For which perfect powers $\ b>1\ $ can the expression $$\frac{b^{p^2}-1}{b^p-1}$$ be a prime number , if $\ p\ $ is prime ?
If $\ b>1\ $ is a perfect power , then $\ b=a^k\ $ for some integers $\ a,k>1\ $.
So we get $$\frac{a^{kp^2}-1}{a^{kp}-1}$$
and if the polynomial $$f(x)=\frac{x^{kp^2}-1}{x^{kp}-1}$$ is reducible we (probably) will get aurifeullian factors. It seems that in most cases, we get an irreducible polynomial. Who can shed some light on this ?
The irreducible factorization of $x^m-1$ (over $\Bbb Z$) is the product of the cyclotomic polynomials $\Phi_d(x)$ over all positive integers $d$ dividing $m$. So the irreducible factorization of $f(x)$ is the product of the cyclotomic polynomials $\Phi_d(x)$ over all positive integers $d$ that divide $kp^2$ but not $kp$.
Assuming for simplicity that $p$ doesn't divide $k$: these integers $d$ are precisely those of the form $bp^2$ where $b$ divides $k$; in particular, there is more than one such integer when $k>1$, and so $f(x)$ will not be irreducible. (The assumption that $p$ doesn't divide $k$ can be weakened to $k$ not being a power of $p$.)