Squares are amazing!

The number can be written as

$XXYY = 1000X + 100X + 10Y + Y = 1100X + 11Y = 11(100X + Y).$

Now, for $XXYY$ to be a perfect square, $100X + Y$ should be a multiple of $11$ and a perfect square. Also, $100X + Y$ is of the form $X0Y$ . Checking for some values of perfect squares ( $16,25,36,49,64,...$ ), $11*64 = 704$ fits our case leading to $X = 7$ and $Y = 4$ and hence, $XXYY = 7744$ .

A more generalized approach would be to limit some values for X and Y. As $100X + Y$ divides $11$ times a perfect square, last digit of $X0Y$ ie., $Y$ , should be one of $0,1,4,5,6$ or, $9$ as $11a \equiv a\ (mod\ 10)$ .

Checking for divisibility of $11$ , $(X + Y) - 0 = 11m$ . As $X$ and $Y$ are two single digit numbers, $m$ can't be $0$ or $>1$ . Hence, $X + Y = 11$ and $Y$ is one of $4,5,6$ or $,9$ . So, $(X,Y)$ has to be one of $(7,4), (6,5), (5,6)$ or $, (2,9)$ . Substituting these values in $X0Y$ , we get $(7,4)$ fits the scenario.