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?

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.
