<s>[INST] Show that $4x^2-yz$ is a perfect square Here is my problem.
$A=xy+yz+zx$, where $x,y,z\in\mathbb{Z}$. It is known that if we add $1$ to $x$, and subtract $2$ from both $y$ and $z$, the value $A$ won't change. Prove that $-A$ is a square of whole number.

My attempt:
\begin{align}
A=xy+yz+zx
\end{align}
and
\begin{align}
A=(x+1)(y-2)+(y-2)(z-2)+(z-2)(x+1)
\end{align}
comparing them we get
\begin{align}
4x+y+z=0.
\end{align}
Now we insert last equation to first equation:
\begin{align}
-A&=-xy-yz-zx\\
&=-x(y+z)-yz\\
&=4x^2-yz
\end{align}
From here I don't know how to show that $4x^2-yz$ is a perfect square.
 [/INST] If we expand, we get
$$A=(x+1)(y-2)+(y-2)(z-2)+(z-2)(x+1)\quad =\quad x y + x z - 4 x + y z - y - z$$
Subtracting the original from this equation should be zero
$$(x y + x z - 4 x + y z - y - z)-(xy+yz+xz)=0\\
\implies 4x+y+z=0\implies z = -4 x - y$$
Substituting $z$ now allows the equation to be viewed more simply
$$xy+y(-4 x - y)+x(-4 x - y)= -4 x^2 - 4 x y - y^2=A\\
\implies A=-(4x+4xy+y^2)$$
$$\therefore -A=(2x+y)^2$$
\begin{align*}
\end{align*}
</s>