Prove that no integer in the sequence $11,111,1111....... $ is a perfect square.

Question

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??

Answer 1

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.

Answer 2

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

Answer 3

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.