Question

Find a four digit number which is a perfect square. The last two digits are equal and the first two digits are equal of the number.

Answer 1

`1000x+100x+10y+y=k^2\qquad (k\in \mathbb{Z})\\ 1100x+11y=k^2\\ 11(100x+y)=k^2`

11 is a prime number, so for
`k^2` to be a perfect square,
`100x+y` must be also equal to 11. But
`11^2=121` which is not even a four digit number. Therefore there is no such number.