Proving an infinite number of primes of the form 6n+1 The proofs given on other sites weren't that clear and used different methods that I have yet to learn.

Prove that there are an infinite number of primes of the form 6n+1.
The hint that was given was:
Let p = p1, p2, ..., pk + 1, where p1 = 2, p2 = 3,...pk are the first k primes.  Show that p is prime.
(p1 means p sub 1, p2 means p sub 2, and pk is p sub k.  Wasn't sure how to write it on this.)

Can someone explain this hint on how they came about of p1 = 2, p2=3, etc, and prove this please?  
Also, how would the proof change if the form changed?  ("Prove that there are infinite number of primes of the form....")



 A: Hint is clearly false. Let me answer a question that was not asked as it used the hint that is provided. I realize that this does not help OP but hopefully it will help the OP in some other problem.
Show that there are infinitely many primes of the form $6n-1$
Suppose not. Let there be only finitely many primes, say $p_1$, $p_2$ $\cdots$, $p_k$. Let 
$$
P = 6 p_1 p_2 p_3 \cdots p_k - 1
$$
Now every prime is either of the form $6n-1$ or $6n+1$ and product of any two numbers of the form $6n+1$ is also of the form $6n+1$. So the question is 
What are the prime dividers of $P$?
They all can't be of the form $6n+1$ since $P$ is of the form $6n -1$. So it must have at least one prime factor of the form $6n-1$. Clearly $p$ is not divisible by any of the primes $p_1$, $p_2$, $\cdots$ $p_k$. So there has to be a prime of the form $6n-1$ which is different from these primes. Hence, there has to be infinitely many primes of the form $6n-1$.
I fully realize that this does not answer OP's question but the method and the hint is similar. Hope this helps
A: Make a list of (supposedly) all the primes, then write the product
$$  Q = p_1 p_2 p_3 \cdots p_k, $$
and write
$$  N = 12 Q^2 + 1. $$
You need to know how to prove this much: if we have a prime $q \equiv 5 \pmod 6,$ and
$$ 3 u^2 + v^2 \equiv 0 \pmod q, $$ then both
$$  u,v  \equiv 0 \pmod q. $$
Proved the general fact at
Prime divisors of $k^2+(k+1)^2$
A: This is a direct conclusion of Dirichlet's theorem on arithmetic progressions.
Or you can prove it directly, but it takes more work. Here's a proof from MATHBLAG:
Let P be a finite set of primes of the form $6k + 1$, and let N be a number that is divisible by every number in P. Assume that N is also divisible by 6. Let p be a prime divisor of $N^2-N+1$.
Note that $(N^2-N+1)(N+1)=N^3+1$. so p divides $N^3+1$, or in other words $N^3 \equiv -1 \pmod{p}$ and so $N^6 \equiv 1 \pmod{p}$.
Recall that the order of N modulo p is the least positive k so that $N^k \equiv 1 \pmod{p}$. The order must divide 6. so k = 1, 2, 3, or 6. But $N^3 \equiv -1 \pmod{p}$, so the order cannot be 1 or 3.
Can the order be 2? If $N^2 \equiv 1 \pmod{p}$ and $N^3 \equiv -1 \pmod{p}$ then $N \equiv -1 \pmod{p}$. This would be bad, because then p would divide both $N+1$ and $N^2-N+1$; but $\gcd(N+1,N^2-N+1) = \gcd(N+1,3) < p$, contradiction.
Thus N has order 6 mod p, and the group of units mod p has order $p-1$, so 6 divides $p-1$, which means that p has the form $6k+1$. Therefore, P does not contain all primes of the form $6k+1$, so the set of primes of this form is infinite.
A: let there are finite number primes in the form $6n-1$
Let they all are $S=\{p_1,p_2,p_3,p_4,p_5.....p_n\}$.
Let a number $Q=6(p_1*p_2*p_3*p_4.......p_n)-1$.
Either Q is prime or composite.
If Q is prime we are done. Since Q is in form 6n-1 and not in S.(Contradiction)
If p is composite then there must be a prime let M (not in S) which is a factor of Q. We know M must be in the form 6n-1 or 6n+1.
so Q=Mq for some q. If m is in form 6n-1 then this contradicts our supposition. If M is in the form 6n+1 then our left-hand side is congruent to 5 mod6 but the right-hand side is congruent to 1 mod 6 which is not possible, and we are done.
A: If we cross out from sequence of positive integers all numbers divisible by $2 $ and all numbers divisible by $3$ then all remaining numbers will be in one of two forms:
$S1(n)=6n−1=5,11,17,..$ or $S2(n)=6n+1=7,13,19,....n=1,2,3,...$ 
So all prime numbers also will be in one of these two forms and ratio 0f number of primes in the sequence $S1(n)$ to number of primes in the sequence $S2(n)$ tends to be $1$. See link http://www.planet-source-code.com/vb/scripts/ShowCode.asp?txtCodeId=13752&lngWId=3
