TITLE: Proof that there are infinitely many primes of the form $6k+1$. Proof verification
QUESTION [7 upvotes]: Theorem. there are infinitely many primes of the form $6k+1$.

I've just proved that there are infinitely many primes of the form $6k+1$.
Could you please check my proof?
At first, I proved that

$$ for \ \  p:prime, \ p \ge 5 \\\ \\ \left(\frac{-3}{p}\right)= \begin{cases} 1,&
 \ p \equiv 1 \pmod  6 \\ 
-1, &  \ p \equiv 5 \pmod  6 \end{cases} $$
  (I will use this lemma for proving Theorem.)

$$$$
NOW assume that there are finite primes of the form $6k+1$.
then we can say $ \ p_1, \ p_2, \cdots, p_k
\ $ :  all the  primes  of the form  $6k+1$.
Let 

$$ n=(p_1\cdot p_2\cdots\ p_k)^2  +3, $$

then (by Fundamental Theorem of Arithmetic) there is prime factor $p$ of $n$.
$$
$$
Id est, $$(p_1\cdot p_2\cdots\ p_k)^2  \equiv -3  \pmod  p  $$  
So, $$p \equiv 1 \pmod  6$$
Thus $$p=p_i \ for \ some \ i=1, \cdots , k$$
This time
$$p=p_i \  \ \ divides  \ \   (p_1\cdot p_2\cdots\ p_k)^2
\\
p=p_i \  \ \ can't \ \ divide  \ \   3
\\
$$
So, $$p=p_i  \ \ can't \ \  divide  \ \ n. \\$$
It is contradiction with "$p$ is prime factor of $n$"
$\therefore \ $ There are infinitely many primes of the form $6k+1$.
$Q.E.D.$
$$
$$
What about my proof? 
After proving, I saw someone's proof, 
BUT he set $$ n=(2p_1\cdot p_2\cdots\ p_k)^2  +3 $$
I don't know why he set $ n=(2p_1\cdot p_2\cdots\ p_k)^2  +3 $, instead of $ n=(p_1\cdot p_2\cdots\ p_k)^2  +3 $.
Is my proof wrong?
$$
$$
Please give me some hand.
Thanks in advance.

REPLY [2 votes]: If we cross out from sequence of positive integers all numbers divisible by $2$ and all numbers divisible by $3$, then all remaining numbers will be in one of two forms:
$S1(n)=6n−1=5,11,17,...$ or $S2(n)=6n+1=7,13,19,....n=1,2,3,...$ So all prime numbers also will be in one of these two forms and ratio 0f number of primes in the sequence $S1(n)$ to number of primes in the sequence $S2(n)$ tends to be $1$. see [link]