Sum of a Sequence of Prime Powers $p^{2n}+p^{2n-1}+\cdots+p+1$ is a Perfect Square

Question

Find all primes p such that $p^{2n}+p^{2n-1}+p^{2n-2}+\cdots+p^{2}+p+1$ is a square for some value of n.

Answer 1

Some partial results. First we show that $p\geq3$ for any $n\geq1$. Assume otherwise i.e. that $p=2$ then $$2^{2n}+2^{2n-1}+...+2+1=x^2\Rightarrow 2^{2n+1}-1=x^2\Rightarrow 2^{2n+1}=x^2+1$$ Since $n\geq1$ then $$2^{2n+1}\equiv0\mod{4}\Rightarrow x^2\equiv3\mod{4}$$ which is impossible as a square is either $1$ or $0$ modulo $4$. So $p$ is an odd prime and clearly $x$ is an odd number as well. Now let $p>2$ and $p\equiv1\mod{4}$ then $$p^{2n}+p^{2n-1}+...+p^2+p+1\equiv2n+1\equiv x^2\mod{4}$$ Since $x$ is odd then $$2n+1\equiv1\mod{4}\Rightarrow n\equiv0\mod{2}$$ When $p\equiv-1\mod{4}$ does not reveal any extra information as $$p^{2n}+p^{2n-1}+...+p^2+p+1\equiv1\equiv x^2\mod{4}$$