guided proof that there are infinitely many primes on the arithmetic progression $4n + 3$

Question

The exercise is to prove this proposition using the following lemmas:

1) Prove Lemma A: every prime ≠ 2 is either of the form $4n + 1$ or $4n + 3$

2) Prove Lemma B: the product of numbers of the form $4n + 1$ is of the form $4n + 1$.

I'll omit the proofs of the lemmas, since I'm not doubtful about their demonstrations. Again I ask if there is any flaw on the following proof.

Proof:

Suppose, by contradiction, that there are finitely many primes of the form $4n + 3$, say, $p_1, p_2, ..., p_k.$ Define $N = 4p_1...p_k - 1 = 4(p_1...p_k - 1) + 3.$ $N$ > $p_k$ and is of the form $4n + 3$, therefore is not prime. So $N$ must be divisible by some prime. Since $N$ is not divisible by any $p_i$ or 2, this prime divisor must be > $p_k$ and consequently, due to lemma A, of the form $4n + 1$. Now, the fundamental theorem of arithmetic states that any non-prime number must be factorable in prime numbers. Since $N$ is not divisible by any $p_i$, there must be primes $f_1, ..., f_l$ on which $N$ is factorable. From the hypothesis and from lemma A follows that these primes must be of the form $4n + 1$ and, by lemma B, so must be $N$, a contradiction.

Answer 1

Your proof is good, however consider the following(slightly shorter) version:

Suppose there are finitely many primes of the form $4n + 3$, call them $p_1, p_2, \dots p_k$. Then consider $$N = 4(p_1p_2\dots p_k) - 1$$ This is still of the form $4n + 3$ since it is congruent to $-1 \equiv 3(\mod 4)$. Note also that none of the $p_i$ divide $N$. Thus, $N$ is either prime or composite. If $N$ is prime, we reach a contradiction since it is a new prime of the desired form. Suppose $N$ is composite, then there must be a prime of the form $4n+3$ dividing it by your lemma, also a contradiction since none of the $p_i$ divide $N$.

This proof is nearly identical to yours, however the thing to note is that you can just leave it in the form $4k - 1$ since this is equivalent to $4(k-1) + 3$.

Answer 2

The proof is sound, depending on the course you may be asked (at the very least should be readily able to) justify the statement

"N is not divisible by any $p_i$, or 2.

Otherwise looks good.