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

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?
