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