Sum of the digits

The rest of the divison of a number by 3 is equal to the rest of the division by 3 of the sum of that number's digits.

$1991 \equiv 2 (mod 3)$

So now we have to check if any number divided by 3 can have rest 2.

$(3k)^2 \equiv 0 (mod 3)$

$(3k+1)^2 = 9k^2 + 6k + 1 \equiv 1 (mod 3)$

$(3k+2)^2 = 9k^2 + 12k + 4 \equiv 1 (mod 3)$

This proves that the rest of the divion of a perfect square by 3 is either 0 or 1.

So there isn't a perfect squarte which digits add up to 1991.