SOLUTION: Find the largest three digit number such that the number minus the sum of its digit is a perfect square.

Let h = the hundreds digit
Let t = the tens digit
Let u = the units (or ones) digit

Then the number = 100h+10t+u
Sum of the digits = h+t+u

The number - sum of digits) = perfect square.

100h+10t+u - (h+t+u) = perfect square

100h+10t+u - h-t-u = perfect square

99h+9t = perfect square

This shows that the solution is independent of u.
Since we are looking for the largest three digit
number, we will therefore take u=9

9(11h+t) = perfect square

So the perfect square must be a square root divisible by 9
So its square root must be divisible by 3.

Since V999 = 31.6... the largest 3-digit perfect square is 31² or 961.
The largest number that does not exceed 31 that is divisible by 3
is 30. so 30 is a candidate for the square root, and 30² or 900 is
a candidate for the perfect square we are looking for:

9(11h+t) = 900

   llh+t = 100

       t = 100-11h

All digits are less than 10, so
  
          t < 10
    100-11h < 10
       -11h < -90
          h > 8.18...

The only digit larger than 8.18... is 9, so h = 9

       t = 100-11(9) = 100-99 = 1

Checking:  919 has sum of digits 9+1+9 = 19 

919-19 = 900 = 30²

Answer: 919

Edwin