# The greatest common divisor $\gcd(x_1x_2\pm y_1y_2, x_1y_2\mp x_2y_1)$

I wonder if someone could shed some light in the following question

Let $(x,y)$ denote the greatest common divisor of $x$ and $y$, and let $x_1,y_1,x_2,y_2$ be integers.

Is the following statement true?

If $(x_1,y_1)=(x_2,y_2)=(x_1^2+y_1^2,x_2^2+y_2^2)=1$, then $$(x_1x_2\pm y_1y_2, x_1y_2\mp x_2y_1)=1$$

If not, what further hypotheses are necessary to guarantee the claim?

Hint  Interpret it in Gaussian integers $$\ \mathbb Z[i].\,$$ We define $$\ a + b\ i\$$ to be primitive if it has no prime factor $$\in\Bbb Z,\,$$ i.e. $$\ (a,b) = 1.\,$$ Then the statement says that the product of two primitive Gaussian integers is primitive if they have coprime norms. Put $$\, \alpha = x_1+y_1\ i,\,$$ $$\, \beta = x_2+y_2\ i.\,$$ Suppose $$\ p\,|\,\alpha\beta\$$ for a prime $$\, p\in \mathbb Z.\,$$ If $$\, p\,$$ is prime in $$\, \mathbb Z[i]\,$$ then $$\ p\,|\,\alpha\$$ or $$\ p\,|\,\beta\$$ contra $$\ldots\,$$ Otherwise we have $$\, p = \pi\pi'\,$$ for a prime $$\, \pi \in \mathbb Z[i]\,$$ hence $$\ \pi\pi'|\,\alpha\beta\ \Rightarrow\ \ldots$$
Alternatively, if you are not familiar with Gaussian integer arithmetic then you may employ the Brahmagupta–Fibonacci identity for composition of squares. It is a consequence of the fact that the norm is multiplicative: $$\ N(\alpha)\ N(\beta) = N(\alpha\beta)\$$ i.e. $$\ \alpha\alpha'\beta\beta' = (\alpha\beta) (\alpha\beta)'\,$$ which, rationally, is
$$(x_1^2 + y_1^2)\ (x_2^2 + y_2^2)\, =\, (x_1x_2\pm y_1y_2)^2 + ( x_1y_2\mp x_2y_1)^2\quad$$
The upper $$\pm$$ signs are from $$\ N(\alpha) N(\beta') = N(\alpha\beta')$$