# 数学代写|数论作业代写number theory代考|Math124

## 数学代写|数论作业代写number theory代考|Wilson’s Theorem

Wilson’s theorem, in number theory, signifies that any prime $p$ divides $(p-1) !+1$, where $n !$ is the factorial notation for $1 \times 2 \times 3 \times 4 \times \cdots \times n$. For example, 7 divides $(7-1) !+1=6 !+1=721$. The conjecture was first published by the English mathematician Edward Waring in Meditationes Algebraicae (1770 ‘Thoughts on Algebra’), where he described it to the English mathematician John Wilson.
After that it was proved by the French mathematician Joseph-Louis Lagrange in 1771. The converse of the theorem is also true; that is, $(n-1) !+1$ is not divisible by a composite number $n$. In theory, these theorems provide a test for primes; in practice, the calculations are impractical for large numbers.

Theorem 5.4.1. Wilson’s Theorem: If $p$ is a prime then $(p-1) ! \equiv-1(\bmod p)$.
Proof. Let us choose $p>3$ and consider the linear congruence $a x \equiv 1(\bmod p)$ where $a$ is any one of $1,2,3, \cdots, p-1$. Therefore $\operatorname{gcd}(a, p)=1$. Hence, it has an unique solution viz $a \tilde{a} \equiv(\bmod p)$ with $1 \leq \tilde{a} \leq p-1$. Because $p$ is prime, $a=\tilde{a} \Leftrightarrow a=1$ or $a=p-1$ provided $a^2 \equiv 1(\bmod p) \Rightarrow(a-1)(a+1) \equiv 0($ $\bmod p)$. Therefore $(a-1) \equiv 0(\bmod p)$ or $(a+1) \equiv 0(\bmod p)$. Now if we delete 1 and $p-1$, then the remaining $2,3, \ldots, p-2$ are set into pairs $a$ and $\tilde{a}$, where $a \neq \tilde{a}$. So if these $\frac{p-3}{2}$ congruences are multiplied, we obtain $2 \cdot 3 \cdots(p-2) \equiv 1($ $\bmod p) \Rightarrow(p-2) ! \equiv 1(\bmod p) \Rightarrow(p-1) ! \equiv(p-1) \equiv-1(\bmod p)$

$p=11$. Divide the integers $2,3,4,5,6,7,8,9$ into $\frac{p-3}{2}$ pairs such as
\begin{aligned} &2 \cdot 6 \equiv 1(\bmod 11) \ &3 \cdot 4 \equiv 1(\bmod 11) \ &7 \cdot 8 \equiv 1(\bmod 11) \ &5 \cdot 9 \equiv 1(\bmod 11) \end{aligned}
Multiplying each pair together we obtain, $9 ! \equiv 1(\bmod 11)$. Hence $10 ! \equiv 1($ $\bmod 11$ ), shows the result is true for $p=11$. An interesting observation is that the converse is also true. Let $n$ be a non-prime required integer. Then $n$ must have a divisor $d$ where $1<d<n$. As $d \leq n-1$, we have $d \mid(n-1)$ !. Now from the condition we have, $n \mid((n-1) !+1)$. Hence combining the conditions, we have $d \mid((n-1) !+1)$. Thus $d \mid 1$ leads to contradiction, showing $n$ is prime. Taking Wilson’s theorem and its converse together we can say that the condition is necessary and sufficient for an integer to be prime. Thus it gives us a condition of testing primality.

## 数学代写|数论作业代写number theory代考|Worked out Exercises

Problem 5.5.1. Find the remainder when 15 ! is divided by 17.
Solution 5.5.1. Since $(17-1) !=16$ !, we have by virtue of Wilson’s theorem $(17-1) ! \equiv-1(\bmod 17)$. Therefore $16 ! \equiv-1(\bmod 17) \equiv 16(\bmod 17) \Rightarrow 15 ! \equiv 1($ mod 17). Hence the remainder is 1.
Problem 5.5.2. Find the remainder when 2(26)! is divided by 29.
Solution 5.5.2. From Wilson’s theorem, we find $28 ! \equiv-1(\bmod 29) \Rightarrow 28 ! \equiv 28($ $\bmod 29) \Rightarrow 27 ! \equiv 1(\bmod 29)$. Here we note that $\operatorname{gcd}(28,29)=1 \Rightarrow 27(26) ! \equiv$ $(1+29)=30(\bmod 29) \Rightarrow 9(26) ! \equiv 10(\bmod 29) \Rightarrow 9(26) ! \equiv(10+29)=9($ $\bmod 29) \Rightarrow 3(26) ! \equiv 13(\bmod 29) \Rightarrow 3(26) ! \equiv(13+29)=42(\bmod 29) \Rightarrow(26) ! \equiv$ $14(\bmod 29)$. Therefore $2(26) ! \equiv 28(\bmod 29)$. Thus, 28 is the remainder.
Problem 5.5.3. Show that $18 ! \equiv-1(\bmod 437)$.
Solution 5.5.3. Note that $437=19 \cdot 23$, where both 19 and 23 are prime numbers. By Wilson’s theorem, we have $18 ! \equiv-1(\bmod 19)$ therefore $19 \mid(18 !+$ 1) holds. So here the only thing we need to show is $23 \mid(18 !+1)$, because $\operatorname{gcd}(19,23)=1$. Further by Wilson’s theorem, we obtain $22 ! \equiv-1(\bmod 23) \equiv 22($ $\bmod 23) \Rightarrow 21 ! \equiv 1(\bmod 23) \equiv 1+23=24(\bmod 23) \Rightarrow 7(20) ! \equiv 8(\bmod 23) \Rightarrow$ $7 \cdot 5 \cdot 19 ! \equiv 2 \equiv 2+23 \equiv 25(\bmod 23) \Rightarrow 7 \cdot 19 \cdot 18 ! \equiv 5(\bmod 23) \equiv 5+23=28($ $\bmod 23) \Rightarrow 19 \cdot 18 ! \equiv 4(\bmod 23) \Rightarrow 19 \cdot 18 ! \equiv(4-23)=-19(\bmod 23) \Rightarrow 18 ! \equiv$ $-1(\bmod 23)$. Therefore $23|(18 !+1) \Rightarrow 437|(18 !+1)$.

Problem 5.5.4. Prove that for $n(>1)$ is prime if and only if $(n-2) ! \equiv 1($ $\bmod n)$.

Solution 5.5.4. By Wilson’s theorem and it’s converse we have, $n$ is prime if and only if $(n-1) ! \equiv-1(\bmod n)$. Hence $(n-1) ! \equiv-1+n=n-1(\bmod n)$. Therefore $(n-2) ! \equiv 1(\bmod n)$, as $\operatorname{gcd}(n, n-1)=1$.

# 数论代考

## 数学代写|数论作业代写number theory代考|Wilson’s Theorem

$1 \times 2 \times 3 \times 4 \times \cdots \times n$. 例如， 7 分 $(7-1) !+1=6 !+1=721$. 这个猜想首先由英国数学家喛德 华.沃林在 Meditationes Algebraicae（1770 年的”代数思想”) 中发表，在那里他向英国数学家约翰.威尔 逊渵述了它。

$p=11$. 除以整数 $2,3,4,5,6,7,8,9$ 进入 $\frac{p-3}{2}$ 对如
$2 \cdot 6 \equiv 1(\bmod 11) \quad 3 \cdot 4 \equiv 1(\bmod 11) 7 \cdot 8 \equiv 1(\bmod 11) \quad 5 \cdot 9 \equiv 1(\bmod 11)$

## 数学代写|数论作业代写number theory代考|Worked out Exercises

$\bmod 23) \Rightarrow 21 ! \equiv 1(\bmod 23) \equiv 1+23=24(\bmod 23) \Rightarrow 7(20) ! \equiv 8(\bmod 23) \Rightarrow$ $7 \cdot 5 \cdot 19 ! \equiv 2 \equiv 2+23 \equiv 25(\bmod 23) \Rightarrow 7 \cdot 19 \cdot 18 ! \equiv 5(\bmod 23) \equiv 5+23=28($ $\bmod 23) \Rightarrow 19 \cdot 18 ! \equiv 4(\bmod 23) \Rightarrow 19 \cdot 18 ! \equiv(4-23)=-19(\bmod 23) \Rightarrow 18 ! \equiv$ $-1(\bmod 23)$. 所以 $23|(18 !+1) \Rightarrow 437|(18 !+1)$.

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