Puzzle given to me by Vidyu

A puzzle was given to me by Vidyu as follows-
There is a four digit number 'aabb' formed by the 2 digits'a' and 'b' which is a perfect square.Find the values of 'a' and 'b'.
There are no such cases in our normal 'decennial' system.i.e numbers to base 10.
It could be in a system with another base.
If the base is 'x',the number can be expressed as ax^3+ax^2+bx+b which can be simplified as
(ax^2+b)(x+1).
If this has to be a perfect square both (x+1) and (ax^2+b) should be squares.
Thus x can have a value 3 or 8.(See below where x=3)
If x=8 we must have (64a+b) must be a square.The only possible values for a and b are a=5 and b=4.as 64*5+4=324 which is a square.
Thus the solution is 5544 in base 8.
5544 in base 8 is equal to 2916 in base 10 which is the square of 54(base 10)
54 in base 10 is equal to 6*8+6 i.e 66 in base 8.
Thus purely considering both numbers in base 8,...5544 is the square of 66
You can check by multiplying 66 by 66 following the rules for all calculations viz multiplication,addition etc for base 8.
6*6=36=4*8+4=44 in base 8.We have to carry forward 4.
6*6+4(carried forward)=36+4=40=50 in base 8.
Thus 6*66=504 and so 66*66=5544.
There is no use in considering the value x=3 as in that case the number should be 0011,which is not a 4 digit number.
As a matter of variation, I cosidered other bases.
In base 9 ,the square of 66 will be 4840  and in base 7 it will be 6501,You can check.