Prove that primes of the form $4k+1$ does divide $m=4q_1q_2\cdots q_n4$? I saw similar questions related to this about the primes of the form $4k+3, 4k+1, \cdots$, but still I had some issues to ask.
To prove that there are infinitely many primes of the form $4k+3$, where $k$ is nonnegative integers, we can do that with proof by contradiction.
By Contradiction:
Let $m = 4q_1q_2\cdots q_n -1$, $m$ is of the form $4k+3$ because it's equivalent to $4k-1$. If we assume that primes $q_1, \cdots, q_n$ are primes of the form $4k+3$, which is equivalent to $4k-1$.
Let $m = 4q_1q_2\cdots q_n -1$, but this number is not of the form $4k+3$ as it's larger than all primes $q_1, \cdots, q_n$ of form $4k+3$. But since $m$ is not prime as it's not one of the primes of form $4k+3$, then we conclude that it's a multiple of primes.
None of the $q_i's$ of the form $4k+3$ divides $m$ as none divides $-1$, so we can conclude that $m$ needs to be the product of prime numbers of the form $4k+1$.
We have a contradiction that $m$ is of two forms $4k+1, 4k+3$.
Question: Why we concluded that since none of the $q_i's$ of the form $4k+3$ divides $m$ as none divides $-1$, so we can conclude that $m$ needs to be the product of prime numbers of the form $4k+1$? Why this form and not any other form specifically?
 A: Rewriting this answer to make it constructive, as per @BillDubuque.
Let $S$ be a finite set of primes of the form $4k+3$. The following argument shows there is another such prime, not in $S$. It  follows that there must be infinitely many such primes.
The product of two numbers of the form $4k+1$ is again of that form (easy to check).
There are only two kinds of odd primes when you work modulo $4$: some of the form $4k+1$, others of the form $4k+3$, since the numbers of the form $4k$ and $4k+2$ are even.
Let $Z$ be the product of all the primes in $S$ and let $m = 4Z+3$. Since when you divide $m$ by $4$ the remainder is $3$,  $m$ cannot be the product only of primes of the form $4k+1$. It must have at least one prime factor  of the form $4k+3$.
Since none of the  primes in $S$ divides $m$ you have found another prime of that form.
This kind of argument does not work to show that there are infinitely many primes of the form $4k+1$. That's true, but you have to work harder to show it.
A: Let's sketch the key ideas of the proof in constructive (vs. contradiction) form. This yields a simple intuitive algorithm for generating primes of the form $4n-1$ - just like Euclid's classical algorithm.
All $\,q_i>1\Rightarrow m = 4q_1q_2\cdots q_n -1>1\,$ thus $\,m\,$ has at least one prime factor.
$\!\!\bmod 4\!:\ m\equiv -1\Rightarrow\,$ some prime factor $\,q\,$ of $\,m\,$ is $\equiv -1\,$ (else by $\,m\,$ odd all prime factors of $\,m\,$ are $\equiv 1\,$ so their product $m\equiv 1,\,$ contra $\,m\equiv -1)$
$q$ is new $\,q\neq q_i$ else $\,q\mid\color{#0a0}{q_i},\ q\mid 4q_1\!\cdot \cdot\, \color{#0a0}{q_i}\!\cdot \cdot\,  q_n-\color{#c00}{\bf 1}\Rightarrow q\mid \color{#c00}{\bf 1}\,$ (i.e. $\,m\,$ is coprime to all $\,q_k)$.  Hence given any finite list of primes $\equiv -1\pmod{\!4}\,$ we can construct a new one, so the number of such primes is not finite,  i.e. there are  infinitely many.
Remark $ $ The key idea is that the sequence $\ m_1 = 3,\,\ \color{#94f}{m_{n}} = \,\color{#c00}{\bf 1} + 4m_1\cdot\cdot\,\color{#0a0}{m_k}\cdot\cdot\, m_{n-1}\,$ is an infinite sequence of coprimes, i.e. $\,{\rm gcd}(m_n,m_k) = 1\,$ when $\,n> k,\,$ because it follows that every common divisor of $\,\color{#94f}{m_n},\,\color{#0a0}{m_k}\,$ must further divide $\ \color{#c00}{\bf 1} = \color{#94f}{m_n} - 4m_1\cdot\cdot\, \color{#0a0}{m_k}\cdot\cdot\, m_{n-1}.\,$ Further if we choose prime factors $\,q_i\,$ of each $\,m_i\,$ then they too remain pairwise coprime (thus unequal) and, further, as above, we can choose $\,q_i\equiv -1\pmod{\!4}$.
This constructive proof is the natural way to proceed (and is the way Euclid proceeded). There is no need to rephrase it as a "trivial" contradiction, i.e. assume it false then prove it true (and doing so not only destroys its algorithmic content but also leads to much confusion for readers not adept at such, e.g. the common false conclusion that the new number must be prime, which is true only in the (contradictory) hypothetical universe of "the naturals with only finitely many primes").
The above constructive proof yields the following simple  algorithm to generate such primes.
Starting with the empty list of primes with product $= \color{#c00}{\bf 1},$ we construct the new prime
$ 4(\ \ )-1 = 4(\color{#c00}{\bf 1})-1 = 3  =: q_1.\ $ Repeating with the singleton list $\, q_1\,$ leads to
$4(q_1)\!-\!1 = 4(3)\!-\!1 = 11 =: q_2.\ $  Repeating with the list $\, q_1,q_2\,$ leads to
$4(q_1q_2 )-1 = 4(3\cdot 11)-1 =  131 =: q_3.\ $
They remain prime $\ 3, 11, 131, 17291, 298995971 \ $ till we reach the sixth element
$$ m = 89398590973228811 = 8779\cdot 10079\cdot 1010341471\qquad$$
where we need to choose a (guaranteed) factor $\,\not \equiv 1\pmod{4},\,$ e.g. the least $= 8779$.
You can find further terms in OEIS sequence A057205
Using $\,4n\!+\!3\,$ vs. $4n-1$ yields $\, 7,31,13,11287,67,\ldots$ (choosing least prime factors)
Generalization $ $ See this answer for a more gweneral result which includes both this result and also Euclid's classical proof as special cases.
A: You should start by saying "assume that there are only finitely many primes of the form $4k+3$.  Call them $q_1,q_2,\ldots q_n$" so we know what we are talking about.  Your second "Let $m=\dots$" is not correct.  $m$ is of the form $4k+3$ by construction.  Because it is greater than all the primes in your list, it cannot be prime.  We conclude that none of the primes of the form $4k+3$ divide $m$ because $m$ is one less than a multiple of them.  $2$ also does not divide $m$ because $m$ is odd.  All the other primes are of the form $4k+1$.
