let say x and y is even then (x,y)$\neq $1 so possibility is that eithr x EVEN ,y ODD or,xODD ,y EVEN now there is also a case that both x and y can be odd but we can not say that (x,y)=1 for all x,y $ if x,y \in S then x^2-y \in s so, (x^2-y,y)=1$ we have to prove that $gcd(x^2-y,x)=(X^2-y,y)=1$ for all integers let say $(x^2-y,x)=d_x and (x^2-y,y)=d_y$ $d_x| x(x-1)-y$ and $d_x|x(x+1)-y$ same for$ d_y $ clearly$(d_x,d_y)=1$ as (x,y)=1 we can say that $d_x=d_y=1$ so it is true for any integer x and y so $S \equiv \mathbb{Z}$