The arithmetic sequence of 4 prime numbers We know that $p, q, r, s$ are primes. And they create an arithmetic sequence.
We are to prove that 
$$18|p-s$$
I tried to solve this problem with completely no success.
I would appreciate any hints or tips.
 A: Note that any prime is of form $6k\pm 1$ for some integer $k$. Now if we have $d$ as our arithmetic increment since it is even there are 3 cases:

Case 1: $d\equiv 2\text{ mod 6}$

at this case if $p=6k+1$ therefore $q=p+d=6k'+3$ and if $p=6k-1$ therefore $r=p+2d=6k''+3$ where both can't be primes so this case is also impossible.

Case 2: $d\equiv 4\text{ mod 6}$

at this case if $p=6k+1$ therefore $r=p+2d=6k'+9$ and if $p=6k-1$ therefore $q=p+d=6k''+3$ where both can't be primes so this case is also impossible.
The only remaining case is where $d\equiv 0\text{ mod 6}$ or $6|d$ which leads to $$6|d\to18|3d=s-p\to 18|p-s$$
A: So $p, q, r, s$ are primes and they create an arithmetic sequence? Does that mean not just that $p < q < r < s$ or $p > q > r > s$ but also that $p - q = q - r = r - s$?
Then set $d = p - q$ or $q - r$ or $r - s$, doesn't matter which, since they're supposed to be the same. This means that $q = p + d$, $r = p + 2d$ and $s = p + 3d$.
Since $-2$ and $2$ are the only even primes, this means that $p, q, r, s$ must all be odd primes. Then $d$ must be even.
But more than that, $d$ must be a multiple of $6$. Suppose $d$ is even but not a multiple of $6$. Further suppose $d \equiv 2 \pmod 6$ and $p \equiv 1 \pmod 6$. Then $q$ is a nontrivial multiple of $3$. Or suppose that $p \equiv 5 \pmod 6$, then $r$ is a nontrivial multiple of $3$. You can work out the remaining possibilities yourself.
Now let's say $d = 6n$. Then $3d = 18n$, proving the assertion $18 \mid p - s$.
A: Without loss of generality, assume $s < r < q < p$.
Let's start by writing primes in the form $18k + c$. All primes are in this form except for $p = 2$:
\begin{align}
18k&+1 \\
18k&+5 \\
18k&+7 \\
18k&+9\\
18k&+11 \\
18k&+13\\
18k&+17 \\
18k&+19\\
18k&+23 \\
18k&+25 \\
&\vdots
\end{align}
And the list goes on infinitely. Basically, any $c$ such that $\gcd(18, c) = 1$ works. But there are several terms of note.
Notice that the difference between $18k + 7$ and $18k + 13$ is $6$. The difference between $18k + 13$ and $18k + 19$ is $6$ as well. Finally, the difference between $18k + 19$ and $18k + 25$ is also $6$. Hence, we have found the terms of the sequence and the common difference as well.
Well then $p = 18k + 25$ and $s = 18k + 7$.
\begin{align}
p-s & = 18k+25-(18k+7)\\
& = 25-7\\
& = 18
\end{align}
Obviously $18 \mid 18 \implies 18 \mid p - s$.
