Amir Hossein wrote:
They just copied this old question but put it in a nice form...
That problem is from AMM.
Amir Hossein wrote:
Determine all integers $n \geq 2$ for which the number $11111$ in base $n$ is a perfect square.
This is equivalent to saying that $n^4+n^3+n^2+n+1$ is a perfect square. So there is $m \in \mathbb{N}$ such that $m^2=n^4+n^3+n^2+n+1.$
Thus $(2m)^2=4n^4+4n^3+4n^2+4n+4.$ However, $(2n^2+n)^2<4n^4+4n^3+4n^2+4n+4<(2n^2+n+2)^2$ and so $(2n^2+n)^2<(2m)^2<(2n^2+n+2)^2$ which leads to $(2m)^2=(2n^2+n+1)^2 \Leftrightarrow 4n^4+4n^3+4n^2+4n+4=(2n^2+n+1)^2.$
The last one becomes $n^2-2n-3=0.$ Hence $n=3.$