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# 数学代写|数论代写Number Theory代考|Fibonacci Numbers and Divisibility

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## 数学代写|数论代写Number Theory代考|Fibonacci Numbers and Divisibility

The Fibonacci numbers have some remarkable divisibility properties, and we will look at several of these properties in this section. We have already mentioned that consecutive Fibonacci numbers are relatively prime (and you are asked to prove this fact in Problem 12.25). Note also that $F_3=2, F_5=5, F_7=13, F_{11}=89, F_{13}=233, F_{17}=1597$ are all prime, so it is tempting to conjecture that $F_p$ is prime whenever $p$ is prime; unfortunately, $F_{19}=4181=37 \cdot 113$, so this conjecture fails.
However, an interesting flip side to this failed conjecture is that for any prime $p$ greater than 5 it is always the case that $p$ divides either $F_{p-1}$ or $F_{p+1}$, but not both (for example, $11 \mid F_{10}=55$, but $11 \nmid F_{12}=$ 144). In order to prove this property we will use Binet’s formula,the binomial theorem, congruences, Fermat’s little theorem, and also Theorem 12.5.
Let $p$ be a prime such that $p>5$. Using Binet’s formula and the binomial theorem, we get
\begin{aligned} F_p= & \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^p-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^p \ = & \frac{1}{2^p \sqrt{5}}\left(1+\left(\begin{array}{l} p \ 1 \end{array}\right) \sqrt{5}+\left(\begin{array}{l} p \ 2 \end{array}\right) \sqrt{5}^2+\left(\begin{array}{l} p \ 3 \end{array}\right) \sqrt{5}^3+\cdots+\left(\begin{array}{l} p \ p \end{array}\right) \sqrt{5}^p\right) \ & -\frac{1}{2^p \sqrt{5}}\left(1-\left(\begin{array}{l} p \ 1 \end{array}\right) \sqrt{5}+\left(\begin{array}{l} p \ 2 \end{array}\right) \sqrt{5}^2-\left(\begin{array}{l} p \ 3 \end{array}\right) \sqrt{5}^3+\cdots-\left(\begin{array}{l} p \ p \end{array}\right) \sqrt{5}^p\right) \ = & \frac{1}{2^{p-1}}\left(\left(\begin{array}{l} p \ 1 \end{array}\right)+\left(\begin{array}{l} p \ 3 \end{array}\right) 5+\left(\begin{array}{c} p \ 5 \end{array}\right) 5^2+\cdots+\left(\begin{array}{l} p \ p \end{array}\right) 5^{\frac{p-1}{2}}\right) \end{aligned}
Looking at this last expression modulo $p$, the first thing we notice is that $2^{p-1} \equiv 1(\bmod p)$ by Fermat’s little theorem. This means that, modulo $p$, we can multiply through by $2^{p-1} \equiv 1$ to cancel the $2^{p-1}$ term. Next, recall from Chapter 5 that $p$ divides each of the binomial coefficients $\left(\begin{array}{l}p \ 1\end{array}\right),\left(\begin{array}{l}p \ 3\end{array}\right),\left(\begin{array}{c}p \ 5\end{array}\right), \ldots,\left(\begin{array}{c}p \ p-1\end{array}\right)$ (see Problem 5.40). Thus, since $\left(\begin{array}{l}p \ p\end{array}\right)=1$, we can conclude that
$$F_p \equiv 5^{\frac{p-1}{2}} \quad(\bmod p)$$
But then, $F_p^2 \equiv 5^{p-1} \equiv 1(\bmod p)$, again by Fermat’s little theorem, and now, since Theorem 12.5 gives us that $F_{p+1} F_{p-1}=F_p^2-1$, we can see that $F_{p+1} F_{p-1} \equiv 0(\bmod p)$; hence, as claimed, $p \mid F_{p+1}$ or $p \mid F_{p-1}$. Finally, it is impossible for $p$ to divide both of these numbers because if it did, then it would also have to divide $F_p$ (by the recurrence relation $F_{p-1}+F_p=F_{p+1}$ ), but then $p$ would also divide $F_{p-2}$ (again by the Fibonacci recurrence relation), and also divide $F_{p-3}$, and so on, until $p$ would divide $F_2=1$.

## 数学代写|数论代写Number Theory代考|Generating Functions

In this section we introduce a powerful tool for finding formulas-such as Binet’s formula-for sequences of numbers. This tool is called generating functions. In his outstanding book about generating functions, generatingfunctionology, Herbert Wilf describes a generating function metaphorically as “a clothesline on which we hang up a sequence of numbers for display.”

Here is the “clothesline” (i.e., the generating function) for the Fibonacci numbers:
$$0+1 x+1 x^2+2 x^3+3 x^4+5 x^5+8 x^6+13 x^7+\cdots$$
How can this be a useful thing to do? All we have really done is hang each Fibonacci number $F_n$ on the clothesline in its position as the coefficient of $x^n$. The reason this turns out to be useful is that this clothesline is now an algebraic object that we can manipulate using the ordinary rules of algebra.

So, a generating function is an infinite series, but try not to think of it as a function. We are not concerned with whether it converges. We will not evaluate it for particular values of $x$ (in fact, there is no domain in sight). To emphasize that our point of view here is purely algebraic, we often refer to these objects as power series.

Let’s begin by doing some algebra on the generating function (i.e., the power series) for the Fibonacci sequence. Let the power series for the Fibonacci sequence be given by
$$f(x)=F_0+F_1 x+F_2 x^2+F_3 x^3+F_4 x^4+\cdots$$

Then, we can use the recurrence relation $F_n=F_{n-1}+F_{n-2}$ and the fact that $F_0=0$ and $F_1=1$ to write
\begin{aligned} f(x)-x & =F_2 x^2+F_3 x^3+F_4 x^4+\cdots \ & =\left(F_0 x^2+F_1 x^3+F_2 x^4+\cdots\right)+\left(F_1 x^2+F_2 x^3+F_3 x^4+\cdots\right) \ & =x^2 f(x)+x f(x) \end{aligned}
Solving now for $f(x)$, we get $\left(1-x-x^2\right) f(x)=x$; and so
$$f(x)=\frac{x}{1-x-x^2}$$

## 数学代写|数论代写Number Theory代考|Fibonacci Numbers and Divisibility

\begin{aligned} F_p= & \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^p-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^p \ = & \frac{1}{2^p \sqrt{5}}\left(1+\left(\begin{array}{l} p \ 1 \end{array}\right) \sqrt{5}+\left(\begin{array}{l} p \ 2 \end{array}\right) \sqrt{5}^2+\left(\begin{array}{l} p \ 3 \end{array}\right) \sqrt{5}^3+\cdots+\left(\begin{array}{l} p \ p \end{array}\right) \sqrt{5}^p\right) \ & -\frac{1}{2^p \sqrt{5}}\left(1-\left(\begin{array}{l} p \ 1 \end{array}\right) \sqrt{5}+\left(\begin{array}{l} p \ 2 \end{array}\right) \sqrt{5}^2-\left(\begin{array}{l} p \ 3 \end{array}\right) \sqrt{5}^3+\cdots-\left(\begin{array}{l} p \ p \end{array}\right) \sqrt{5}^p\right) \ = & \frac{1}{2^{p-1}}\left(\left(\begin{array}{l} p \ 1 \end{array}\right)+\left(\begin{array}{l} p \ 3 \end{array}\right) 5+\left(\begin{array}{c} p \ 5 \end{array}\right) 5^2+\cdots+\left(\begin{array}{l} p \ p \end{array}\right) 5^{\frac{p-1}{2}}\right) \end{aligned}

$$F_p \equiv 5^{\frac{p-1}{2}} \quad(\bmod p)$$

## 数学代写|数论代写Number Theory代考|Generating Functions

$$0+1 x+1 x^2+2 x^3+3 x^4+5 x^5+8 x^6+13 x^7+\cdots$$

$$f(x)=F_0+F_1 x+F_2 x^2+F_3 x^3+F_4 x^4+\cdots$$

\begin{aligned} f(x)-x & =F_2 x^2+F_3 x^3+F_4 x^4+\cdots \ & =\left(F_0 x^2+F_1 x^3+F_2 x^4+\cdots\right)+\left(F_1 x^2+F_2 x^3+F_3 x^4+\cdots\right) \ & =x^2 f(x)+x f(x) \end{aligned}

$$f(x)=\frac{x}{1-x-x^2}$$

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