Finding all four-digit perfect squares of the form $XXYY$

Question

Can you find a four digit number of the form $XXYY$ using only mathematical tools (without a computer) where the first two digits are same ($XX$) and the last two digits are same ($YY$), and the number $XXYY$ is a perfect square. How many such numbers can exist (again without using a computer) ?

$XXYY = (AB)^2$

Answer 1

$XXYY$ is divisible by $11$, so $AB$ must also be. So $\frac{XXYY}{11}=X0Y$ is also divisible by $11$. That means that X and $Y$ must add up to $11$. So $\frac{X0Y}{11}=\frac{Y+(11-Y)\cdot100}{11}=100-9Y$ must be a perfect square. Now you can easily go through the digits for $Y$ and check that only $Y=4$ works.