Problem 470

If $x$ and $y$ satisfy $2x^{2}+8y=26$, then $x^{2}+4y=13$ and so $4y=13-x^{2}$. Since $x$ and $y$ are integers, then $4y$ is even and so $13-x^{2}$ is even, which means that $x$ is odd. Since $x$ is odd, we can write $x=2q+1$ for some integer $q$. Thus, $4y=13-x^{2}=13-(2q+1)^{2}=13-(4q^{2}+4q+1)=12-4q^{2}-4q$. Since $4y=12-4q^{2}-4q$, then $y=3-q^{2}-q$. Thus, $x-y=(2q+1)-(3-q^{2}-q)=q^{2}+3q-2$. When $q=4$, we obtain $x-y=q^{2}+3q-2=4^{2}+3 \cdot 4-2=26$. We note also that, when $q=4, x=2q+1=9$ and $y=3-q^{2}-q=-17$ which satisfy $x^{2}+4y=13$.