Is it true that every pair of twin primes >3 is of the form 6n plus or minus 1?

Question

Is it possible to have twin primes whose center is not divisible by 6?

Answer 1

It is too easy to prove. Observe that all primes nos. can only be of the form $6n\pm1$ except for $2$ and $3$. So if one of them is $6n+1$ other will be $(6n+1)\pm2$ but since $6n+3$ isn't prime so the other prime will be of form $6n-1$ and vice versa.

Answer 2

A twin prime is two consecutive primes that differ by 2.

Primes greater than 2 must be odd, so the greater prime of a twin prime pair must be of the form 6n+1, 6n+3, or 6n+5.

Since 6n+3 is not prime, the greater twin prime must be 6n+1 and the lesser 6n-1.

Answer 3

Think about the remainders:

• Any number $p$ of the forms $6k$ or $6k+3$ is divisible by $3$, hence not prime (unless $p=3$).

• Any number $p$ of the form $6k$, $6k+2$, or $6k+4$ is divisible by $2$, hence not prime (unless $p=2$).

So this means that if $p$ is prime $>3$, then either $p=6k+1$ or $p=6k+5$; note that that second case can be rewritten as $p=6(k+1)-1$.

Do you see how to continue from here? HINT: assuming $p,q$ are a twin prime pair with $p=q+2$, show that (1) $p$ can't have the form $6k+1$, since $q$ must be prime, and (2) $q$ can't have the form $6k+5$, since $p$ must be prime.

Answer 4

Take any integer $n$ greater than $3$, and divide it by $6$. That is, write $n=6q+r$

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

$\implies$ If the remainder is $0, 2$ or $4$, then the number $n$ is divisible by $2$, and cannot be prime.

$\implies$ If the remainder is $3$, then the number $n$ is divisible by $3$, and cannot be prime.

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

$\implies1(\mbox{and $n = 6q + 1$ is one more than a multiple of $6$})$ or

$\implies5(\mbox{and $n = 6q + 5 = 6(q+1) - 1$ is one less than a multiple of $6$})$

Remember that being one more or less than a multiple of $6$ does not make a number prime. We have only shown that all primes other than $2$ and $3 ($which divide $6)$ have this form.