Expressing products as the sum of two squares

Question

Express $(1+x^2)(1+y^2)(1+z^2)$ as the sum of two squares.

I have no clue where to start, I know that we should not post just for solution but a hint will definitely help.

Answer 1

By Lagrange's identity $$ (a^2+b^2)(c^2+d^2) = (ad-bc)^2 + (ac+bd)^2 \tag{1}$$ it follows that numbers that are a sum of two squares form a semigroup.
A straightforward proof of $(1)$ is to consider that $a^2+b^2$ is the norm of $a+bi$ in $\mathbb{C}=\mathbb{R}[i]$.
The norm is multiplicative and $(a-bi)(c+di)=(ac+bd)+(ad-bc)i$. An equivalent alternative is to exploit Binet's theorem $\det(AB)=\det(A)\det(B)$ through the matrices $$ A = \begin{pmatrix}a & -b \\ b & a\end{pmatrix},\qquad B = \begin{pmatrix}c & d \\ -d & c\end{pmatrix},\quad AB=\begin{pmatrix}ac+bd & ad-bc \\ bc-ad &ac+bd\end{pmatrix}.\tag{2} $$ One way or another, since $$ (i+x)(i+y)(i+z)=(xyz-x-y-z)+i(xy+xz+yz-1)\tag{3} $$ it follows that: $$ \boxed{\phantom{\sum_{i=0}^{10}}(1+x^2)(1+y^2)(1+z^2)=\color{red}{(xyz-x-y-z)^2+(xy+xz+yz-1)^2}.\phantom{aa}}\tag{4} $$

Answer 2

Hint:

$$1+x^2=(1+ix)(1-ix)$$

Do the same with other factors.

Finally, rearrange the product this way:

$$[(1+ix)(1+iy)(1+iz)]\cdot[(1-ix)(1-iy)(1-iz)]$$