Peter of Madras College, St Andrew's
employed an exhaustive search to find the smallest perfect
square that ends in 9009 and came up with the answer 1503.
I first noted that the last digit has to be a 3 or a 7 for the
square to end in 9. Noting that the last two digits of $x^2$ are
only affected by the last two digits of $x$. I then systematically
went through all the squares.
I kept a record of the numbers tried in two tree diagrams starting
from the units digits 3 and 7. If any of these produced a number
that ended in 09 then I marked that as the next branch point on the
diagram.
I then went on to further generations looking for numbers ending in
009, and then finally the next generation looking for numbers
ending in 9009. I found that there are no numbers with 3 digits or
less whose squares end in 9009 and the four digit numbers are 1503,
6503, 2753, 7753, 2247, 7247, 3497 and 8497.
Alternatively suppose $x^2 = 100a + 10b
+ c$ where $a$, $b$ and $c$ are whole numbers, $a \geq 1$ and $b$
and $c$ are between 0 and 9 inclusive.