If $x$ and $y$ are integers with $2 x^{2}+8 y=26$, a possible value of $x-y$ is
(A) -8
(B) 26
(C) -16
(D) 22
(E) 30
If $x$ and $y$ satisfy $2 x^{2}+8 y=26$, then $x^{2}+4 y=13$ and so $4 y=13-x^{2}$.

Since $x$ and $y$ are integers, then $4 y$ is even and so $13-x^{2}$ is even, which means that $x$ is odd.

Since $x$ is odd, we can write $x=2 q+1$ for some integer $q$.

Thus, $4 y=13-x^{2}=13-(2 q+1)^{2}=13-\left(4 q^{2}+4 q+1\right)=12-4 q^{2}-4 q$.

Since $4 y=12-4 q^{2}-4 q$, then $y=3-q^{2}-q$.

Thus, $x-y=(2 q+1)-\left(3-q^{2}-q\right)=q^{2}+3 q-2$.

When $q=4$, we obtain $x-y=q^{2}+3 q-2=4^{2}+3 \cdot 4-2=26$.

We note also that, when $q=4, x=2 q+1=9$ and $y=3-q^{2}-q=-17$ which satisfy $x^{2}+4 y=13$.

We can also check that there is no integer $q$ for which $q^{2}+3 q-2$ is equal to any of $-8,-16,22$, or 30. (For example, if $q^{2}+3 q-2=-16$, then $q^{2}+3 q+14=0$, and this quadratic equation has no integer solutions.)

ANSWER: (B)