###### tags: `108上助教` `數學`
# 數學導論HW4、5
## ch3-5
#### proof
$\mbox{We prove by contradiction.}\\
\mbox{Suppose there are finitely many primes of the form }4n+3\\
\mbox{Write as } p_1, p_2, \ldots , p_N\\
\mbox{(i.e. }\ p_i = 4n+3\mbox{, } n \in \mathbb{N} \mbox{ , for } i = 1, 2, \ldots , N)$

$\mbox{Let }S = 4p_1 p_2\cdots p_N -1\\
\therefore S \mbox{ is of the form } 4n+3\\
\mbox{Since } S > 1 \mbox{ , we can get that}\\
S \mbox{ is either a prime or the product of prime numbers.}$

$\because p_i < S \mbox{ , for } i = 1, 2, \ldots N\\
S \mbox{ is not a prime.(If } S \mbox{ is a prime , then }S \mbox{ is a new prime of the form }4n+3)\\
\mbox{So we suppose }S = q_1 q_2 \cdots q_k \mbox{ , }\\
\mbox{note that } 2 \nmid S \mbox{ , so }q_i \mbox{ is a prime of the form } 4n+1 \mbox{ or }4n+3 \mbox{ , for }i = 1, 2, \ldots k$

$\because \dfrac{S}{p_i} = 4p_1 p_2 \cdots p_{i-1}p_{i+1} \cdots p_N - \dfrac{1}{p_i} \mbox{ , for }i = 1, 2, \ldots N\\
\therefore p_i \nmid S \mbox{ , for } i = 1, 2, \ldots N\\
\Rightarrow q_i \mbox{ is of the form } 4n+1 \mbox{ , for }i = 1, 2, \ldots N$

$\mbox{But let } q_i = 4s+1\mbox{ , }q_j = 4t+1 \mbox{ , }s, t \in \mathbb{N}\\
q_i q_j = (4s+1)(4t+1) = 4(4st + s + t) + 1 \mbox{ , is still the form }4n+1\\
\Rightarrow S \mbox{ is the form of } 4n+1\\
\mbox{This is a contradiction.}$

$\mbox{Hence there are infinitely many primes of the form }4n+3.\ \Box$