if $ax|n$ and $ax+1$ is prime does $ax+1|a^{n}-1$? Are there any $a,x,n$ such that $ax|n$ and $ax+1$ is prime but $a^{n}-1$ is not a multiple of $ax+1$, apart from $a=x=n=1$? 
I had an answer to a related question earlier: Can $x^{n}-1$ be prime if $x$ is not a power of $2$ and $n$ is odd?
$n^{a}-1=(n-1)(n^{a-1}+n^{a-2}+\cdots+1)$
 A: Since $ax|n$ and $ax+1$ is prime, for some $b\in\mathbb{Z}$ we have:
$$ n=bax = b(p-1) $$
where $p=ax+1$.
Since $ax+1$ is prime, $a$ and $p$ are coprime. Applying Fermat's Little Theorem gives:
$$a^n-1 \equiv  a^{b(p-1)}-1 \equiv (a^{p-1})^b - 1 \equiv 1^b - 1 \equiv 0 \mod{p} $$
Thus $a^n-1 =k(ax+1)$ for $k\in\mathbb{Z}$. Requiring that $a^n-1\not=0\Rightarrow a \not=1$ and $a\not=-1$ when $n$ is even.
Therefore $(ax+1)|(a^n-1)$ whenever $a\not=1$ or $a\not=-1$ when $n$ is even.
A: Let's denote: $ax+1=p$ , where $p$ is prime number.We may write :
$p-1 | n \Rightarrow p | a^n-1$
So we know that $n=k_2(p-1)$ , for some integer $k_2$
Fermat little theorem states that :
$a^p\equiv a \pmod p$ , therefore:
$a^p-a=k_1p\Rightarrow a(a^{p-1}-1)=k_1p\Rightarrow$
$\Rightarrow a^{p-1}-1=\frac{k_1p}{a}$
Since we have denoted $n=k_2(p-1)$ we may write :
$a^n-1=a^{k_2(p-1)}-1=(a^{p-1})^{k_2}-1=$
$=(a^{p-1}-1)(\sum_{i=0}^{k_2-1} a^{(p-1)i})=\frac{k_1p}{a}(\sum_{i=0}^{k_2-1} a^{(p-1)i})$
This means that: $p|a^n-1 \Rightarrow ax+1 |a^n-1$
