Diophantine Equation Equaling a Square

Question

I was wondering if someone could explain the conditions for which the general Diophantine equation

$$m^2 = n^k + n^{k-1} + ... + n^1 + 1$$

where $m^2 \in \mathbb{Z}^+$ is a perfect square. For instance, in the case where $k = 4$, $$n^4 + n^3 + n^2 + n + 1$$ is a perfect square if and only if $n=3$. To show this, we bound the expression for cases where $n$ is even and odd. Is there a well-known process for solving the more general case where $k$ is some arbitrary positive integer?

Answer 1

There can be no general method that works for all $k$ since there is no solution for $k=2$. To see this note that:

$$n^2 < n^2+n+1 < n^2 + 2n+1 = (n+1)^2$$

$$n < \sqrt{n^2+n+1} < n+1$$

Answer 2

Above equation shown below:

$m^2 = n^k + n^{k-1} + ... + n + 1$

For $k=3$ we have:

$m^2=(n^3+n^2+n+1) ---------(1)$

$(m,n)=(20,7)$ & $(m,n)=(2,1)$ are solution's to equation $(1)$