Proving an infinite number of primes of the form 6n+1

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

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 $k^2+(k+1)^2$

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.

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.