Say $A=\mathbb{Z}[i]$ the Gaussian integers with multiplication and addition of complex numbers. Now we define the Norm $N$ on $\mathbb{Z}[i]$ by: $\forall a=x+yi\in\mathbb{Z}[i], N(a)=x^2+y^2$. Now suppose $x=a+bi, y=c+di, m=N(x), n=N(y)$, where $gcd(a,b)=gcd(c,d)=gcd(m,n)=1$. I have to prove - for $xy = (ac-bd)+(ad+bc)i, gcd(ac-bd, ad+bc)=1$. I tried assuming by contradiction that there's a prime number $p$ that divides both $ac-bd, ad+bc$ - therefore $p$ divides both $(a-b)(c+d)$ and $(a+b)(c-d)$ but wasn't able to come up with a contradiction. Are there any $gcd$ arithmetic rules I can apply here? any hint would help.
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If $p$ is a prime integer such that $p|xy$ then either (case 1) $p$ is a prime Gaussian integer, in which case $p|x$ or $p|y$ contradicting $\gcd(a,b)=1$ or $\gcd(c,d)=1$, or (case 2) $p=\alpha\bar{\alpha}$ where $\alpha$ is a prime Gaussian integer.
Again in case 2 we cannot have $p|x$ or $p|y$ as we cannot have $\gcd(a,b)=1$ or $\gcd(c,d)=1$. Therefore we must have $\alpha|x$ and $\bar{\alpha}|y$ (or vice versa). But then $$p=\alpha\bar{\alpha}|x\bar{x}=N(x),\qquad p=\alpha\bar{\alpha}|y\bar{y}=N(y),$$ contradicting $\gcd(n,m)=1$.
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$\begingroup$ why does $p$ a prime gaussian number dividing $x$ or $y$ leads to a contradiction? Does $p$ dividing $x$ implies $p$ dividing a and b? $\endgroup$– GBANov 9, 2020 at 16:25
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$\begingroup$ @GalBenAyun Exactly. $p$ is an integer, and an integer divides $a+ib$ if and only if $p$ divides $a$ and $p$ divides $b$: $$p(u+iv)=pu+ipv,$$ so $p(u+iv)=a+ib \implies pu=a$ and $pv=b$. $\endgroup$– tkfNov 9, 2020 at 17:33