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Actually, I'm solving some exercises from the book "Finite Field" by Rudolf Lidl et al.

There is an exercise for which the idea is missing to solve it:

Let $r$ be the least period of the Fibonacci sequence in the finite field $F_q$ i.e. the sequence with $s_0= 0, s_1= 1$, and $s_{n+2}=s_{n+1}+s_n$, for $n \geq 0$. Let $p$ be the characteristic of $F_q$. Prove that $r=20$ if $p = 5$, that $r$ divides $p-1$ if $p =$ +/- 1 mod 5 and that $r$ divides $p^2-1$ in all other cases.

Could anyone help me with a good idea / a good proposition?

JohnD
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    I think you need the explicit formula for $f_n$ in term of $\phi^n=(\frac{1+\sqrt{5}}{2})^n$ for $p \ne 2,5$ it is an element of $\Bbb{F}{p}$ or $\Bbb{F}{p^2}$ depending on $(\frac{5}{p})=(\frac{p}{5})$ – reuns Apr 11 '19 at 20:12
  • See https://en.wikipedia.org/wiki/Pisano_period – lhf Apr 12 '19 at 02:12
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    This old answer of mine describes the basic idea. You need the law of quadratic reciprocity to decide whether the golden ratio is an element of $\Bbb{F}p$ or $\Bbb{F}{p^2}$. Then you need to deduce that in the latter case, the conjugate of the golden ratio must be the Frobenius conjugate. After that you are basically done. – Jyrki Lahtonen Apr 12 '19 at 03:16

1 Answers1

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My favorite Fibonacci technique is the matrix formulation, which is well worth knowing and easily proved: $$ A^n= \begin{pmatrix}1&1\\1&0\end{pmatrix}^n= \begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix} $$

The first part of the question follows easily. Indeed, $A^2=A+I$ and so $A^5=5A+3I \equiv 3I \bmod 5$. Therefore, $A^{20} \equiv 3^{4} I \equiv I \bmod 5$. Since $A^4 = 3A + 2I \not\equiv I \bmod 5$, the period mod $5$ is $20$.

lhf
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  • Thanks a lot for your answer! I got the second part of the question, but the third one (in all other cases) is difficult...could you show me a hint for this? – JohnD Apr 12 '19 at 09:41