Had the same Idea. First tried just iterating through every number in the range but my solution was timing out except for the first 2 and last 2 cases. Then I just noticed that the closest squares (in low ranges) appear always after more than twice the previous perfect square added to the current position (e.g. square 1: 1 + 2*1 = 3, the next square is 4: 4+2*2 = 8, the next square is 9: 9+3*2 = 15, the next square is 16 and so on for 25). By doing this you skip a lot of useless numbers and the solution does't time out.

int main() {

int t;
cin >> t;


for(int a0 = 0; a0 < t; a0++){
    int a;
    int b;
    int count = 0;
    cin >> a >> b;
    int range;

    for(range = a; range <=b; ++range){


       int temp = sqrt(range);

        if(temp*temp == range)

        {

            count++;
            range += temp*2 ;
        }



    }


    cout<<count<<"\n";
}


return 0;

}