Find prime numbers $p,q$ such that $p^n+p^{n-1}+...+p+1=q^2+q+1$

Question

Let $p,q$ are prime numbers and $n$ is a even number such that : $p^n+p^{n-1}+...+p+1=q^2+q+1$

Find $p,q$?

I think :

$p^n+p^{n-1}+...+p+1=q^2+q+1\Rightarrow p^n+p^{n-1}+...+p=q(q+1)\Rightarrow p|q(q+1)\Rightarrow p|q$ or $p|q+1$

If $p|q$ then $p=q$. I can solve this

If $p|q+1$, I don't know how to do next !?

Answer 1

Partial answer: Suppose $p\neq q$. $q\neq 2$ and $q$ doesn't divide $n$. Assume

$$p^{n}+\cdots +p+1=q^2+q+1$$

with $n$ an even number. Using geometric sum, we have

$$\frac{p^{n+1}-1}{p-1}=q^2+q+1$$

and therefore

$$p^{n+1}-1=(p-1)(q^2+q+1)$$

If we take modulo $q$, we obtain

$$p^{n+1}-1\equiv p-1 \mod q$$

and thus, since $p\neq q$

$$p^n\equiv 1\mod q$$

Let $m$ be the multiplicative order of $p$ in the group $\mathbb{Z_q}$, then $m$ divides $n$ and $q$ (Lagrange), hence, since $n$ and $q$ are relatively prime $m=1$. This implies

$$p\equiv 1\mod q\ .$$

At this point, we return to the original equality

$$p^{n}+\cdots +p+1=q^2+q+1$$

If we take modulo $q$ at both sides, we get

$$1+1+\cdots+1\equiv 1 \mod q$$

where we sum $1$ $n+1$ times, i.e.

$$n+1\equiv 1\mod q$$

and so $n\equiv 0\mod q$, but this is absurd since $q$ doesn't divide $n$.