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Show that every prime $p>3$ is either of the form $6n+1$ or of the form $6n+5$, where $n=0,1,2, \dots$

Martin Sleziak
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  • My own solution: According to the division algorithm a number can be of the form $6n$, $6n+1$, $6n+2$, $6n+3$, $6n+4$ or $6n+5$. –  Sep 14 '11 at 08:04
  • alvoutila, is this really your homework? What division algorithm are you referring to? – Dan Brumleve Sep 14 '11 at 08:08
  • My own solution: According to the division algorithm( There exist unique integers q and r such that a = bq + r and 0 ≤ r < |b|, where |b| denotes the absolute value of b)a number can be of the form $6n$, $6n+1$, $6n+2$, $6n+3$, $6n+4$ or $6n+5$. If $6n=2*3 = 2k $ => number is even. If $6n+1$ => Number is either product of primes or prime. If $6n+2=2(3n+1)=2k$ => number is even. If $6n+3=3(2n+1)$=> number is divisible by three. If $6n+4=2(3n+2)$ => number is even. If $6n+5$ => number is either product of primes or prime. Thus every prime $p>3$ is either of the form $6n+1$ or of the form $6n+5$. –  Sep 14 '11 at 08:15
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    alvoutila, your solution is correct. – Dan Brumleve Sep 14 '11 at 08:18
  • What if I say that every prime number is of the form 4n+1 or 4n+3 (by division algorithm) – Vidyanshu Mishra Nov 08 '16 at 05:49
  • @THELONEWOLF.: Then I will tell you that you're wrong, since $2$ is neither. And if you insist on prime numbers larger than $2$, then I will tell you that every prime number is in the form of $2n+1$... And hey, you know what? Even in the form of $n$... My point being $4n+1,3$ is by no means an "improvement" of $6n+1,5$. – barak manos Nov 22 '16 at 15:38
  • @Barak Manos I excluded 2 – Vidyanshu Mishra Nov 22 '16 at 15:40
  • Not in the comment though – Vidyanshu Mishra Nov 22 '16 at 15:40

5 Answers5

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Every integer is of the form $6n$ or $6n+1$ or $6n+2$ or $6n+3$ or $6n+4$ or $6n+5$ for some integer $n$. This is because when we divide an integer $m$ by $6$, we get a remainder of $0$, $1$, $2$, $3$, $4$, or $5$.

If an integer $m>2$ is of the form $6n$ or $6n+2$ or $6n+4$, then $m$ is even and greater than $2$, and therefore $m$ is not prime.

If an integer $m>3$ is of the form $6n+3$, then $m$ is divisible by $3$ and greater than $3$, and therefore $m$ is not prime.

We have shown that an integer $m>3$ of the form $6n$ or $6n+2$ or $6n+3$ or $6n+4$ cannot be prime. That leaves as the only candidates for primality greater than $3$ integers of the form $6n+1$ and $6n+5$.

Comment: In fact, it turns out that there are infinitely many primes of the form $6n+1$, and infinitely many primes of the form $6n+5$. Showing that there are infinitely many of the form $6n+5$ is quite easy, it is a small variant of the "Euclid" proof that there are infinitely many primes. Showing that there are infinitely many primes of the form $6n+1$ requires more machinery. But your question did not ask for such a proof.

André Nicolas
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    http://en.wikipedia.org/wiki/Dirichlet's_theorem_on_arithmetic_progressions – Dan Brumleve Sep 14 '11 at 08:19
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    @Dan Brumlewe: The reference to Dirichlet's theorem is nice, but that's fairly heavyweight stuff. For $6n+5$ one needs basically nothing, and for $6n+1$ some information about quadratic residues. – André Nicolas Sep 14 '11 at 08:32
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Copied from Are all primes (past 2 and 3) of the forms 6n+1 and 6n-1?

[Considering] $n = 6q + r$

where q is a non-negative integer and the remainder $r$ is one of $0, 1, 2, 3, 4$, or $5$.

  • If the remainder is $0, 2$ or $4$, then the number $n$ is divisible by $2$, and can not be prime.
  • If the remainder is $3$, then the number $n$ is divisible by $3$, and can not be prime.

So if n is prime, then the remainder r is either

  • $1$ (and $n = 6q + 1$ is one more than a multiple of six), or
  • $5$ (and $n = 6q + 5 = 6(q+1) - 1$ is one less than a multiple of six).
grg
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Stat-R
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2

This is elementary algebra. For what value(s) of $n$ is $6n$ prime? $6n+2$? $6n+3$? $6n+4$? Are there any other possibilities besides these and the two that you already mentioned?

Brian M. Scott
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$6$ divides $6n$, $2$ divides $6n+2$, $3$ divides $6n+3$, $2$ divides $6n+4$, and there are no other cases.

Dan Brumleve
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Indeed, all primes greater than $3$ are in the form of $6n-1$ and $6n+1$. I've studied this a few years ago. Here's a basic visual proof of that using a sieve and isolation method that I used:

First, list down all the numbers in 6 columns:

$$\begin{array}{c|c|c|c|c|c} 1&2&3&4&5&6\\ 7&8&9&10&11&12\\ 13&14&15&16&17&18\\ \end{array}$$(the list will be infinite)

Then, we cross out the column of $2, 3, 4$, and $6$ as they are all composite. so we are just left out with two columns of $1$ and $5$. Using Algebraic progression with a difference of $6$, Column 1 generates the prime path of $$6n+1\quad[7,13,19,\ldots,\infty]$$ and column 5 generates the prime path of $$6n-1\quad[5,11,17,23,\ldots,\infty]$$

Thus, by isolation and sieve method, we can see that all primes must be in the form $6n-1$/$6n+1$.

grg
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Yolito Maghinay de Gracia
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