<s>[INST] Prove that $|x-2y|$ is the squaring number with $x^2-4y+1=2(x-2y)(1-2y)$ For $x,y$ are positive integer, statisfying  $x^2-4y+1=2(x-2y)(1-2y) \,\,\ (1)$. Prove that $|x-2y|$ is a square number.
I try to compute, from $(1)$ We have $(1) \Leftrightarrow x^2+4xy-2x-8y^2+1=0$ and $x=\dfrac{1}{4} \left(x+\sqrt{3x^2-4x+2}\right)$ or $x=\dfrac{1}{4} \left(x-\sqrt{3x^2-4x+2}\right)$. I can't show $|x-2y|$ is a square number.
Another way, I see $x=1$ and $y=0$ is a solution of $(1)$.
 [/INST] Continuing from what you've already stated,
$$\begin{equation}\begin{aligned}
x^2 + 4xy - 2x - 8y^2 + 1 & = 0 \\
x^2 - 2x + 1 & = 8y^2 - 4xy \\
(x - 1)^2 & = (2^2)y(2y - x)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This shows $y \mid (x-1)^2$, so $\gcd(x, y) = 1$. Furthermore,
$$\gcd(y, 2y - x) = \gcd(y, (2y - x) - 2y) = \gcd(y, -x) = 1 \tag{2}\label{eq2A}$$
Can you finish the rest (note $y$ must also be a square number)?
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