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.

$(2y-1)^2 - 4 = (2y-1)^2 - 2^2 = (2y-1+2)(2y-1-2) = (2y+1)(2y-3)$
Note that $2y+1$ and $2y-3$ are always distinct integers. Hence proving their product cannot be a square is accomplished by showing they're coprime (no prime factors in common) and that they're not both squares at the same time.
$\mathrm{gcd}(2y+1, 2y-3) =\mathrm{gcd}(2y+1, (2y+1)-(2y-3)) = \mathrm{gcd}(2y+1, 4) = 1$ (the last part is trivially observing that one is odd, the other even). Hence $2y+1$ and $2y-3$ are coprime.
Now note that both $2y+1$ and $2y-3$ are odd with a difference of $4$. The minimum difference between two odd squares is $3^2 - 1^2 = 8$. So they cannot both be squares.
Therefore $(2y+1)(2y-3) = (2y-1)^2 - 4$ cannot be a square.
