TITLE: Integers of the form 89, 889, 8889,... not perfect squares
QUESTION [5 upvotes]: Show that an integer of the form 89, 889, 8889,... is not a perfect square.
I know a perfect square will have an odd number of divisors so my first attempt would be assuming that an integer of the form 89, 889, 8889,... is a perfect square and then show that 89, 889, 8889,... all have an even number of divisors. I do not know how to begin proving that 89, 889, 8889,... all have an even number of divisors. Any hints or suggestions would be greatly appreciated.
Thanks.

REPLY [4 votes]: Suppose $8\cdot10^n+1=x^2$ for some $n$.
Then $(x+1)(x-1)=2^{n+3}5^n$
Note that for all $n\geq3$, we have $5^n>2^{n+3}$+2.
Since only one of $x+1,x-1$ can be a multiple of $5$, $5^n$ must divide one of them and we have $x+1\geq5^n$, $x-1\leq2^{n+3}$ or $x-1\geq5^n$, $x+1\leq2^{n+3}$ which are both impossible for $n\geq 3$.
So we need to only check $n=1,2$ and $81=9^2$ while $801$ is not a square.
However $81<89$ so it does not correspond to one of the listed number. So no listed number is a square.