Question

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

Answer 1

Let the numbers be $11,111,1111,.......$

To disprove that last two digits cannot be odd for any square

As the last digit is one the last digit of number must be either $1or9$

Let number be $xy$ then $y=1or9$

When $y=1$ then tens digit is $(x+x)$ (i.e. even)

When $y=9$ then tens digits is $9x+8+9x$

$=18x+8$ (i.e. even)

So we can say that a square cannot end with two odd digits.

$\therefore 11,111,1111,......$ cannot be a square.