Proof: not a perfect square Let $y$ be an integer.
Prove that
$$(2y-1)^2 -4$$
is not a perfect square.
I Found this question in a discrete math book and tried solving it by dividing the question into two parts:
$$y = 2k , y = 2k + 1$$
But that got me nowhere.

For $y\le-1$, $(2y-1)^2-4$ is between consecutive squares $(2y)^2$ and $(2y-1)^2$.
For $y\in\{0,1\}$, $(2y-1)^2-4$ is negative, so not a   square.
For $y\ge2$, $(2y-1)^2-4$ is between consecutive squares $(2y-2)^2$ and $(2y-1)^2$.
