Show that if each number {b, a + b, 2a + b, ...., (n-1)a + b} is prime then every prime p <= n, must divide a, i.e. p | a

Note that

__if we had__ \(\displaystyle \{b,\,a+b,\,\ldots,\,(n-1)a+b\}=\{0,\,1,\,2,\ldots,\,n-1\}\!\!\!\pmod n\) , then if \(\displaystyle q\) is any prime dividing \(\displaystyle n\) we'd have that

\(\displaystyle q=(ka+b)\!\!\!\pmod n\iff q=(ka+b)+xn\,,\,\,for\,\,\,some\,\,\,k,\,x\in\mathbb{Z}\,,\,\,0\leq k<n-1\) .

But \(\displaystyle q\mid xn\,,\,q\mid q\Longrightarrow q\mid (ka+b)\) , and this is impossible since \(\displaystyle ka+b\) is a prime \(\displaystyle \forall k=0,1,\ldots,n-1\) \(\displaystyle \Longrightarrow \{b,\,a+b,\,\ldots,\,(n-1)a+b\}\neq\{0,\,1,\,2,\ldots,\,n-1\}\!\!\!\pmod n\)

and since there are \(\displaystyle n\) different numbers in \(\displaystyle \{b,\,a+b,\ldots,\,(n-1)a+b\}\) then there must exist \(\displaystyle 0\leq k,j<n-1\,,\,\,k\neq j\) , s.t.

\(\displaystyle ka+b=ja+b\!\!\!\pmod n\Longrightarrow (k-j)a=0\!\!\!\pmod n\Longrightarrow\) any prime dividing \(\displaystyle n\) also divides \(\displaystyle a\) (hmmm

**...why**?? Just a little explanation more...(Giggle)).

Tonio