1 /*思路：利用x1+x2=n;x1*x2=m;两式求出(x1-x2)^2=n*n-4*m;判断该数能否求出整数解,
 2 如果能的话再将该数与m相加求x1,如果x1为整数,则说明方程有整数解,反之,无整数解
 3 */
 4 #include <bits/stdc++.h>
 5 using namespace std;
 6 int main()
 7 {
 8     int n,m;
 9     int a,b,c;
10     while(cin>>n>>m)
11     {
12         if(n==0&&m==0)
13             break;
14         else
15         {
16             a=n*n-4*m;
17             b=sqrt(a);
18             if(b*b!=a)
19                 cout<<"No"<<endl;
20             else
21             {
22                 c=b+n;
23                 if(c%2==0)
24                     cout<<"Yes"<<endl;
25                 else cout<<"No"<<endl;
26             }
27         }
28     }
29     return 0;
30 }