I guess if it is true that $5^k \mid f(5^k)$, where $f(n)$ denotes the $n$-th Fibonacci's number. I have tried to prove it by induction on $k$, but nothing. Have you got any ideas?
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Please, please do specify where exactly the Fibonacci sequence starts. Do you define $f(0)=f(1)=1$, or do you define $f(1)=f(2)=1$? – Dan Shved May 30 '14 at 10:17
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$f(0)=1, f(1)=1$ so $5\mid f(5)$ – Nisba May 30 '14 at 10:19
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1@DanShved $f(0)=0$, $f(1)=1$, $f(2)=1$, $f(3)=2$, $f(4)=3$, $f(5)=5$. If the definition was with $f(0)=f(1)=1$, the assertion would fail at $k=1$. – egreg May 30 '14 at 10:21
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Sorry! Your are right, I mean that definition – Nisba May 30 '14 at 10:23
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Hint $\ $ It is very easy using the matrix representation of fibonacci numbers. Then the inductive proof amounts simply to lifting the exponent $\,\color{#0a0}{5^k\mapsto 5^{k+1}}\,$ using the Binomial Theorem, e.g.
$$\begin{eqnarray} &&\left[\begin{array}{rr}f_6&f_5\\f_5&f_4\end{array}\right] &=& \left[\begin{array}{rr}8 &5\\5&3\end{array}\right] &=&\, 3I + \color{#0a0}5A\\ \Rightarrow\, &&\left[\begin{array}{rr}f_{26}\!\!&\!\!f_{25}\\f_{25}\!\!&\!\!f_{24}\end{array}\right] &=& \left[\begin{array}{rr}8 &5\\5&3\end{array}\right]^{\large \color{#c00}5} \!\!&=& (3I + 5A)^{\large \color{#c00}5}\! =\, 3^{\large \color{#c00}5} I + {\color{#c00}5}(5A)(\cdots) =\, 3^5 I + \color{#0a0}{5^2} B\end{eqnarray}$$
Bill Dubuque
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1Very creative! Just to clarify, we use induction to prove that $\left[\begin{matrix}F_{5^k+1} & F_{5^k}\F_{5^k} & F_{5^k-1}\end{matrix}\right] = 3^{5^{k-1}}I + 5^kA_k$ right? And from this the required result follows by comparing corresponding matrix elements on both sides. – Anant May 30 '14 at 21:23
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1@Anant Right. The induction step follows simply by taking the fifth power of that equation, as in the example I gave. The key idea is that the representation $,f_n = (M^n)_{1,2},$ as a power allows us to use that standard "lifting the exponents" for such powers to lift the congruence up from the base case. – Bill Dubuque May 30 '14 at 21:54
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Yes, the statement is correct, provided the Fibonacci sequence is initialized with $f(0)=0$ and $f(1)=1$. That is, $f(n)$ is divisible by $5^k$ if and only if $n$ is divisible by $5^k$.
Hint: Use the explicit formula for the $n$th term, namely:
$$f(n)=\frac{\binom n1+5\binom n3+5^2\binom n5+5^3\binom n7+\cdots}{2^{n-1}}.$$
This formula for $f(n)$ can be derived by applying the binomial theorem to the formula
$$f(n)=\frac{(1+\sqrt5)^n-(1-\sqrt5)^n}{\sqrt5\cdot2^n}.$$
bof
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It is true. We have $F_{5} = 5.$ If we write $\sqrt{5}F_{n} = \alpha^{n}-\beta^{n}$ where $\alpha$ and $\beta$ are the roots of $x^{2}-x-1$ and $\alpha > \beta,$ then it follows by induction that $\alpha^{5^{k}}- \beta^{5^{k}} = \sqrt{5}5^{k} c_{k}$ for some integer $c_{k},$ though the way I see this requires working in the ring of algebraic integers $\mathbb{Z}[\omega],$ where $\omega = e^{\frac{2 \pi i}{5}}.$ You seem to want to find your own proof, so I don't give all the details. The key point is that $u^{5} -v^{5} = (u-v) \prod_{i=1}^{4}(u - \omega^{i}v)$ and that $\prod_{i=1}^{4} (1-\omega^{i}) = 5.$
Geoff Robinson
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is it also true that the highest power 5 which divides $f(5^k)$ is $k$? – happymath May 30 '14 at 10:47
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@happymath: Yes, I believe that is the case, and you can build that into an inductive proof. The point is that $u - \omega^{i}v \equiv (1- \omega^{i}v)$ (mod $(1- \omega)^{2}$), so the above product is divisible by $(1-\omega)^{4}$ but no higher power of $1- \omega.$ – Geoff Robinson May 30 '14 at 11:31