Infinitely number of primes in the form $4n+1$ proof Question: Are there infinitely many primes of the form $4n+3$ and $4n+1$?

My attempt: Suppose the contrary that there exist finitely many primes of the form $4n+3$, say $k+1$ of them: $3,p_1,p_2,....,p_k$

Consider $N = 4p_1p_2p_3...p_k+3$, $N$ cannot be a prime of this form. So suppose that $N$=$q_1...q_r$, where $q_i∈P$

Claim: At least one of the $q_i$'s is of the form $4n+3$:

Proof for my claim: $N$ is odd $\Rightarrow q_1,...,q_r$ are odd $\Rightarrow q_i \equiv 1\ (\text{mod }4)$ or $q_i ≡ 3\ (\text{mod }4)$

If all $q_1,...q_r$ are of the form $4n+1$, then $(4n+1)(4m+1)=16nm+4n+4m+1 = 4(\cdots) +1$

Therefore, $N=q_1...q_r = 4m+1$. But $N=4p_1..p_k+3$, i.e. $N≡3\ (\text{mod }4)$, $N$ is congruent to $1\ \text{mod }4$ which is a contradiction.

Therefore, at least one of $q_i \equiv 3\ (\text{mod }4)$. Suppose $q_j\equiv 3\ (\text{mod }4)$

$\Rightarrow$ $q_j=p_i$ for some $1\leq i \leq k$ or $q_j =3$

If $q_j=p_i≠3$ then $q_j$ | $N = 4p_1...p_k + 3 \Rightarrow q_j=3$ Contradiction!

If $q_j=3$ ($\neq p_i$, $1\leq i \leq k$) then $q_j | N = 4 p_1...p_k + 3 \Rightarrow q_j=p_t$ for some $1 \leq i \leq k$ Contradiction!

In fact, there must be also infinitely many primes of the form $4n+1$ (according to my search), but the above method does not work for its proof. I could not understand why it does not work. Could you please show me?

Regards
Suppose $n&gt;1$ is an integer.  We define $N=(n!)^2 +1$. Suppose $p$
is the smallest prime divisor of $N$. Since $N$ is odd, $p$ cannot
be equal to $2$. It is clear that $p$ is bigger than $n$
(otherwise $p \mid 1$).  If we  show that $p$ is of the form
$4k+1$ then we can repeat the procedure replacing  $n$ with  $p$
and we produce an infinite sequence of primes of the form $4k+1$.

We know that $p$ has the form $4k+1$ or $4k+3$.    Since $p\mid N$
we have $$ (n!)^2 \equiv -1 \ \ (p)  \ . $$   Therefore $$
(n!)^{p-1} \equiv (-1)^{ \frac{p-1}{2} } \ \ (p) \ . $$
Using Fermat's little Theorem  we get  $$ (-1)^{ \frac{p-1}{2} } \equiv 1 \ \ (p) \ . $$
If $p$ was of the form $4k+3$ then $\frac{p-1}{2} =2k+1$ is odd
and therefore we obtain $-1 \equiv 1 \ \ (p)$ or $p \mid 2$ which
is a contradiction since $p$ is odd.
Actually there is an even more elegant proof that shows that this holds true for all primes>2:

∀ primes,p ∈P &amp; p>2,∃ k∈Q| k=(p-1)/4  or k= (p+1)/4

Proof:
By definition every 4th integer, k ≥ 4, must have 2^2 (2 squared) as a factor.  This means that every other even integer > 2 must have 2^2 as a factor.

For all primes p>2, the prime itself is odd and therefore p-1 and p+1 are both even.  

Since p-1 and p+1 represent consecutive even integers, one and only one of them must have 2^2 as factors and thus (p±1)/4 is an integer.  ∎

This applies to all primes greater than 2.