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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?

Origin — Elementary Number Theory — Jones — p28 — Exercise 2.6

Martin Sleziak
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CodeKingPlusPlus
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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.http://en.wikipedia.org/wiki/Primes_in_arithmetic_progression – s.b Sep 13 '12 at 06:26
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    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 09:02
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    "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 Sep 13 '12 at 22:25

6 Answers6

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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.

Community
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André Nicolas
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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 04:08
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    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 04:08
  • 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 04:35
  • We have produced a "big" number $N$. Some of the prime divisors of $N$ could be of the form $3k+1$, no good for us. But at least one is of the form $3k+2$. – André Nicolas Sep 13 '12 at 04:41
  • How do you know that this one divisor $3k+2$ is none of the $q_i$? I thought all of the $q_i$ were of the form $3$(some integer) + $2$ – CodeKingPlusPlus Sep 13 '12 at 04:46
  • Because $N=3q_1\cdots q_n+2$. If $p$ was one of the $q_i$, it would divide $3q_1\cdots q_n$. But $p$ divides $N$, so $p$ would divide the difference, which is $2$. That's impossible. This part of the argument is basically the same as in the ordinary "infinite number of primes" proof. – André Nicolas Sep 13 '12 at 04:53
  • One minor correction is needed: since OP does not mention "odd" anywhere, you should do so in your first sentence: you cannot allow $q_i=2$ for any $i$. – Marc van Leeuwen Sep 13 '12 at 05:19
  • @MarcvanLeeuwen: Thanks, corrected. – André Nicolas Sep 13 '12 at 05:41
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    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 09:09
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    @joriki Not that I can see that huge a difference between "assume these are all prime numbers of the given form, then arrive at a contradiction by producing a new one" and "take any finite set of primes of the given form and produce a new one" – Arthur Sep 13 '12 at 21:38
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    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
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$3P - 1$ 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.

Will Jagy
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  • This doesn't answer the question.. (at least I don't see how) – picakhu Sep 13 '12 at 14:34
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    @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
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Let $N > 2$ be an integer. Early on in Chapter 10 of these notes, on p120 as Theorem 121, 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 $H$.)

Martin Sleziak
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Pete L. Clark
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  • @Dwayne: I'll leave your edit as it is for now (thanks for taking an interest), but there is a reason I refer to my notes by Chapter and Section rather than page number: many of these notes are being edited and reuploaded as time passes, so a reference to a page number is less likely to stay correct than a reference to a Chapter. (A better solution would be for me to implement some kind of version-control...Maybe someday.) – Pete L. Clark Jan 06 '14 at 02:58
  • Thanks — no problem — please go ahead and change as it as you update. – Dwayne E. Pouiller Jan 06 '14 at 05:07
  • @Dwayne: Well, that's the problem. I have approximately 2000 pages of lecture notes on various topics that I tinker with at various times...e.g. when someone sends me an email pointing out a typo. I don't then have any easy way to go back and update the references I may have made to them on this site. – Pete L. Clark Jan 06 '14 at 08:09
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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.

Brian M. Scott
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Try squaring P and adding 1. (The square of a number of form 3n+2 is of the form 3n+1)

Syd Henderson
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An intuitive proof is simply this: Note that least prime factors are governed by combinatorial units, the smallest of which is $6$ elements.

$3, 2, n_{1}, 2, n_{2}, 2, 3$

Apart from multiples of $3$, all odd numbers are $n_{1}$, and $n_{2}$. These are potentially prime. If either one is prime, the odd composite less than it or greater than it is a multiple of $3$.

michaelmross
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