The question is:
Suppose $p_1<p_2<…<p_{15}$ is a sequence of prime numbers in arithmetic progression, with common difference $d$. Prove that $d$ is divisible by $2,3,5,7,11$ and $13$.
Let $p_n = a+(n-1)d$, by the definition of an arithmetic sequence. It is easy to see that $a$ is odd, and I can prove that $d$ is even easily enough.
I could then show $d$ is divisble by 3 by using the technique of letting $d=3x+c$, and $a=3y+d$, and by going through each possible value of c and d, d not equal to zero, I showed that the only solution where a and d were still prime was when $c=0$ or $3$.
While I could do the same thing for showing d is divisible by 5,7,11 and 13, this is a very laborious process that takes quite a long time. Are there any faster ways of proving this? I feel there is a way to do it using modular arithmetic, but I'm not very competent with regards to modular arithmetic, so I can't think of anything.
Best Answer
Hint: if $q$ is a prime and $d$ is not divisible by $q$, then $p_i \equiv p_j \mod q$ only if $i-j$ is divisible by $q$. There are only $q-1$ nonzero residue classes mod $q$...