Problem 429

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 $x^2 + 4y = 13$.

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