What is the square root?

Since the first subtrahend consists of $1$ digit, yet is being subtracted from a $2$ digit number, we deduce that the first digit of the square root is $3$ . Since the second subtrahend consists of $2$ digits, and must be of the form $(60+b)b$ for some digit $b$ , we deduce that the second digit $b$ of the square root must be $1$ , and also that the first two digits of the square must be $10$ .

Since the $10$ s digit of the final subtrahend is $3$ , this means that the $10$ s digit of the square must be $3$ . This means that the last two digits of the square must be $36$ , and that the last two digits of the square root must be one of $06$ , $44$ , $56$ or $94$ . Since $\sqrt{10000000} = 3162.27786$ , the third digit of the square root must be at least $6$ , to ensure that the square is an $8$ -digit number. Thus there is only one possible option for the square root, namely $\boxed{3194}$ , and $3194^2 = 10201636$ :

$\begin{array}{r|rrrr} & 3 & 1 & 9 & 4 \\ & 10 & 20 & 16 & 36 \\ 3 & 9 \\ & 1 & 20 \\ 61 & & 61 \\ & & 59 & 16 \\ 629 & & 56 & 61 \\ & & 2 & 55 & 36 \\ 638 & & 2 & 55 & 36 \\ & & & & 0 \end{array}$

The first digit is at most 3 because its square is a single digit and it is at least 3 because there are 2 digits from which it is to be subtracted. In turn, the first 2 digits of the number being square-rooted is between 10 and 15. The second digit of the square root can not be a 0, can be a 1 as (3 20+1) 1 is a two digit number and can not be a 2 as (3+20+2)*2 is a three digit number and the second subtrahend is two digits. Now, we are in some trouble as there is no limit on the next digit, except that it is at least 7 or the number being square-rooted would not be 8 digits. Now it is necessary to do some searching, which can be simplified by only looking at the squares of the integers of 70 through 99 looking for one with squares with a tens digit of 3. There is only one such: 94. Therefore, the entire square root is 3194.