Prove that $11,111,111.....$ is not a perfect square I know there are many other same related questions, but as my profile says I don't have a mentor, I just wanted you guys to check my solution.
First I noticed that,
$11$ = $10 + 1$,
$111$ = $10^2+10+1$,
Similarly, $\underbrace{111\cdots111}_{n\,\text{times}}$ = $10^{n-1}+10^{n-2}+...+10+1$,
which  can easily be evaluated as it is a finite geometric series.
$$\sum_{i=0}^{n-1} 10^i=\frac{10^n-1}{9}$$ for $n \ge
2$.
After this, I checked perfect square modulo $4$,
Let $j$ be any integer, then if $j$ is even, then $j=2k \Rightarrow j^2 = 4k^2 \equiv 0 \mod 4$
and if $j$ is odd then, $j = 2k+1 \Rightarrow j^2 = 4k^2+4k+1 \equiv 1 (\mod4)$
Then, Since $\underbrace{111\cdots111}_{n\,\text{times}} = \frac{10^n-1}{9} \Rightarrow (9) (\underbrace{111\cdots111}_{n\,\text{times}}) = {10^n-1}$ $\equiv$ $2^n-1$ $(\mod 4)$
$\Rightarrow (\underbrace{111\cdots111}_{n\,\text{times}}) \equiv 2^n-1 \mod 4$
Since for $n \ge 2, 2^n$ is always a multiple of $4$. Hence,
$\Rightarrow (\underbrace{111\cdots111}_{n\,\text{times}}) \equiv -1 \mod 4$
$\Rightarrow (\underbrace{111\cdots111}_{n\,\text{times}}) \equiv 3\mod 4$
But a perfect square leaves remainder $0$ if it is even and $1$ if it is odd. So, clearly for
$n \ge 2,$  $(\underbrace{111\cdots111}_{n\,\text{times}})$ is not a perfect square.
I know there are other questions related to this but I just wanted to verify if my solution was correct. Thank you and suggestions would be appreciated!
base 3:  102 * 102 = 11111
base 7:   26 *  26 = 1111
These disprove the assertion.
In decimal, any number 11 modulo 100 is 3 mod 4, and is thus not square.
base 3:  102 * 102 = 11111
base 7:   26 *  26 = 1111
These disprove the assertion.
In decimal, any number 11 modulo 100 is 3 mod 4, and is thus not square.