Prove that in an arithmetic progression of 3 prime numbers the common difference is divisible by 6 Here's the question from the book:

Three prime numbers $p, q$, and $r$, all greater than 3, form an arithmetic progression:
  $$\begin{align} p&=p \\ q&=p+d \\r&=p+2d \end{align}$$
  Prove that $d$ is divisible by 6.  

Now here's what I have done:
For $d$ to be divisible by 6 it has to be divisible by 2 and 3. Now for proving divisibility by 2; $d$ has to be even, or else one of $p, q, r$ will be even. Hence it is divisible by 2.
Can anyone help me find how to prove that $d$ is divisible by 3?
 A: Hint $\ $ Primes $> 3$ have the form $\,6k\pm1.\,$ If  $\,p\not\equiv q\pmod 6\,$ then there are $2$ possibilities  
$${\rm mod}\ 6\!:\quad \begin{array}{rrr}p & q & r\\ -1 & 1 & 3\\ 1 & -1 & -3\end{array}\ \ \qquad\quad $$
thus $\,3\mid r,\,$ contra $\,3 < r\,$ is prime. Therefore $\ p\equiv q\pmod 6,\, $ so $\ 6\mid q-p = d$
Remark $\ $ Note that the proof does not require that $\,p,q,r\,$ are primes $> 3.\,$ Rather, it requires only the weaker hypothesis that they are all  $\equiv \pm1 \pmod 6,\,$ i.e. all coprime to $\,6.$
A: This is the special case $k=3$, of the following fact. 

Let $p,p+d, \dots, p+(k-1)d$ be an arithmetic progression of length $k$ of primes with $p>k$, then $q\mid d$ for each prime $q \le k$ (that is $\prod_{q \le k}q \mid d$).  

To see this just note that if $d \not\equiv 0 \mod q$ then $0,d, \dots, (k-1)d$ is a complete set of residues modulo $q$. Thus for some $j$ we have $j d \equiv -p \mod q$, whence $q \mid p +jd$, a contradiction to $p+jd $ being prime (note $p+jd > q$).
A: HINT:
$$p(p+d)(p+2d)=p(p^2+3pd+2d^2)$$
Now as $3$ is prime, if $3\nmid d;(d,3)=1$
$\implies d\equiv\pm1\pmod3\implies d^2\equiv1$
If $3\nmid p,p^2\equiv1$ as well (we don't need $p$ to be prime!)
Then $(p+d)(p+2d)=p^2+3pd+2d^2\equiv1+0+2\cdot1\equiv0\pmod3\iff3|(p+d)(p+2d)$
$\implies3$ must divide at least one of $p+d,p+2d$ each of which $>3$
