How to prove that $(1+x^2)(1+y^2)(1+z^2)$ is a square of an integer?

Question

$x,y,z \in \Bbb Z$, if $xy+yz+xz=1$ then prove that $(1+x^2)(1+y^2)(1+z^2)$ is a square of an integer

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

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

Answer 1

Let me try. $$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx) = x^2 + y^2 + z^2 + 2.$$

So, $$(1+x^2)(1+y^2)(1+z^2) = 1 + (x^2+y^2+z^2) + (x^2y^2+y^2z^2 + z^2x^2) +(xyz)^2 = 1 + (x+y+z)^2-2 + 1 - 2xyz(x+y+z) + (xyz)^2 = (x+y+z-xyz)^2.$$

Answer 2

Use Brahmagupta–Fibonacci identity, $$(1+x^2)(1+y^2)=(1\pm xy)^2+(x\mp y)^2$$

Again, $$\{(1-xy)^2+(x+y)^2\}(1+z^2)=\{(1-xy)-z(x+y)\}^2+\{(1-xy)z+x+y\}^2$$

Do you see the destination?