Let $x, y, z$ be different prime numbers with $x, y, z > 3$. Prove that if $x + z = 2y$, then $6 | (y - x)$. I have troubles to prove the following task:
Let $x, y, z$ be different prime numbers with $x, y, z > 3$. Prove that if  $x + z = 2y$, then $6 | (y - x)$.
The only idea I have is that every prime number $> 3$ divided by $6$ has remainder $1$ or $5$.
But I do not have any idea how to prove this statement?!
Thank you for any help!
 A: Every prime $p>3$ has the form 
$$p = 6k\pm1. \tag{1} $$
The condition 
$$
x+z=2y
$$
tells us that $x$ and $z$ cannot have opposite signs in the $\pm1$ term in $(1)$. (Otherwise $y$ would be a multiple of $3$ and hence composite.)
Therefore we have
$$
x = 6k-1 \quad\mbox{ and }\quad z = 6j-1 
\qquad\Rightarrow\qquad y=3(k+j)-1=6m-1,
\tag{2}
$$
or
$$
x = 6k+1 \quad\mbox{ and }\quad z = 6j+1 
\qquad\Rightarrow\qquad y=3(k+j)+1=6m+1.
\tag{3}
$$
In either case $(2)$, $(3)$ we see that the difference $y-x=6(m-k)$ is divisible by $6$.
A: Let $d=y-x$. Then $$d=x-y=-y-x+2y=-y-x+(x+z)=z-y$$ Thus, $(x,y,z)=(x,x+d,x+2d)$, three primes with an equal difference. If $3\not\mid d$, then at least one of these must be divisible by $3$; impossible, since they are prime, thus, $3\mid d$. The same reasoning works to show $2\mid d$. Those two combined give $6\mid d$.
A: You're idea is fine. The possibilities for $x+z$ modulo $6$ are:
$2$ when $x\equiv z\equiv 1$ $(\operatorname{mod}6)$
$4$ when $x\equiv z\equiv 5$ $(\operatorname{mod}6)$
$0$ when $z\equiv 1, x\equiv 5$ (or the other way around) $(\operatorname{mod}6)$
the possibilities for $2y$ are 
$2$ when $y\equiv 1$
and
$4$ when $y\equiv 5$
we must have that these are equal and either all $3$ have remainder $1$ or all $3$ have remainder $5$.
So $6|(y-x)$
A: $6| y-x$ if and only if $3|y-x$ and $2|y-x$.
1) $2|y-x$.  
$y$ and $x$ are both primes larger than $2$ so $x$ and $y$ or both odd.  So $y-x$ is even. 
2) $3|y-x$.
$y > 3$ and $y$ is prime so $y \equiv \pm 1 \mod 3$.  $x > 3$ and $x$ is prime so $x \equiv \pm 1 \mod 3$. and similarly $z \equiv \pm 1 \mod 3$.
Either $x \equiv y \mod 3$ or $x \equiv -y \mod 3$.  
If $x \equiv -y \mod 3$ then $2z \equiv x+y \equiv -y+y \equiv 0 \mod 3$.  But $2z \equiv 2*\pm 1 \equiv \mp 1 \not \equiv 0 \mod 3$.
So $x \equiv y \mod 3$.  So $y-x \equiv 0 \mod 3$ and $3|y-x$.
