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?
Best Answer
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 $\,6.$