# Is that true that all the prime numbers are of the form $6m \pm 1$?

Is that true that all the prime numbers are of the form $6m \pm 1$ ?

If so, can you please provide an example?

Thanks in advance.

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– Jonas Meyer May 27 '11 at 19:37

## 5 Answers

This is true of all prime numbers except for $2$ and $3$. The reason is that numbers with remainders $0$, $2$ and $4$ modulo $6$ are divisible by $2$, and numbers with remainders $0$ and $3$ modulo $6$ are divisible by $3$, so other than $2$ and $3$ themselves, all prime numbers must have remainder $1$ or $5$ modulo $6$.

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even for 2 and 3 we have 2 modulo 6 = 2 and 3 modulo 6 = 3 but that we can't generate them with this formula – obounaim May 27 '11 at 11:46
@oussama: Yes, this is what I was trying to say: $2$ and $3$ are primes, but you can't generate them with this formula, since their remainders modulo $6$ are $2$ and $3$, respectively, not $1$ or $5$. A number divisible by $2$ or $3$ can only be prime if it is $2$ or $3$, since otherwise it would be divisible by a number other than itself. – joriki May 27 '11 at 11:54
can we use a bigger number than 6 (that verify this propriety) to generate prime numbers in a faster way. sorry for my english – obounaim May 30 '11 at 15:48
@oussama: You can do the same thing with any number; for each natural number $n$, primes (except for the prime factors of $n$ themselves) can only have remainders modulo $n$ that are coprime to $n$. For instance, all primes except $3$ and $5$ have remainder $1$, $2$, $4$, $7$, $11$, $13$ or $14$ modulo $15$. – joriki May 31 '11 at 0:35

The classes $1$ and $5$ are the two reduced residue classes modulo $6$, i.e., they are the congruence classes modulo $6$ which are relatively prime to $6$. Any sufficiently large prime number has to lie in one of these reduced residue classes, since otherwise it would share a common factor with $6$. In this case it is pretty clear that "sufficiently large" means any prime $p \geq 5$.

More generally, let $d$ be any positive integer. Then by the same reasoning all but finitely many primes must lie in one of the $\varphi(d)$ reduced residue classes modulo $d$. Since there are $d$ residue classes modulo $d$ in all, this shows that the "probability" that a large number is prime is at most $\frac{\varphi(d)}{d}$. It is not hard to show that for all $\epsilon > 0$, there exists $N \in \mathbb{Z}^+$ such that $\frac{\varphi(d)}{d} < \epsilon$. This shows that the "probability that a large number is prime" is zero -- or, when this probability business is made rigorous by recasting it in terms of density -- that the prime numbers have zero density inside the positive integers:

$\lim_{n \rightarrow \infty} \frac{ \pi(n)}{n} = 0$.

(This is a weak consequence of the Prime Number Theorem, but as we saw, one certainly need not know PNT in order to prove it.)

This argument is given in $\S 3$ of these notes from an undergraduate number theory course. The argument seems not to be entirely standard, but I think it is an appealingly direct way to show that the primes have density zero. There is also a funny irony here in that the way to see the above claim about the $\varphi$ function is to take $N = p_1 \cdots p_k$ for large $k$ (the argument is given in $\S 3$ of this other handout). Thus, in order to show that the primes have density zero, we are using the fact that there are infinitely many of them. Well, if we happened not to know that, then we would divide the proof into two cases: in case there were only finitely many primes, their density would clearly be zero!

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All integers $n\ge 3$ take one of these forms

$$6k,6k\pm 1,6k\pm 2=2\left( 3k\pm 1\right) ,6k\pm 3=3(2k\pm 1),\qquad k=1,2,3,\dots$$

where $6$ is the product of the first two primes ($2$ and $3$). Since $6k,6k\pm 2,6k\pm 3$ are not primes, we are left with $p=6k\pm 1$. So, all primes $p>3$ are of the form $$6k\pm 1\qquad k=1,2,3,\dots .$$

Added: as examples (asked for in the edited question), we can take the prime number $17$, which is of the form $17=6k-1=6\cdot 3-1$ and the prime number $31$, of the form $31=6k+1=6\cdot 5+1$.

Taking the product of the first three primes $30=2\cdot 3\cdot 5$ this argument generalizes immediately to

$$p=30k\pm r,\qquad p\geq 13,$$

with $r=1,7,11,13$.

And by an easier argument we could also prove that if a positive integer is not of the form $4k\pm 1$, ($k=1,2,3,\dots$), then it is not a prime.

This generalizes the

Proposition. If a number $p>2$ is a prime, then

$$p=4k\pm 1.$$

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@Américo: You forgot 1 and 29. – TonyK May 27 '11 at 15:22
@TonyK: $17$ and $31$ were given as examples of primes $p>3$ which satisfy $p=6k\pm 1$. – Américo Tavares May 27 '11 at 15:28
Américo: what Tony is referring to is your statement about primes of the form p=30k+-r - you forgot r=1 in your list of residue classes mod 30... – Steven Stadnicki May 27 '11 at 19:42
@Steven, @TonyK: Thanks! That's right. Corrected. – Américo Tavares May 27 '11 at 19:52
@Américo: Perhaps remove $17$ too, to avoid redundancy... – TonyK May 29 '11 at 18:13

We know that every integer has one of the forms 6K+r, with 0 ≤ r < 6. Other than 2, each integer of the form 6K, 6K+2 or 6K+4 has a proper factor 2. Other than 3, each integer of the form 6K or 6K+3 has a proper factor 3. Thus, any prime other than 2, 3 has one of the forms 6K+1, 6K+5. Numbers of the latter type can be rewritten as 6L-1, where L = K-1.

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In fact, more is true. The primes are equally distributed between these two residue classes, as per the stronger version of the Dirichlet's theorem on arithmetic progressions -- a prominent theorem in analytic number theory.

More generally, given any positive integer $N$, the primes are equally distributed among the residue classes that are relatively prime to $N$. In the case of $N =6$, your surprise arises because there are only two such residue classes: $\pm 1$. The same is true for $N =4$ and might be less surprising for you there.

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