# Proving that there are infinitely many primes with remainder of 2 when divided by 3

I need to prove that there are infinitely many primes with remainder of 2 when divided by 3. I started out similarly to Euclid's classic proof of an infinite number of prime numbers:

Suppose there is only a finite set of prime numbers with remainder of 2 when divided by 3, then we can write their product as:

$$P = q_1 \cdot q_2 \cdots q_r, \qquad \text {for some integer } r, =$$ $$(3q_1+2)\cdot(3q_2+2)\cdots(3q_r+2),$$ for integers $q_r$.

This is where I am stuck. I do not know how to get to a similar contradiction as Euclid did when he considered $P$+$1$ and how the $q_i$'s could not divide $P$+$1$ since they divided $P$. (If they divided $P$+$1$ then they would divide $P$ and $1$, where dividing $1$ is the contradiction). Any ideas on how I can get to a similar contradiction?

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Use Dirichlet's theorem on primes in Arithmetic progressions. Well it is an overkill but you don't have to do anything.en.wikipedia.org/wiki/Primes_in_arithmetic_progression –  s.b Sep 13 '12 at 6:26
This approach is not analogous to Euclid's classic proof. Euclid did not assume that there is only a finite set of prime numbers. –  joriki Sep 13 '12 at 9:02
"It is often erroneously reported that Euclid proved this result by contradiction, beginning with the assumption that the set initially considered contains all prime numbers, or that it contains precisely the n smallest primes, rather than any arbitrary finite set of primes. Although the proof as a whole is not by contradiction, in that it does not begin by assuming that only finitely many primes exist, there is a proof by contradiction within it: that is the proof that none of the initially considered primes can divide the number called q above." –  Pedro Tamaroff Sep 13 '12 at 22:25

Let $q_1,q_2,\dots,q_n$ be odd primes of the form $3k+2$. Consider the number $N$, where $$N=3q_1q_2\dots q_n+2.$$ It is clear that none of the $q_i$ divides $N$, and that $3$ does not divide $N$.

Since $N$ is odd and greater than $1$, it is a product of one or more odd primes. We will show that at least one of these primes is of the form $3k+2$.

The prime divisors of $N$ cannot be all of the shape $3k+1$. For the product of any number of (not necessarily distinct) primes of the form $3k+1$ is itself of the form $3k+1$. But $N$ is not of the form $3k+1$. So some prime $p$ of the form $3k+2$ divides $N$. We already saw that $p$ cannot be one of $q_,\dots,q_n$. It follows that given any collection $\{q_1,\dots,q_n\}$ of primes of the form $3k+2$, there is a prime $p$ of the same form which is not in the collection. Thus the number of primes of the form $3k+2$ cannot be finite.

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See Euclidean Proofs Of Dirichlet’s Theorem for a nice discussion of when these variants of Euclid's argument work. –  lhf Sep 13 '12 at 4:08
The simplest ones are $4k+3$ and $3k+2$. After one has some information about quadratic congruences, there are a few other accessible ones. –  André Nicolas Sep 13 '12 at 4:08
Could you explain more about the 3k+1 part? Aren't all the primes of the form $3q+2$ since the primes must have a remainder of 2 when divided by $3$? –  CodeKingPlusPlus Sep 13 '12 at 4:35
Note that this proof, unlike the one in the question, is analogous to Euclid's proof, since it infers the existence of a further prime without assuming that the given primes are all the primes. –  joriki Sep 13 '12 at 9:09
One nice advantage of using $3P-1$ rather than $3P+2$ is that the prime $2$ isn't a special case and can be treated like any other $3k+2$ prime. –  Erick Wong Sep 13 '12 at 21:42

$3P - 1$ is either a prime itself or has at least one prime factor $q \equiv 2 \pmod 3.$ The reason for this is that the product of any number of primes that are $\equiv 1 \pmod 3$ is again $\equiv 1 \pmod 3.$ So, since $3P - 1 \equiv 2 \pmod 3,$ we know there is some $q$ as described.

Note that $$\gcd(3P-1,P) = 1,$$ so the new $q$ cannot divide $P,$ and therefore is not equal to any of your original list of primes.

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This doesn't answer the question.. (at least I don't see how) –  picakhu Sep 13 '12 at 14:34
@picakhu Hint: it's a big hint for the inductive step. For details see Andre's answer, which uses $\rm\:3P+2\:$ vs. $\rm\:3P-1.\:$ The point is that its both coprime to the other $\rm\:p_i\:$ since they divide $\rm\,P,\,$ and, being $\equiv -1\pmod 3\:$ it must have at least one prime factor $\equiv -1\pmod 3\$ (else all prime factors are $\equiv 1\:$ so their product $\rm\:3P-1 \equiv 1\pmod 3,\:$ contradiction). $\$ –  Bill Dubuque Sep 13 '12 at 14:53
@Bill Thanks! I hope it helps others too. –  picakhu Sep 13 '12 at 15:00

Let $N > 2$ be an integer. Early on in Chapter 10 of these notes, I explain how Euclid's proof of the infinitude of primes can be very slightly modified to prove the following generalization:

There are infinitely many prime numbers $p$ such that $p$ is not of the form $kN+1$.

When $N = 3$, this means that there are infinitely primes which are either of the form $3k$ or $3k+2$. Since there is only one prime of the form $3k$, this answers your question.

(The notes go on to mention that essentially the same argument proves that for any proper subgroup $H$ of $(\mathbb{Z}/N\mathbb{Z})^{\times}$, there are infinitely many primes $p$ such that the reduction of $p$ modulo $N$ does not lie in $H$.)

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+1, esp. for the notes! –  Bill Dubuque Sep 17 '12 at 14:16
Break it into two cases depending on the parity of $r$. Start by showing that if $r$ is even, $P$ has a remainder of $1$ when divided by $3$, while if $r$ is odd, it has a remainder of $2$. In the first case show that $P+1$ must have a prime factor of the right form; in the second case, consider $P+3$ instead.