Find the four digit number? Find a four digit number which is an exact square such that the first two digits are the same and also its last two digits are also the same.

We know that the only square that has last two digits as same is $12^2=144$. For a four digit perfect square it should either be $(50-12)^2$ or $(50+12)^2$ or $(100-12)^2$  
$38^2= 1444 $
$62^2= 3844$
$88^2= 7744 $
As per the question we know $88^2$ is the perfect square that has first 2 digits & last 2 digits as the same.
