Prove there are infinitely many primes equivalent to $5 \pmod 6$. I know that if $n$ is equivalent to $5 \pmod 6$ then there is a prime factor of $n$ which is as well. I am having trouble converting this to a meaningful proof.
 A: Consider the product $Q$ of the first $k$ primes $\{p_i\}$ equivalent to $5 \bmod 6$, if necessary multiplying by $5$ again so that $Q\equiv 5 \bmod 6$.
Now the prime factors of $Q{+}6$ will not include any of the above primes but must include at least one other prime of the form $5 \bmod 6$. Therefore there is no finite list of such primes.
A: You can just adapt Euclid's proof that there are infinitely many primes to suit your purpose.
Take any finite list of positive primes $p$ satisfying $p \equiv 5 \pmod 6$. Since $p \equiv -1 \pmod 6$, if the list contains an even amount of primes, duplicate one of the primes in the list and multiply them all together to obtain $N \equiv 5 \pmod 6$.
The number $N + 6$ must be either prime or composite, though it is coprime to all the primes in the list. But if it's prime, it means your finite list of primes satisfying the congruence is incomplete. But if it's composite, one of its prime factors must also be of the form $6k + 5$, but it's not any of the primes in your list, which also means the list must be incomplete.
For example: $\{11, 17\}$, but we need to duplicate one of them to have an odd number of terms. Then $11^2 \times 17 + 6 = 2063$, which is prime and congruent to $5 \pmod 6$, so our list of primes of this form becomes $\{11, 17, 2063\}$. Then $11 \times 17 \times 2063  + 6 = 385787 = 29 \times 53 \times 251$. Whoa! That's three primes of the form $6k + 5$. Our list then becomes $\{11, 17, 29, 53, 251, 2063\}$, but since that's an even number of terms, we need to duplicate one of them. And so on and so forth.
A: This community wiki is to elaborate the comment from Thomas Andrews.
What you can do with a lot of these "infinitely many primes of the form $r \pmod m$" proofs is take a finite list of primes of that form, multiply them together, maybe also multiply them by $r$, add $m$ or subtract some other number, in order to obtain yet another number of the form $r \pmod m$. This other number must either itself be prime, or be a multiple of a prime of the form $r \pmod m$ which was not on the list.
The most obvious way for $5 \pmod 6$ is to take a finite list of primes of that form, multiply them together, then multiply them by 6 and subtract 1. Thus we get a number which is also $5 \pmod 6$, but it's coprime to the primes in our finite list. If it happens to be prime, bam! our list of primes of that form is shown to be incomplete. But if it's composite, and you

know that if $n$ is equivalent to $5 \pmod 6$ then there is a prime factor of $n$ which is as well

that means this composite number must be divisible by some prime of the form $5 \pmod 6$. But since this composite number is coprime to the primes in our list, its prime factor of the form $5 \pmod 6$ isn't in our list, so our finite list of primes of this form is shown to be incomplete anyway.
The only reason to use the first $k$ primes of this form as your finite list is to keep the numbers small, at first, anyway. Let's say 5 is the only prime of this form. But $6 \times 5 - 1 = 29 \equiv 5 \pmod 6$, which is prime. Then $6 \times 5 \times 29 - 1 = 869 = 11 \times 79$, of which the smaller prime factor satisfies the congruence. And so on and so forth.
A: Any prime $p>3$ is either $\equiv 1\pmod{6}$ or $\equiv 5\pmod{6}$. There are an infinite number of primes, so let us try to derive a contradiction from the assumption that just a finite number of primes are $\equiv 5\pmod{6}$. Let $\chi(n)$ be the multiplicative function that equals $1$ if $n\equiv 1\pmod{6}$, $-1$ if $n\equiv 5\pmod{6}$ and zero otherwise. We have
$$ L(\chi,1)=\sum_{n\geq 1}\frac{\chi(n)}{n}=\sum_{n\geq 0}\left(\frac{1}{(6n+1)}-\frac{1}{(6n+5)}\right)=\frac{\pi}{2\sqrt{3}}\tag{1} $$
Let $\mathcal{P}_1$ be the set of primes $\equiv 1\pmod{6}$ and $\mathcal{P}_5$ the set of primes $\equiv 5\pmod{6}$.
By Euler's product
$$ L(\chi,1) = \prod_{p\in\mathcal{P}}\left(1-\frac{\chi(p)}{p}\right)^{-1}= \prod_{p\in\mathcal{P}_1}\left(1+\frac{1}{p}\right)^{-1}\prod_{p\in\mathcal{P}_5}\left(1-\frac{1}{p}\right)^{-1} \tag{2}$$
if $\mathcal{P}_5$ were finite, the last product would be zero, since for any $s>1$ the identity $\prod_{p\in\mathcal{P}}\left(1+\frac{1}{p^s}\right)=\frac{\zeta(s)}{\zeta(2s)}$ holds. It isn't so by $(1)$, hence $\mathcal{P}_5$ is infinite.
A: The only prime congruent to 2 modulo 6 is 2 and the only prime congruent to 3 modulo 6 is 3. All other primes must either be congruent to 1 modulo 6 or 5 modulo 6. We want to prove that there are infinitely many primes congruent to 5 modulo 6.
Assume there are finitely many primes congruent to 5 modulo 6. Let $q_1, q_2, \ldots, q_s$ be the list of all primes such that $q_i \equiv 5 (\mod 6)$ for all $i$ with $1 \leq i \leq s$. Consider $N = 6q_1q_2\ldots q_s - 1$. Observe $N \equiv 5(\mod 6)$. The Fundamental Theorem of Arithmetic tells us that $N$ has a unique prime factorization. Let $p_1p_2\ldots p_t$ be the prime factorization of $N$. Observe since $2\mid 6$ and $3\mid 6$ then $2\nmid N$ and $3\nmid N$. Thus for all $j$ with $1 \leq j \leq t, \, p_j\neq 2$ and $p_j \ne 3$. So either $p_j \equiv 1(\mod 6)$ or $p_j \equiv 5(\mod 6)$. We know from the theory of congruence that if $a \equiv b( \mod n)$ and $c \equiv d(\mod n)$ then $ac \equiv bd(\mod n)$. Moreover if $a \equiv 1(\mod n)$ and $b \equiv 1(\mod n)$ then $ab \equiv 1(\mod n)$. So we cannot have $p_j \equiv 1(\mod n )$ for all j with $1 \le j \le t$. Therefore there exists $j$ with $1 \leq j \leq t$ such that $p_j \equiv 5(\mod 6)$. So $p_j = q_i$ for some $i$ with $1 \le i \le s$,
which implies that $p_j \mid 6 q_1q_2\ldots q_s$. Since $p_j \mid N$ we must have that $p_j \mid 6q_1q_2\ldots q_s - N$. Which further implies $p_j \mid 1$. This is a contradiction. Hence there is no finite list of primes congruent to 5 modulo 6. So there are infinitely many primes congruent to 5 modulo 6.          
