Given information is turned into Code which results to the o/p of the problem.
#include<iostream>
using namespace std;
int main(){
for(int i=101;i<300;i++)
{
if(((i*i-i-2)%16==0)&&((i*i+2*i-3)%25==0))
cout<<i;
}
return 0;
}
O/p :- 226
i.e. only one (1) number is there.