What is the proof to the fact that all prime numbers are 1 above or below a 6 multiple? I was just having an argument with my friend and I dunno how we got here. But he suddenly said all primes are 1 above or below a multiple of 6.
At first I tried a lot of primes but couldn't disprove this. I tried googling but the stuff is too complicated for me.
Is there a simple to understand proof for this statement?
$p \equiv \pm 1 \pmod{6}$, where $p$ is prime.
As pointed out by the answers. I forgot to mention that p > 3. I never checked 2 and 3 when talking to my friend. Somehow thought of them as corner cases.
 A: Given any integer $p > 3$, we know by (a modified version of) the Division Algorithm that there exist unique integers $q \geq 0$ and $r \in \{-1,0,1,2,3,4\}$ such that:
$$
p = 6q + r
$$
Now suppose that $p$ is prime. Then observe that $r \notin \{0,2,4\}$, since otherwise $p$ would be even (contradicting the fact that $p \neq 2$).
Likewise, observe that $r \neq 3$, since otherwise $p = 6q + 3$ would be divisible by $3$ (contradicting the fact that $p \neq 3$).
So $r = \pm 1$, as desired. $~~\blacksquare$
A: $3$ divide $p^2-1=(p-1)(p+1)$ then $3$ divide $p+1$ or $3$ divide $p-1$ (because $3$ is prime), as $2$ divide $p+1$ and $p-1$, then $6=2\times 3$ divide $p+1$ or $p-1$.
A: Hint $\ $ Primes $>3\,$ are coprime to $\,2,3\,$ so coprime to $\,6.\,$ The integers $\,n\,$  coprime to $6$ are  those of form $\,6q\!\color{#c00}{+\!1},\ 6q\!+\!5 = 6(q\!+\!1)\color{#c00}{-\!1},\,$ since $\,2\mid 6q\!+\!r,\ r\in\{0,2,4\},\,$ and $\, 3\mid 6q\!+\!3,\,$ exhausting all possible cases, since, by the Division Algorithm, $\ n = 6q+r\,$ for unique remainder $\, 0\le r \le 5.$
A: If $p$ is a prime >3, then since $p$ is not divisible by 3, we must have $p−1$ or $p+1$ divisible by 3. And both are even, so one must be divisible by 6.
More generally, if you are neither divisible by 2 nor 3, then you neighbor a multiple of 6. 
A: First, every number is within a distance six of a multiple of six. Primes bigger than 2 are odd so must be an odd distance from a multiple of six otherwise they would be even. That means primes are a distance of one or three from a multiple of six. 
However, multiples of six are divisible by 3, so numbers a distance of three from multiples of six are also divisible by 3. That means that 2 (being even) and 3 are the only primes not a distance of one from a multiple of six. Thus, we conclude, all primes bigger than 4 are plus or minus one from a multiple of six.
