Answer :
from the well known theorem that, primes are multiple of 6 ±1 ( eg 5,7,11,13,17,19...)
and one of them has [tex]-1[/tex] and other has $+1$ from the multiple of 6
let , $p=6n-1$, so $p+2=6n+1$
$\implies p+1=6n$
$\therefore 6|(p+1)$
QED
from the well known theorem that, primes are multiple of 6 ±1 ( eg 5,7,11,13,17,19...)
and one of them has [tex]-1[/tex] and other has $+1$ from the multiple of 6
let , $p=6n-1$, so $p+2=6n+1$
$\implies p+1=6n$
$\therefore 6|(p+1)$
QED