A strange phenomena?

Consider that 21^2=441,

Also (2+1)^2=(4+4+1).

i.e. The sum of square of the digits of a number is equal to the sum of digits of the number squared.

Now try 22^2=484,

Also (2+2)^2=(4+8+4)
Actually 11,12,13,31,10,20,30 also share the same property.

To find out why those number has this property, we can do some simple algebra:
For 2 digits, suppose the number is X(1) X(2), and the square of this number is S(1)S(2)S(3)
that has the property which The sum of square of the digits of a number is equal to the sum of digits of the number squared,
i.e. (10X(1)+X(2))^2=S(1)*100+S(2)*10+S(3) AND
(X(1)+X(2))^2=S(1)+S(2)+S(3)

Or 99X(1)^2+18X(1)*X(2)=99S(1)+9S(2)
Or 11X(1)^2+2X(1)*X(2)=11S(1)+S(2)
(which X(1),X(2),S(1),S(2),S(3) must be natural number)

Is that given any X(1),X(2);we can always find suitable pairs of S(1),S(2)?
Not necessarily, because All number are within the range of 0-9,
therefore, 108>11S(1)+S(2)>10 or 108>11X(1)^2+2X(1)*X(2)>10