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Let $F_n$ be the $n$th Fibonacci number. I've been playing with Fibonacci numbers for fun and I think that the following property should holds: there is no prime $p$ such that $F_{p^2}\equiv 1$ mod $p^2$, and (at the same time) $F_{p^2-1}\equiv 0$ mod $p^2$.

I've tried to formally prove this property but I could not succeed. I know that $F_{p-1}$ and $F_{p+1}$ divide $F_{p^2-1}$, and that if $p\neq2,5$, then either $p$ divides $F_{p-1}$ or $F_{p+1}$. But is this of any help? I would be glad if you could provide any suggestion or ideas on how to prove it, or if this sound reasonable or not.

W4cc0
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    It is easiest to see if you know finite fields. Then the closed formula that works in the integers also works in finite fields (except for,$2,5.$) – Thomas Andrews Nov 22 '24 at 01:04
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    Have you found primes satisfying the first congruence, but not the second? the second, but not the first? – Gerry Myerson Nov 22 '24 at 02:21
  • @GerryMyerson No I didn't find them. You say that neither of the conditions is ever satisfied? – W4cc0 Nov 22 '24 at 12:38
  • @ThomasAndrews You mean the Euler-Binet Formula? – W4cc0 Nov 22 '24 at 12:40
  • Yes, Euler Binet. Basically, in the field of order $p^2$ with $p\neq2,5,$ there is a $\sqrt5.$ So the Euler-Binet formula makes sense there. But this only gets you to $\pmod p,$ not $\pmod{p^2}.$ So I might have spoken too soon. @W4cc0 – Thomas Andrews Nov 22 '24 at 15:09
  • The other way to look at it is as linear algebra, $$\begin{pmatrix}F_n\F_{n+1}\end{pmatrix}=\begin{pmatrix}0&1\1&1\end{pmatrix}^n\begin{pmatrix}0\1\end{pmatrix} $$ But I'm not sure that will really help. – Thomas Andrews Nov 22 '24 at 15:20
  • @ThomasAndrews I actually thought of the matrix way, but could not get anything from it. – W4cc0 Nov 22 '24 at 15:24
  • Of course, for,$p\neq2,5$ prime, $5$ is a square modulo $p^2$ iff $p\equiv1,4\pmod 5.$ Then Euler-Binet works in the group of units of $R=\mathbb Z/p^2\mathbb Z.$ Let's just look at the case $p=11.$ – Thomas Andrews Nov 22 '24 at 15:52
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    @ThomasAndrews By the way, this seems to be a Fibonacci-Wieferich prime, according to https://math.stackexchange.com/questions/1365104/fibonacci-equiv-1-mod-p2 which seems be unknown if they exist. But here we have many additional conditions. Thus, according to the cited question for $p<2\cdot 10^14$ there are no solutions – W4cc0 Nov 22 '24 at 15:59
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    Then we can take $\sqrt5=48,$ and you get $$F_n\equiv (85^n-37^n)\cdot48^{-1}\pmod {121}.$$ But $a^{p^2-p}\equiv1\pmod {p^2}$ for $(a,p)=1.$ So what this shows is that $F_{n+p^2-p}\equiv F_n\pmod{p^2}.$ This would seem to mean, for $p\equiv1,4\pmod p,$ your conjecture is true iff it is true that $F_p\equiv0\pmod{p^2}.$ – Thomas Andrews Nov 22 '24 at 16:13
  • I'm saying nothing about the conditions, W4. But if you can't even find an example where one of the conditions is satisfied, what is the point of asking whether they are ever both satisfied? – Gerry Myerson Nov 22 '24 at 20:55
  • @GerryMyerson the point is that two conditions are usually better then one if you want to prove something... If something can be said on both conditions, then maybe one can try to weaken the hypothesis by deleting one of the conditions... – W4cc0 Nov 24 '24 at 15:06

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