Notice that: (1) if $x,y \geq 0$ then all negative number $\notin S$ (2) if $gcd(x,y)>1$ then $1 \notin S$ So we can guess the solutions are: $(x,y)=(-1,0),(-a,b),(-a,-b)$ or their symmetrics (where $a,b \in \mathbb{N} \wedge gcd(a,b)=1$) Let's prove all the solutions above satisfied the condition 1. For $(x,y)=(-1,0)$ ($(0,-1)$ can be prove in the similar way) $(-1)(-1)+0=1$, $(-1)(-1)+1=2$, $(-1)(-1)+2=3$, ...... $\therefore$ all positive untegers $\in S$ $(-1)(1)+0=1$, $(-1)(2)+0=-2$, $(-1)(3)+0=-3$, ...... $\therefore$ all negative untegers $\in S$ $\implies$ $S=\mathbb{Z}$ 2. For $(x,y)=(-a, \pm b)$ ($(a,-b)$ can be prove in the similar way) $(-a)(-a)+a=a^2+a$, $(-a)(-a)+(a^2+a)=2a^2+a$, $(-a)(-a)+(2a^2+a)=3a^2+a$,...... $\therefore$ $ma^2+a \in S$ $\forall m \in \mathbb{N}$ $(\pm b)(\pm b)+a=b^2+a$, $(\pm b)(\pm b)+(b^2+a)=2b^2+a$, $(\pm b)(\pm b)+(2b^2+a)=3b^2+a$,...... $\therefore$ $nb^2+a \in S$ $\forall n \in \mathbb{N}$ $(\pm b)(\pm b)+ma^2+a=ma^2+b^2+a$, $(\pm b)(\pm b)+ma^2+b^2+a=ma^2+2b^2+a$, $(\pm b)(\pm b)+ma^2+2b^2+a=ma^2+3b^2+a$,...... $\therefore$ $ma^2+nb^2+a \in S$ $\forall m,n \in \mathbb{N}$ From McNuggest Theorem we know $\forall N>(ab-a-b)+a=ab-b$, N can be written as the form $ma^2+nb^2+a$ since $gcd(a^2,b^2)=1$ so $N \in S$ for $N$ suffi large choose $N$ suffi large we have $(-a)N+(aN \pm 1)= \pm 1 \in S$ and thus $(-1)(1)+1=0 \in S$ from the privious case we know $S=\mathbb{Z}$, Done!