When $p(n,m)=\sum\limits_{k=0}^{m}n^{k}$, find $(n,m)$ that $p(n,m)$ is square. $n \in \mathbb N,n>1$.

Question

When $p(n,m)=\sum\limits_{k=0}^{m}n^{k}$, there are no other solution exept $(n,m)=(7,3),(3,4)$ that $p(n,m)$ is square. $n,m \in \mathbb N,n,m>1$.

I got when $m$ is square number, $(1,m)$ is solution.

When $m=2$, $n^2 < n^2+n+1<(n+1)^2$

and I got $(7,3)$ and $(3,4)$ for $m=3$ and $4$.

I couldn't find when $m=5$. I didn't think that there are only this solution, but I thought $m=5,6,7,8$ has no solution.

Answer 1

The solution is given in Ribenboim's book on Catalan's conjecture, where all Diophantine equations $$y^2=1+x+x^2+\cdots +x^m$$ are studied.

For example, if $m=3$, then only $(x,y)=(1,2)$ and $(x,y)=(7,20)$ are integer solutions, just as you have found.