$\frac{q^p-1}{q-1}$ squarefree?

Question

Is $\frac{q^p-1}{q-1}$ always squarefree with $q,p$ prime and $p>2$ and $(q,p)=(3,5)$ excluded?

This is a follow up of $3^p-2^p$ squarefree?

I know the case $q=2$ (Mersenne) and $q=3$ are still open, but is there a similar/generalised conjecture for the other prime $q$?

Answer 1

The following PARI/GP-code finds examples :

? forprime(q=2,100,forprime(p=3,30,if(issquarefree((q^p-1)/(q-1))==0,print([q,p]
))))
[3, 5]
[53, 23]
[53, 29]
[67, 3]
[71, 23]
[79, 3]
?

So, for example $$\frac{53^{23}-1}{53-1}$$ is not squarfree.

Answer 2

You can construct counterexamples easily enough. For instance, say we want an example divisible by $29^2$. Noting that $\varphi(29^2)$ is divisible by $7$ we first find an element of order $7$ $\pmod {29^2}$, I got $645$ with a little calculating. Then we find a prime $q\equiv 645 \pmod {29^2}$, I think $22511$ is the least. Then we get $$\frac {22511^7-1}{22511-1}\equiv 0 \pmod {29^2}$$