TITLE: Help understand the proof of infinitely many primes of the form $4n+3$
QUESTION [17 upvotes]: This is the proof from the book:

Theorem. There are infinitely many primes of the form $4n+3$. 

Lemma. If $a$ and $b$ are integers, both of the form $4n + 1$, then the product $ab$ is also in this form.
Proof of Theorem:
Let assume that there are only a finite number of primes of the form $4n + 3$, say 
$$p_0, p_1, p_2, \ldots, p_r.$$
Let $$Q = 4p_1p_2p_3\cdots p_r + 3.$$
Then there is at least one prime in the factorization of $Q$ of the form $4n + 3$. Otherwise, all of these primes would be of the form $4n + 1$, and by the Lemma above, this would imply that $Q$ would also be of this form, which is a contradiction. However, none of the prime $p_0, p_1,\ldots, p_n$ divides $Q$. The prime $3$ does not divide $Q$, for if $3|Q$  then $$3|(Q-3) = 4p_1p_2p_3\cdots p_r,$$ which is a contradiction. Likewise, none of the primes $p_j$ can divides $Q$, because $p_j | Q$ implies $p_j | ( Q - 4p_1p_2\cdots p_r ) = 3$, which is absurd. Hence, there are infinitely many primes of the form $4n +3$.   END
From "however, none of the prime ...." to the end, I totally lost!  
My questions:

Is the author assuming $Q$ is prime or is not?
Why none of the primes $p_0, p_1,\ldots, p_r$ divide $Q$? Based on what argument?  

Can anyone share me a better proof?   
Thanks.

REPLY [3 votes]: I have seen many proofs of this here, but I think a $basic$ proof is still lacking. I attempt one below:
(1) First note that $any$ integer may be written in the form $4k, 4k+1, 4k+2$, and $4k+3$. This is the result of the $Division$ $Algorithm$.
(2) $Any$ number is either a prime or a product of primes - Fundamental Theorem of Arithmetic (FTA)
(3) Also note this $lemma$: a product of two or more integers of the form $4n+1$ is also of the same form. To show this is simple:
Take two numbers of form $4n+1$, say ${N}_1=4m+1$ and ${N}_2=4m'+1$. Straight multiplication gives $16mm'+4(m+m')+1$ = $4[4mm'+(m+m')]+1$ = $4k+1$, where $k=4mm'+(m+m')$, i.e the product is the same $form$ as its multiplicands.
Now we have all we need. As already mentioned in many answers here, we use a modified Euclidian proof of the infinitude of primes, which is a proof by contradiction (It is a good idea to familiarize oneself with the said Euclidian proof before proceeding)
In anticipation of a contradiction, we assume there are only a finite number of primes of the form $4k+3$. We list them as follows ${q}_1,{q}_2,...,{q}_s$. Similar to Euclidian Proof, we form a positive integer $N=4{q}_1{q}_2....{q}_s-1$.
Note the following about $N$:
(1) It is odd (because $odd*even$=$even$ and $even-1$ = $odd$)
(2) $N$ may be written as $N=4({q}_1{q}_2....{q}_s-1)+3$, i.e. $N$ is of the form $4k+3$
(3) Per FTA, we may write $N$ as a product of primes, say $N={r}_1{r}_2...{r}_t$. But remember all these primes must be odd (i.e. ${r}_i=2$ is excluded).
(4) And the only form that $every$ ${r}_i$ may take is either $4k+1$ or $4k+3$.
(5) This is a $crucial$ point: $N$ $cannot$ contain only primes of the form $4k+1$. If this were the case, by the lemma above, $N$ would be of the form $4n+1$, which is clearly not the case - $N$ is of the form $4n+3$. Therefore, we conclude that $N$ must contain at least $one$ factor of the form $4k+3$.
(6) Lastly, we show that if $N$ had a factor of the form $4k+3$, it leads to the anticipated contradiction:
Let's refer to this prime factor of $N$ as ${q}_i$. Since all the prime factors of the form $4k+3$ are limited to the list ${q}_1,{q}_2,...,{q}_s$ by assumption, ${q}_i$ must belong in this list. But this implies that ${q}_i$ divides $N$:
${q}_i$|$N$ = ${q}_i$|$4{q}_1{q}_2....{q}_s-1$
$\rightarrow$ ${q}_i$|$1$
And we have arrived at a contradiction. Therefore we conclude there exist infinitely many primes of the form $4k+3$