I'm looking for a 4 digit square number where its first two digits are a square as are the last two.
What is its square root?
Incidentally, I'm not interested in the square of zero.
The answer must be (10x+y) squared, which equals 100x^2+(20xy+y^2). If x is 5 or more, the left two digits won't be a square, because (29xy+y^2) will be greater than 100. If x is 3 or less, the square will either be three digits long, or not start with a square.
This allows only x=4, and y must be less than 2: 41 is the only possible number to check, and it works!
No tries at all
