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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!

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  • $\begingroup$ You don't need to prove its divisible by six. You can prove it's divisible by 2 (as they are prime you can prove that if neither is 2 than both are odd so y-x is even) and that it is divisible by 3. Which... is the heart of the problem. But easier than showing it is divisible by 6. $\endgroup$ – fleablood Apr 25 '17 at 19:34
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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 $6$.

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  • $\begingroup$ If you could replaced the word 'must' with 'can' - that would be much better. $\endgroup$ – usiro Apr 26 '17 at 20:39
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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 $6\mid d$.

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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 $6|(y-x)$

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

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