Find all possible sum of digits of perfect squares This question was asked in an Iberoamerican Math Olympiad. I formed a conjecture that all numbers not equivalent to $3$ or $6$ $\mod 9$. But there are no squares with the sum of digits $2$. This is the only work from my side, any hints and help would be appreciated.
 A: Hint 
$$9999999..9^2=9999..9800001 \\
19999..9^2=3999..9600..01 \\
29999...9^2=8999..94000..01 \\
9999...95^2=9999..9000.025$$
Edit:
Faster solution if $a$ is a digit, then
$$(10^n-a)^2=10^{2n}-2a\times 10^n+a^2$$
Now
$$10^{2n}-2a\times 10^n=999....9xy0000$$ 
where $xy=100-2a$. Now setting $a=1,2,5,9$ gives you exactly the formulas you want.
A: Consider a number $d$ greater than $9$. Then its $k{+}1$ digits $\{d_k,d_{k-1},\ldots, d_1, d_0\}$ represent the sum $\sum_0^k 10^i\cdot d_i$. The digit sum $s(d) = \sum_0^k d_i$ and the difference between them is $d-s(d) = \sum_0^k (10^i-1) d_i$. It is apparent that all the terms $(10^i-1)$ are divisible by $9$ and thus the remainder when divided by $9$ of $d$ and $s(d)$ is the same. 
The digit sum can be iterated until the value is less than $10$, which I will show as $r(d)$. This value again has the same remainder as $d$ when divided by $9$, and for $r(d)<9$ it actually is the remainder.
Now consider $d^2$. Writing $d=9j+t$ with $t=r(d)$, we have $d^2 = 81j+18jt + t^2$. Now the first two terms are divisible by $9$, so to find $r(d^2)$ we only need to find $r(t^2)$. 
But $t\in \{1,2,3,4,5,6,7,8,9\}$ so $t^2\in \{1,4,9,16,25,36,49,64,81\}$ and thus $r(d^2)=r(t^2) \in\{1,4,9,7,7,9,4,1,9\}=\{1,4,7,9\}$
There is also one special case: the digit sum of the perfect square $0$ is of course $0$.
Thus there are no repeated digit sums of squares that are any of $\{2,3,5,6,8\}$.
