Can You Prove This Integer Property in Real Numbers?

[anemone]

Here is this week's POTW:

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The real numbers $x,\,y$ and $z$ are such that $x^2+y^2=2z^2$, $x\ne y,\,z\ne -x,\,z\ne -y$.

Prove that \(\displaystyle \frac{(x+y+2z)(2x^2-y^2-z^2)}{(x-y)(x+z)(y+z)}\) is an integer.

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[anemone]

Congratulations to the following members for their correct solution::)

1. kaliprasad
2. Opalg

Solution from kaliprasad:

Spoiler

We have $4x^2- 2y^2 -2z^2 = 4x^2 - 2y^2 - (x^2 +y^2) = 3(x^2-y^2) = 3(x-y)(x+y)\cdots(1)$ (by the given relation)
Now the numerator

$(x+y+2z)(2x^2-y^2-z^2)$
$= \frac{3}{2}(x+y + 2z)(x+y)(x-y)$
$= \frac{3}{2}((x+y)^2 + 2z(x+y))(x-y)$
$= \frac{3}{2}(x^2+y^2+ 2xy + 2z(x+y))(x-y)$
$= \frac{3}{2}(2z^2 + 2xy + 2z (x+y))(x-y)$ using $x^2 + y ^2 = 2z^2$
$= 3(z^2+ xy + zx + zy)(x-y)$
$= 3(z+ x)(z+ y)(x-y)$

Hence $\dfrac{(x+y+2z)(2x^2-y^2-z^2)}{(x-y)(x+z)(y+z)} = \dfrac{3(x-y)(x+z)(y+z)}{(x-y)(x+z)(y+z)} = 3$, which is an integer.

Alternate solution from Opalg:

Spoiler

Let $$A = (x+y+2z)(2x^2 - y^2 - z^2) = (2x^3 - y^3 - 2z^3) + (2x^2y - 2y^2z - z^2x) + (4x^2z - y^2x - z^2y),$$ $$B = (x-y)(x+z)(y+z) = (x^2y - y^2z + z^2x) + (x^2z - y^2x - z^2y).$$ If $x^2 + y^2 = 2z^2$ then $$x^3 = x\cdot x^2 = x(2z^2-y^2) = 2z^2x - y^2x,$$ and similarly $$y^3 = 2z^2y - x^2y,$$ $$2z^3 = x^2z + y^2z.$$ Substitute those expressions for $x^3$, $y^3$, $2z^3$ into $A$ to get $$\begin{aligned}A &= 2(2z^2x - y^2x) - (2z^2y - x^2y) - (x^2z + y^2z) + (2x^2y - 2y^2z - z^2x) + (4x^2z - y^2x - z^2y) \\ &= (3x^2y - 3y^2z + 3z^2x) + (3x^2z - 3y^2x - 3z^2y) = 3B. \end{aligned}$$ The other conditions of the problem say that $B\ne0$, so we can conclude that $\dfrac AB = 3.$