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# 数学代写|数论代写Number Theory代考|MAST90136 Fermat’s Theorem for Prime Moduli

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## 数学代写|数论代写Number Theory代考|Fermat’s Theorem for Prime Moduli

A first immediate question is, Can we reduce an exponent which is larger than a given modulus $n$ in the same way we can reduce numbers in the base, that is, by dividing by $n$ and taking the remainder? This turns out to be not true, and here is a simple example:
Example 5.1. Question: What is $2^9$ modulo 5 ?
Correct solution method: $2^9=512$, and $512(\bmod 5)=2$.
Incorrect solution method: Reducing the exponent 9 modulo 5 gives a new exponent of $4.2^4=16$ and $16(\bmod 5)=1$, which evidently is wrong!

Hence we see that in general we cannot reduce exponents by the modulus the way we can do with numbers in the base. This is an issue we need to deal with since we could be asked to compute, say, $2^{90}$ modulo 5 . Surely there is a better and faster way to obtain this least non-negative residue than to multiply out the value $2^{90}$ (which, by the way is a number with 28 decimal digits). Fortunately, there is indeed a reduction method for exponents, which we asked you in Problem $4.17$ to form a conjecture about by looking at data. It may help to look back at that problem now.

## 数学代写|数论代写Number Theory代考|Euler’s Function and Euler’s Theorem

Working about a century after Fermat, Euler saw a way to generalize Fermat’s Theorem to congruences involving an arbitrary modulus $n$. The key idea was to find what quantity with respect to $n$ is the analog of $p-1$ in Fermat’s Theorem. To this end, Euler introduced a function which takes in a positive integer $n$ and returns the number of elements in the range ${1,2, \ldots, n}$ which are relatively prime to $n$. He denoted this function using the Greek letter phi ( $\phi$, pronounced “phee”), so it is often referred to as “the Euler $\phi$ function” as well as “Euler’s Function.” We note immediately that if $p$ is prime, then $\phi(p)=p-1$ since every non-zero element of $\mathbb{Z}_p$ is relatively prime to $p$. This function will be the key to generalizing Fermat’s Theorem to arbitrary moduli. Let us look at some examples.

Example 5.4. (a) It is easy to check that $\phi(2)=1, \phi(3)=2$, $\phi(4)=2, \phi(5)=4, \phi(6)=2, \phi(7)=6, \phi(8)=4, \phi(9)=6$, $\phi(10)=4$, and so on.
(b) Computing $\phi(20)$, we have that the non-zero elements of $\mathbb{Z}_{20}$ which are relatively prime to 20 are ${1,3,7,9,11,13,17,19}$, so $\phi(20)=8$. Looking ahead to our next result, let us observe that whereas $\phi(4) \phi(5)=2 \cdot 4=8=\phi(20), \phi(2) \phi(10)=1 \cdot 4=4 \neq$ $\phi(20)$. This is resulting from the fact that 4 and 5 are relatively prime, but 2 and 10 are not.
(c) Computing $\phi(120)$ “by hand” would be possible but difficult. The next theorem tells us how to break down a “$\phi$-calculation” into smaller parts.

Our next theorem tells us how compute Euler’s Function for any positive integer $n$ by computing it for each of $n$ ‘s prime-power factors and then multiplying these answers together. This is what we did in Example 5.4, Part (b) by computing $\phi\left(2^2\right)$ and $\phi(5)$ to get $\phi(20)$ Again, the key is that 4 and 5 are relatively prime.
Theorem 5.3. (a) If $p$ is prime and if $k$ is a positive integer exponent, then
$$\phi\left(p^k\right)=p^k-p^{k-1}=p^{k-1}(p-1) .$$
(b) If $m_1$ and $m_2$ are relatively prime, then $\phi\left(m_1 m_2\right)=\phi\left(m_1\right) \phi\left(m_2\right)$.

## 数学代写数论代写Number Theory代考|Euler’s Function and Euler’s Theorem

(c) 计算 $\phi(120)$ “手工”是可能的，但很困难。下一个定理告诉我们如何分解” $\phi$-计算”分成更小的部分。

$$\phi\left(p^k\right)=p^k-p^{k-1}=p^{k-1}(p-1) .$$
(b) 如果 $m_1$ 和 $m_2$ 是相对质数，那么 $\phi\left(m_1 m_2\right)=\phi\left(m_1\right) \phi\left(m_2\right)$.

## MATLAB代写

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