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I would like to know if my approach is valid. I want to show that there are infinitely many primes of the form $4m+1$.

Assume the contrary, and denote each of these primes by $p_{i}$ with $i \in \{1, 2, \ldots, k\}$ for some $k$. Let $P$ be the product of all of them and note that $P \equiv 1\mod 4$. Let $Q$ be some product of primes only of the form $4m+3$. Then $P+Q$ cannot have prime divisor of the form $4m+1$, else it is some $p_{i}$ making $Q = P-mp_{i}$ divisible by $p_{i}$. Then $P+Q$ is a product of primes only of the form $4m+3$. Hence, $P+Q$ is either $1$ or $3$ modulo $4$ (since this number is congruent to $3^{r} \equiv (-1)^{r} \mod 4$, with $r$ being the number of prime factors). However, $P$ is $1$ modulo $4$ and $Q$ is either $1$ or $-1$ modulo $4$, making the sum $P+Q$ either $0$ or $2$ modulo $4$. Contradiction.

I believe this is valid, but wanted a second opinion. Looking this problem up online, there seems to be a common "trick" to define a number $N = 4P^{2}+1$, where $P$ is defined as above. I understand that method, but want to know if the one above works as well. Thank you.

Freddie
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  • all you know is that $P+Q$ is even : it is the sum of two odd numbers, and it doesn't have any $p_i \equiv 1 \bmod 4$ as a prime divisor since $P +Q \equiv Q \not\equiv 0 \bmod p_i$, – reuns Jun 19 '16 at 00:12
  • $P+Q$ is obviously even. It is true that under our assumptions all odd prime divisors of $P+Q$ are of the form $4k+3$, but that does not lead to a contradiction. – André Nicolas Jun 19 '16 at 00:12
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    But the question stays interesting to me : is there any elementary argument for showing there are infinitely many primes $\equiv 1 \bmod 4$, or infinitely many primes $\equiv 3 \bmod 4$, without using the Dirichlet L-function ? – reuns Jun 19 '16 at 00:15
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    I believe you can use $N$ as defined above in the last paragraph. Then $N$ mustn't have any prime $p \equiv 1 \mod 4$ as a divisor. Then $N$ only has prime divisors that are $3$ modulo $4$. Note that $(2P)^{2} \equiv -1 \mod q$, where $q$ is one of these primes. This can only happen if $q$ is $1 \mod 4$. Contradiction. The proof of this last fact is a little more complicated, but is easily found online. – Freddie Jun 19 '16 at 00:21
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    @user1952009 Define $P$ as above. Then, $4P^2+1$ is an odd integer, so , it is not divisible by $2$. Moreover, as $-1$ is not a quadratic residue modulo primes of the form $4k+3$, it needs to be divided by a prime number of the form $4k+1$. – Emre Jun 19 '16 at 00:21
  • @Emre : Yes, I found those 2 elementary proofs. for the primes $\equiv 1 \bmod 4$ : http://math.stackexchange.com/questions/244915/infinitely-number-of-primes-in-the-form-4n1-proof using $4(\prod_{p \equiv 1 \bmod 4} p_i)^2+1)$, and for the $\equiv 3 \bmod 4$ http://math.stackexchange.com/questions/21333/help-understand-the-proof-of-infinitely-many-primes-of-the-form-4n3 using $4 (\prod p )+3$. The argument should work for $\equiv 1 \bmod q$ (by considering $q (\prod_p )^{\phi(q)}+1$ ?) but not for $\equiv a \bmod q, a \ne 0$ – reuns Jun 19 '16 at 00:24
  • @user1952009 Yes, as I know, this argument, or quadratic residues in general, are not capable of proving the result in every case. – Emre Jun 19 '16 at 00:41

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It does not work, beacuse, you are ignoring the fact that $2$ might be a divisor of $P+Q$, which is not $\equiv 3\pmod4$

Emre
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