I recently stumbled upon a fact that all prime numbers past $3$ are of the form either $6n-1$ or $6n+1$.
Is it true? at least for numbers less than $10^9$.
And does it cover all primes?
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$\begingroup$All prime numbers past $3$ are of one of those two forms.
Think of it this way:
All integers are of one of this forms:
$$\begin{cases}6n-2 & \Rightarrow& 2·(3n-1) \\ 6n -1 \\6n & \Rightarrow & 2·3·n \\ 6n+1 \\ 6n+2 & \Rightarrow& 2·(3n+1) \\ 6n+3 & \Rightarrow & 3·(2n+1)\end{cases}$$
Note that all other than $6n-1$ and $6n+1$ can be expressed as a product of two integers bigger than $1$. So a prime number cannot be of any form other than $6n\pm 1$.
(That doesn't mean that all numbers of the form $6n\pm 1$ are prime)
$\endgroup$ 8 $\begingroup$Hint : By Division Algorithm, any integer can be represented as one of : $6n, 6n+1, 6n+2, 6n+3, 6n+4, 6n+5$ and notice below
$2 | 6n+2$
$3 | 6n+3$
$2 | 6n+4$
$\endgroup$ 1 $\begingroup$Indeed all primes greater than 3, are in the form of 6n-1 and 6n+1. I've studied on these 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 :
1-------- 2---------3--------4 ---------5--------6
7-------- 8-------- 9--------10--------11--------12
13-------14--------15------- 16--------17------- 18.....[ 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 [ 7,13,19,..infinite ] and column 5 generates the prime path of 6n-1. [ 5,11,17,23...infinite..] Thus , by isolation and sieve method, we can see that all primes > 3 must be in the form of 6n-1 and 6n+1...
hope this help a little...thanks..yolito, you can email me at [ ]
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