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

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