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!

$6| y-x$ 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 $3|y-x$.
