Let $n=10a+b$ . We have that $1<=a<=9$ and $0<=b<=9$ , so $0<=a+b<=18$ . The perfect squares for that range is $1,4,9$ and $16$ . For the case $1$ , $10$ is unique because you can't raise $b$ and lower $a$ and $a+b$ remain as $1$ . For the case $4$ the higher value is $40$ , so we have $4$ possibilities because $1<=a<=4$ . For the case $9$ , $90$ is the higher value, $1<=a<=9$ so we have $9$ possibilities. Finally, for the case $16$ , $97$ is the higher value, as we lower $a$ and raise $b$ we will have only $3$ possibilities because $b$ already started at $7$ . Summing all possibilities: $1+4+9+3=17$ .

The required set is

{10, 13, 18, 22, 27, 31, 36, 40, 45, 54, 63, 72, 79, 81, 88, 90, 97}

Here is some Mathematica Code:

Select[Range[10, 99], 
 MemberQ[{1, 4, 9, 16}, Total[IntegerDigits[#]]] &]

It is entirey based on casework as you said :)

just write a c code..and its done ;)

$1 \rightarrow 10 \Rightarrow 1$

$4 \rightarrow 13, 22, 31, 40 \Rightarrow 4$

$9 \rightarrow 18, 27, ..., 90 \Rightarrow 9$

$16 \rightarrow 97,88,79 \Rightarrow 3$

total: $\boxed{17}$

You can solve this problem by exhaustion (listing all the possibilities). It is useful to note that any 2-digit multiple of 9 (except 99) will have its digit sum as 9 which is a perfect square (i.e. 18, 27, 36, 45, 54, 63, 72, 81, 90). 99 is the exception since its digits sum is 18 which is not a perfect square.

include <iostream>

#include<math.h>
using namespace std;
bool isPerfectSquare(long long n){
    long long squareRootN=(long long)round((sqrt(n)));

    if(squareRootN*squareRootN == n) {
        return true; 
    }
     else {
        return false; 
    }}

int sum(int x){
    int re=(x%10);
    x/=10;
    re+=x;
    return (re);
}  
int main() {
    int s=0;
    for (int i=10;i<=99;i++){
       if (isPerfectSquare(sum(i))==1) s+=1;

    }
    cout<<s;
    return 0;
}