solution
\(y^2-x^2=ax+b\)
\(y^2=x^2+ax+b\)
When \ (x ^ 2 + ax + b \) when perfect square \ (Ans = inf \)
\ (x \ leq y \) might Order \ (y = x + t \ )
\(x^2+2xt+t^2=x^2+ax+b\)
\(2xt-ax=b-t^2\)
\(x\times(2t-a)=b-t^2\)
\(x=\frac{b-t^2}{2t-a}\)
Enumeration, to find what makes \ (x \) is a natural number \ (t \) , count the number that is \ (Ans \)
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define int long long
using namespace std;
int A,B,Ans;
signed main()
{
scanf("%lld%lld",&A,&B);
if(A%2==0&&A*A/4==B){
puts("inf");
return 0;
}
int L1=sqrt(B),L2=A/2;
if(L1>L2) swap(L1,L2);
for(int i=max(L1-1,0ll);i<=L2+1;++i){
if(i*2==A||((B-i*i)<0&&(2*i-A)>0)||((B-i*i)>0&&(2*i-A)<0)) continue;
if((B-i*i)%(2*i-A)==0) ++Ans;
}
printf("%lld\n",Ans);
return 0;
}