Prove that no integer in the sequence $11,111,1111....... $ is a perfect square. My problem is well described in title. I know that it can be proved as follow:
Since every number in the sequence is of the form $4x+3$ and perfect square does not exist in such form so none  is a perfect square.
But I need to prove in a different way, a way different from modular arithmatic. Any ideas??
 A: first assume $1111111111...11$ is a perfect square :
it is odd number like $(2k+1)^2$
so 
$$1111111...1111=(2k+1)^2 \\
111111...111=4k^2+4k+1\\111111...111-1=4k(k+1)=8q\\
111111...110=8q \\q=\frac{111111...110}{8}=\frac{55555.555}{4} $$this is paradox .because $q$ must be a natural number .
so, there is no perfect square in $11111...1111$ numbers
A: Assume that it is a perfect square. Then, since the last digit is $1$, the number is of the form $(10n+1)^2$ or $(10n+9)^2$. But then, in either case, the tens digit would be even---a contradiction.
A: Proof by contradiction:
Suppose that one of the elements in the sequence is a perfect square.
Let $n$ denote the root of that element.
The unit digit of $n$ must be either $1$ or $9$.
Observe (or calculate it manually if you don't trust me) that:


*

*$\not\exists{n}\in\{01,11,21,31,41,51,61,71,81,91\}\text{ such that the last $2$ digits of $n^2$ are $11$}$

*$\not\exists{n}\in\{09,19,29,39,49,59,69,79,89,99\}\text{ such that the last $2$ digits of $n^2$ are $11$}$


Any other digits of $n$ surely have no impact on these last $2$ digits.
Therefore no element in the sequence is a perfect square.
