$\forall n,n'\in\mathbb{Z}$ we want to have $n^2+pn+q \neq 2n'^2+rn'+s$ \[n(n+p) - n'(2n'+r)\neq s-q\] If $p$ is odd we choose $r$ even and $s=q+1$ If $p$ is even then $p=2p'$ and we have \[n^2+2np'-2n'^2-n'r\neq s-q\] Modulo $4$ we have \[2np'-n'r\neq s-q+1 (4)\] Taking $r$ even and $s=q$ and it's done.

I didn't quite get your point... how does your reasoning go?

If $p$ is odd, choosing $r$ even and $s=q+1$ gives that $\{f(x)|x \in \mathbb{Z} \} \cap \{g(x)|x \in \mathbb{Z} \} = \emptyset$ since if there is $n, n'$ such that $f(n)=g(n')$ then $n^2+pn+q = 2n'^2+rn'+s \Rightarrow n(n+p) - n'(2n'+r)= s-q \Rightarrow n(n+p) - n'(2n'+r)=1 $ but it's not possible since $n(n+p)$ and $n'(2n'+r)$ are even. If $p$ is even ($p=2p'$), choosing $r$ even and $s=q$ and we have $\{f(x)|x \in \mathbb{Z} \} \cap \{g(x)|x \in \mathbb{Z} \} = \emptyset$ since if there is $n, n'$ such that $f(n)=g(n')$ then $n^2+2np'-2n'^2-n'r = 0$ and OUPS I have made a mistake taking $n$ and $n'$ odd, I need to correct my solution If $n$ and $n'$ are odd we see that we have $n^2+2np'-2n'^2-n'r \equiv 1+2np'-2-n'r (4)$ and can not equals $0$ since $r$ is even I will post end of my solution in few minutes