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

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.
