Proof of infinitely many primes of the form 4k-1 I've found many answers to this question but I haven't been able to complete the proof.
This is what I have so far:
Let as suppose that there exists a finite number of primes of the form $4k-1$. Let $\{p_1, p_2, ..., p_n\}$ be the set of all of them.
I consider the product of all of them: $p_1p_2...p_n = (4k_1-1)(4k_2-1)...(4k_n-1)$. This product is of the form $4k\pm1$.
-If $p_1p_2...p_n$ is of the form $4k+1$, I multiply it by $p_1$ and get $(4k_1-1)(4k+1) = 4^2kk_1+4(k_1-k)-1 = 4k'-1$.
-If $p_1p_2...p_n$ is of the form $4k-1$, I multiply it by $p_1p_2$ and get $(4k_1-1)(4k_2-1)(4k-1) = 4^3kk_1k_2+4^2(kk_1 + kk_2 + k_1k_2)-4(k+k_1+k_2)-1 = 4k''-1$.
How am I doing? Do $(4k'-1)$ and $(4k''-1)$ have a divisor of the form $(4r-1)$? If so, how can I prove it?
Thanks! Any help is kindly appreciated.
 A: Using a polynomial with a smaller coefficient than that in the comments:
Let $\Pi$ be the product of any finite set of $4k-1$ primes. Since $\Pi$ is odd, $2\Pi+1\equiv-1\bmod4$. This forces $2\Pi+1$ to have a $4k-1$ prime factor but $2\Pi+1$ cannot have any factors in the set used to construct $\Pi$.
A: Thanks for all the suggestions, you are all very kind. I'm answering with the proof following @BarryCipra suggestion.
Let as suppose that there exists a finite number of primes of the form $4k−1$. Let $\{p_1,p_2,...,p_n\}$ be the set of all of them. $4p_1p_2...p_n-1$ is of the same form (that's trivial), but not divisible by any of the $p_i$'s.
I consider now the prime factorization of $4p_1p_2...p_n-1$. As this is an odd number, all of its prime factors are odd, so non of them is of the form $4k$ or of the form $4k+2$, so they are of the form $4k\pm1$. If all of them were of the form $4k+1$, then $4p_1p_2...p_n-1$ would be of the same form, but it isn't, so at least one of them has to be of the form $4k-1$, and it's not in the set $\{p_1,p_2,...,p_n\}$, which is a contradiction.
