proof of $3k+1$ has infinite many primes by directly using the proof of there exist infinite primes Explain why you cannot directly adapt the proof that there are infinitely many primes  to show that there are infinitely many primes in the arithmetic progression 3k+1,k=1,2,...
In my opinion ,I feel that the proof that there are infinitely many primes has nothing to do with the arithmetic progression 3k+1 ...  ,but I don't know how to explain it in formal mathematics language..
 A: "nothing to do with" is a bit strong.  The problem is that if $\{p_i\}$ is the collection of known primes of the form you want then, while $P=3\prod {p_i}+1$ is indeed a number of the form $3n+1$ it does not readily follow that $P$ has to be divisible by a new prime of that form. It's clear that $P$ is not divisible by any of the old ones, true, but it could happen that $P$ is the product of evenly many primes of the form $3n+2$.  
Note that the same argument applied to primes of the form $3n+2$ works fine.   
A: "the" usual proof that there are infintely many primes (there are many proofs for this) consists in making the hypothesis that there are finitely many primes $p_1, ..., p_n$, and deriving a contradiction by showing that $p_1...p_n + 1$ must also be a prime.
Mimick this proof : suppose there are finitely many primes having the form $3k+1$ for some $k$, say $q_1, ..., q_n$. Can you directly deduce that $q_1...q_n + 1$ is also prime having the form $3k+1$ ?
A: There are infinitely many primes of the form n^2 + 1 where n=0 (mod 3)
Any prime is in the form 3k+1 or 3k+2
It is obvious that n^2 + 1 can only be in the form 3k + 1
So there are infinitely many primes in the form 3k + 1
