$\exists x \exists z \forall y(2y-z=4x)$
Where $x$, $y$ and $z$ are integers.
Rearranging, I got
$z=2y-4x$
Since $y$ and $x$ are integers, $2y-4x$ would also be an integer for any given value of $y$.
Since $z$ is also an integer, shouldn't the predicate logic statement be true as there would always exist an integer $z$ which equals the integer $2y-4x$?
The explanation on why it was false is here:
False. For every x and z, there is an integer y = 4|x| + |z| + 1, and thus 2y = 8|x| + 2|z| + 2 > 4x + z.
However I don't understand how that shows that it must be false.
As user @Thomas Andrews hinted in the comment, order of mixed type quantifiers matters, see a previous post discussing similar subtle FOL issue. Had your $\forall y$ moved to any previous position then your reasoning is fine. But now you're clearly given a concrete counterexample construction you cannot avoid. Hope you can see that now...