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I have to proof if true or wrong: There are infinite prime numbers of the form 4k+3.

I want to proof: Yes, this is true. My ideas:

1) Assume - as a contradiction - that there are only infinite prime numbers. Let $p_k$ be the highest of them of the form 4k+3. Let $p_1 = 3, p_2 = 5, ...$ the sequence of prime numbers.

So define $N := 4*(3*5*...*p_k) - 1$. Then we have

a) N divided by 4 has remainder 3 and 3 divided by 4 has remainder 3, because: $ N \equiv -1 \equiv 3 (mod 4)$.

b) Then I'd like to show that N does have a prime-factor of the form 4k+3 which is higher than $p_k$ << here I don't know how to show this.

I'd appreciate any comments on whether my ideas are correct or not.

Vazrael
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  • http://math.stackexchange.com/questions/21333/help-understand-the-proof-of-infinitely-many-primes-of-the-form-4n3 – lab bhattacharjee May 06 '14 at 15:21
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    You are already there. the prime factors of the odd number $N$ cannot all be congruent to $1$ modulo $4$, otherwise $N$ itself would be. Thus $N$ has a prime factor congruent to $3$ modulo $4$. And $N$ is not divisible by any of the primes up to $p_{k}$. – Andreas Caranti May 06 '14 at 15:32

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