TITLE: If $n > 1$ and all $n$ positive integers $a, a + k, \cdots , a+ (n - 1)k$ are odd primes, show every prime $<n \mid k$.
QUESTION [6 upvotes]: Background: This is from Rosen 5th edition, $3.2.15$ Number Theory.
This is an important proof because $3$ following problems require it to be correct.

If $i=0$, and $j=p$ then this proof is wrong and $p\mid (i-j)$ and $p$ may not divide $k$. Can this proof be modified so that it is correct or do I not understand it?

REPLY [3 votes]: Assume that $p\nmid k$ for some prime $p<n.$ Then $kx\equiv-a\pmod p$ for some integer $0\leqslant x<p$ so $a+kx\equiv0\pmod p$ and since each of the numbers $a+ik,$ with $0\leqslant i<n,$ is a prime then $a+kx=p.$ Now notice that for each $1\leqslant i<n$ we have $\gcd(a,i)=1$ for otherwise the number $a+ik=\gcd(a,i)\left(\frac{a}{\gcd(a,i)}+\frac{i}{\gcd(a,i)}k\right)$ would not be a prime. Thus $a\geqslant n.$ Therefore we have $n\leqslant a\leqslant a+kx=p<n,$ which is a contradiction. Thus the primorial $(n-1)\#$ divides $k.$