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?
Thanks in advance, Guillermo
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')$