\right)+\frac{2}{5}. \end{align*}

At this point, considering that $p\mid \binom{p}{j}$ for $0