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# 数学代写|数论代写Number Theory代考|Math204A Fermat Numbers

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## 数学代写|数论代写Number Theory代考|Fermat Numbers

In this section our discussion is based on particular numbers of the form $F_{n}=$ $2^{2^{n}}+1$ where $n$ is a non negative integer. They are called Fermat numbers, named after the French mathematician Pierre de Fermat who first studied numbers of this form. Here we have also discussed few basic properties and primality of Fermat numbers. Fermat’s first conjecture was that all the numbers of these type were prime. But in 1732 Euler showed that the number $F_{5}=2^{2^{5}}+1$ is composite. Then it become a question whether there are infinitely many prime of this form or not. Thus to start our discussion we have given the following definition of Fermat numbers as well as Fermat primes.

Definition 10.6.1. An integer is of the form $F_{n}=2^{2^{n}}+1, n \geq 0$ is called Fermat number. If $F_{n}$ is prime then it is called Fermat prime.

For example we can see that the first five Fermat numbers $F_{0}=3, F_{1}=$ $5, F_{2}=17, F_{3}=257$ and $F_{4}=65,537$ all are primes. But the conjecture that Fermat numbers are primes fails to $F_{5}$. Here $F_{5}=2^{2^{5}}+1=641 \times 6700417$ is composite. In fact we have an elementary proof that $641 \mid F_{5}$ due to G.Bennet.
Theorem 10.6.1. The Fermat number $F_{5}$ is divisible by 641 .
Proof. Let us choose $u=2^{7}$ and $v=5$ then we have, $1+u v=1+2^{7} \cdot 5=641$. It is seen that, $1+u v-v^{4}=1+v\left(u-v^{3}\right)=1+\left(2^{7}-5^{3}\right) v=1+3 v=2^{4}$. Now
\begin{aligned} F_{5} &=2^{2^{5}}+1 \ &=2^{4} \cdot 2^{28}+1 \ &=2^{4} \cdot u^{4}+1 \ &=\left(1+u v-v^{4}\right) u^{4}+1 \ &=(1+u v)\left[u^{4}+(1-u v)\left(1+u^{2} v^{2}\right)\right] \end{aligned}
This shows that 641 divides $F_{5}$ as $1+u v=641$.
We are now going to prove some basic properties of Fermat numbers in our next theorem.

## 数学代写|数论代写Number Theory代考|Worked out Exercises

Problem 10.7.1. Show that $2^{2^{n}}+5$ is composite for each integer $n(>0)$.
Solution 10.7.1. For any positive integer $n, 2^{n}$ is even. Then we have, $2^{2^{n}} \equiv 1($ $\bmod 3)$. This shows that, $2^{2^{n}}+5 \equiv 1+5 \equiv 0(\bmod 3)$. Since $3 \mid 2^{2^{n}}+5$, therefore $2^{2^{n}}+5$ is composite.

Problem 10.7.2. Prove that every Fermat number $F_{n}$ is either a prime or a pseudoprime.

Solution 10.7.2. We need to verify, if $F_{n}$ is composite then it is pseudoprime. For that, we need to prove $\left(2^{2^{n}}\right)^{2^{n}+1} \equiv 2\left(\bmod F_{n}\right)$. We know that,
\begin{aligned} & 2^{2^{n}} \equiv-1\left(\bmod F_{n}\right) \ \Rightarrow &\left(2^{2^{n}}\right)^{2^{2^{n}}-n} \equiv(-1)^{2^{2^{n}}-n}\left(\bmod F_{n}\right) \ \Rightarrow & 2^{2^{2^{n}}} \equiv 1\left(\bmod F_{n}\right) \ \Rightarrow &\left(2^{2^{n}}\right)^{2^{n}+1} \equiv 2\left(\bmod F_{n}\right) . \end{aligned}

## 数学代写|数论代写Number Theory代考|Fermat Numbers

$1+u v-v^{4}=1+v\left(u-v^{3}\right)=1+\left(2^{7}-5^{3}\right) v=1+3 v=2^{4}$. 现在
$$F_{5}=2^{2^{5}}+1=2^{4} \cdot 2^{28}+1=2^{4} \cdot u^{4}+1 \quad=\left(1+u v-v^{4}\right) u^{4}+1=(1+u v)\left[u^{4}+(1-u v)\left(1+u^{2} v^{2}\right)\right]$$

## 数学代写|数论代写Number Theory代考|Worked out Exercises

$2^{2^{n}} \equiv-1\left(\bmod F_{n}\right) \Rightarrow \quad\left(2^{2^{n}}\right)^{2^{2^{n}}-n} \equiv(-1)^{2^{2^{n}-n}}\left(\bmod F_{n}\right) \Rightarrow 2^{2^{2^{n}}} \equiv 1\left(\bmod F_{n}\right) \Rightarrow \quad\left(2^{2^{n}}\right)^{2^{n}+1} \equiv 2\left(\bmod F_{n}\right)$

## MATLAB代写

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