# Co-primes in Gaussian integers

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.

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 $$\gcd(n,m)=1$$.
• 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?
• @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 $pv=b$.