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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.

Mathemagician314
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Lee
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    Eliminate $x$ from your last equation using $x = -\frac{y+z}{4}$. You'll get $-A=(\frac{y-z}{2})^2$ – nrynn Nov 28 '20 at 08:57
  • Once you found that the necessary and sufficient condition is $4x+y+z=0$, you can substitute values to determine $-A$, and hence guess the expression. E.g. with $ x = -1, y = 2, z = 2$, we get $ -A = 0$, With $ x = -1, y = 4, z = 0 $, we get $ -A = 4$. After listing enough cases, you should be able to guess that $ -A = ((y-a) / 2 )^2$, and then can work on the algebraic manipulation. – Calvin Lin Nov 28 '20 at 09:02
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    @nrynn I got it, thanks. and since $y+z$ is even, so $y-z$ is even too. – Lee Nov 28 '20 at 09:04
  • https://math.stackexchange.com/questions/419766/number-of-solution-for-xy-yz-zx-n/713998#713998 – individ Dec 23 '20 at 07:49

1 Answers1

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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*}

poetasis
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