[tex]1000x+100x+10y+y=k^2\qquad (k\in \mathbb{Z})\\
1100x+11y=k^2\\
11(100x+y)=k^2
[/tex]
11 is a prime number, so for [tex]k^2[/tex] to be a perfect square, [tex]100x+y[/tex] must be also equal to 11. But [tex]11^2=121[/tex] which is not even a four digit number. Therefore there is no such number.
