<s>[INST] 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.
 [/INST] $(2y-1)^2-4=4(y^2-y)-3$  If it were a perfect square it would be $=c^2$, where c is an integer.  Solve for $y$ in $4(y^2-y)-3-c^2=0$ and get $y=\frac{4\pm \sqrt{16+16(3+c^2)}}{8}=\frac{1\pm \sqrt{4+c^2}}{2}$.
However $c^2+4$ cannot be a square, unless $c=0$ (where $y$ is not an integer). Assume $c^2+4=b^2$ so $b=c+a$ with $(c+a)^2=c^2+2ac+a^2$.    $2ac+a^2=4$ has no possible integer solutions.   ($a=1$ LHS is odd, $a\gt 1$ LHS $\gt 4$).
Therefore no possible integer $y$.
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