// assumes n,root and n+i as pythagoras triplets where root=sqrt((n+i)^2-n^2) i.e root=sqrt(i*(2*n+ i)). Passes first 3 out 5 test cases

#include

#include<math.h>

using namespace std;

int main() {

unsigned int n;

cin>>n;

```
int i;
double root;
if(n>2){
for(i=1;i<=(pow(2,32)-1);i++){
root= sqrt(i*(2*n+ i));
if(!(root-int(root))){ //checks if root is an integer
break;
}
}
cout<<root<<" "<<n+i<<endl;
}
else
cout<<-1;
return 0;
```

}