15

I wish to prove that there are infinitely many primes of the form $3k+1$, $k\in\mathbb{N}$.

Proving that there are infinitely many primes of the form $3k+2$ is simple using a proof analogous to that of Euclid. It goes as follows:

Suppose there are finitely many primes of the form $3k+2$ and let $S=\{p_1,p_2,...,p_j\}$ denote the finite set of all such primes. Now consider the number $n=3p_1p_2...p_j+2$.

$n$ is clearly of the form $3k+2$. If $n$ is prime, we have a contradiction. If $n$ is composite, it must be divisible by a prime number $q$ of the form $3k+2$ (One can list the possibilities to convince oneself of this). But clearly $gcd(n,p)=1$ for every $p\in S$, which implies that $q\notin S$ - a contradiction. Thus our assumption that the number of such primes is finite must be false.

The problem is that when one attempts to apply the same proof to primes of the form $3k+1$, one runs into the difficulty that $(3k+2)^2=9k^2+12k+4=3(3k^2+4k+1)+1$, and thus a composite number of the form $3k+1$ needn't necessarily be divisible by a prime of the same form.

Is there any way of adjusting the proof so that it works for primes of the form $3k+1$ or does the problem call for a completely different proof-method?

Heinrich Wagner
  • 2,897
  • 11
  • 24

2 Answers2

8

The usual proof of this uses the fact that any prime dividing a number of the form $f(n)=n^2+n+1$ has to be congruent to $0$ or $1$ modulo $3$. If $p\mid f(n)$ and $p\ne3$ then $n^3\equiv1\pmod p$ but $n\not\equiv1\pmod p$. This will contradict Lagrange's theorem applied to the multiplicative group $\Bbb F_p^\times$ of integers not divisible by $p$ modulo $p$.

Now do the usual trick of considering the prime factorisation of $f(n)$ when $n$ has a lot of prime factors you want to avoid.

Angina Seng
  • 158,341
4

The above answer may be too concise for some people to follow. We will give a longer answer here.

Suppose we have a finite list of primes $p_{1}, \ldots, p_{n}$, all of the form $3k+1$. We wish to show that there exists another prime $p$, not on the list, that is also of the form $3k+1$.

Define $x = 3\cdot p_{1} \cdots p_{n}$, and $t = x^{2} + x + 1$.

Note that none of $3, p_{1}, \ldots, p_{n}$ divide $t$. (In fact, they each leave a remainder of $1$ when dividing $t$.)

Now, let $p$ be a prime factor of $t$. (Incidentally, note that if $t$ itself is prime, then $t$ is already in the form $3k+1$. But we cannot count on this happening.)

Since none of $3, p_{1}, \ldots, p_{n}$ divide $t$, then $p$ is not on this list.

Hence $p \nmid x$. Therefore we may apply Fermat's Little Theorem, to give $x^{p-1} \equiv 1 \mod p$. The idea now is that we will show that the order of $x$ in $\mathbb{Z}_{p}$ is $3$, and hence we will have $3\mid (p-1)$.

First, show that $x^3 \equiv 1 \mod p$:

$x^{3} - 1 = (x-1)(x^{2} + x + 1) = (x-1)\cdot t \equiv 0 \mod p$, since $p\mid t$.

Hence $x^{3} \equiv 1 \mod p$.

So the possible order of $x$ in $\mathbb{Z}_{p}$ is at most $3$. Now, the order cannot be $2$, since $2\nmid 3$. That leaves only the possibility that the order of $x$ in $\mathbb{Z}_{p}$ is $1$, i.e., that $x^{1} = x \equiv 1 \mod p$.

By way of contradiction, suppose $x \equiv 1 \mod p$. Then $t = x^{2} + x + 1 \equiv 1^{2} + 1 + 1 \equiv 3 \mod p$.

So we would have both $t \equiv 0 \mod p$ (since by assumption, $p\mid t$), and also $t \equiv 3 \mod p$. That is, $p\mid t$ and $p\mid (t-3)$. But this is not possible, unless $p = 3$. (To spell this out, though it is probably not necessary: if $p$ divides both $t$ and $t-3$, then $p$ must also divide their difference, which is $3$. Hence $p\mid 3$. But the only prime with this property would be $p = 3$ itself.)

But we know that $p$ is not in the list $3, p_{1}, \ldots, p_{n}$. In particular, $p \ne 3$. Thus we have a contradiction, and hence $x \not\equiv 1 \mod p$.

So, finally, we have that the order of $x$ in $\mathbb{Z}_{p}$ is exactly $3$. Hence $3\mid (p-1)$ (see above). So $p-1 = 3m$, for some positive integer $m$. Hence $p = 3m + 1$, and we are done.

Reinhold Burger
  • 41