Does $7$ divide $2 x^2 - 4y^2$ for all $x,y$? Does $7$ divide $2 x^2 - 4y^2$ for all $x,y \in \mathbb{Z}$?
 A: Let $x = 7a + 1$ and $y = 7b + 1$ for integers $a,b$. Then,
$$\begin{align}2a^2 - 4b^2 &\equiv 2\cdot1^2 - 4\cdot1^2\pmod 7\\
&= 5\pmod7\\
&\equiv 5 \pmod 7\end{align}$$
So it is not generally true, because we can always generate integer pairs (though not all of them) that give a remainder of $5$ when divided by $7$ as shown above. Using the above result, we see that $(x,y) = (1, 1), (8, 1), (1, 8), (8,8)$ are all counter examples.
In fact, you can play around with the constant term in $x$ and $y$ to see how it affects the remainder.

However, if $x = 7a + 2$ and $y = 7b + 4$, then we see that
$$\begin{align}2x^2 - 4y^2\equiv2\cdot2^2-4\cdot4^2\pmod7\\
= -56\pmod7\\
\equiv 0 \pmod7\end{align}$$
Hence, the result will hold true if $x = 7a + 2$ and $y = 7b + 4$, for integers $a,b$.
A: ...suppose the result holds true, then, 
$2x^2-4y^2=0 \pmod7$ 
since $2$ does not divide $7$, we have $x^2-2y^2=0 \pmod7$, but $2$ is congruent to $9$ modulo $7$, so, $x^2-9y^2=0 \pmod7$ or 
$x=3y \pmod7$, so the result holds true $iff$ $x=3y+7k$, for some integer $k$
