# An expression in rational numbers $x, y,$ and $z$: Why is it a square of a rational number?

Let $$\,x,y,z\in\mathbb Q\,$$ satisfy $$\,xy+yz+zx=1$$. Given this, I would like to prove that $$\big(1+x^2\big)\big(1+y^2\big)\big(1+z^2\big)$$ is the square of a rational number $$n$$.
That is, you can write ... let's call that $$E(x,y,z)$$. You may say that $$E(x,y,z)=n^2$$ with $$n\in\mathbb Q$$.

I tried to factor it out, it just got worse. I tried to prove some sort of relationship, and I ended up where I started.

The only thing that seems to work is to write two of them in terms of the other, but that also gets me nowhere.

Hint: $$1+x^2=xy+yz+zx+x^2=(x+y)(x+z)$$.

• Oh. Thanks for the hint and for not telling me the whole solution. I got to use my brain a bit more. CHEERS! Sep 19, 2020 at 15:49

$$1+x^2+y^2+z^2+x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2$$

$$=1+(x+y+z)^2-2(xy+yz+zx)+(xy+yz+zx)^2-2xyz(x+y+z)+x^2y^2z^2$$

$$=(x+y+z)^2-2xyz(x+y+z)+x^2y^2z^2$$

$$=(x+y+z-xyz)^2$$

If I do the obvious thing, and write $$z=\frac{1-xy}{x+y}$$ I get $$1+z^2=\frac{(x+y)^2+(1-xy)^2}{(x+y)^2} =\frac{1+x^2+y^2+x^2y^2}{(x+y)^2} =\frac{(1+x^2)(1+y^2)}{(x+y)^2}.$$ I think the rest is clear...