Sum of the digits of a perfect square Prove that the sum of the digits of a perfect square can't be 2, 3, 5
  , 6, or 8.


I'm completely stumped on this one, how would I go about proving it?
The sum of the digits of a number is invariant mod $9$. So this is the same as asking what are and what are not the squares mod $9$?

Let's consider: the squares mod $9$ are $1,4, 7$ (coming from $1,2,4$). The complement of these are $2,3,5,6,8$, which are exactly the parts from your question. These are the numbers which do not appear as squares mod $9$, and thus do not appear as sums of digits of squares.
If your square number had the sum of digits equal to 3 or 6, then it must be a multiple of 3. But a square that is a multiple of 3 must be a multiple of 9. Therefore the sum of its digits must be a multiple of 9. 3 or 6 are not multiples of 9.

Therefore 3 and 6 are impossible.