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.
There is a minor issue. Not every prime divisor of your $n$ is necessarily of the form $6k+1$. (Because, as you noted, the theorem with the Légendre-symbol is only valid for $p\geqslant5$.) You would have to argue that $2$ and $3$ do not divide $n$.
This can be fixed by letting $n=(2p_1,\cdots,p_k)^2+3$ as you noted, or alternatively by observing that $x^2\equiv1\pmod8$ for odd $x$, which means your $n$ cannot be a power of $2$, hence has an odd prime divisor.