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In $\Bbb Z[i]$ find $\gcd(2-7i,2+11i).$ Also find $x,y\in \Bbb Z[i]$ such that $(2-7i)x+(2+11i)y=\gcd(2-7i,2+11i).$

I tried solving the problem as follows: Let $d=\gcd(2-7i,2+11i).$ So, $$d|2-7i,d|2+11i\implies d|4\implies d|2(2-7i)-4\implies d|14\implies d|2\implies d|7\implies d|1\implies d$$ is a unit in $\Bbb Z[i].$ So, $\gcd(2-7i,2+11i)=1.$

However, I don't get how to find $x,y\in \Bbb Z[i]$ such that $(2-7i)x+(2+11i)y=\gcd(2-7i,2+11i).$ Of course I can verify that, $$\frac{11}{36}(2-7i) + \frac{7}{36}(2+ 11i) = 1$$ but $\frac{11}{36},\frac{7}{36}\notin\Bbb Z[i].$

Also I dont know how is this is a duplicate of the link: How to calculate GCD of Gaussian integers?, since there the user asks for a way to compute the gcd of two elements in $\Bbb Z[i]$ but that's not the case here. I know how to calculate the gcd but I can't find any suitable integers $x,y\in\Bbb Z[i]$ such that $(2-7i)x+(2+11i)y=\gcd(2-7i,2+11i)=1.$ Any help regarding this issue will be greatly appreciated.

Thomas Finley
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  • You need to solve for $x=x_1+ix_2$, $y=y_1+iy_2$ in integers $x_1,x_2,y_1,y_2$. – Dietrich Burde Jun 08 '24 at 08:42
  • @DietrichBurde But $\frac{11}{36},\frac{7}{36}$ are not in $\Bbb Z[i]$. – Thomas Finley Jun 08 '24 at 08:42
  • Yes, indeed - see my comment. – Dietrich Burde Jun 08 '24 at 08:42
  • @DietrichBurde But then it is an underdetermined system. It's very difficult to solve without using numerical analysis – Thomas Finley Jun 08 '24 at 08:43
  • You only need the Euclidean algorithm. I obtained $x=-22+11i$ and $y=-16$. Then $x(2-7i)+y(2+11i)=1$. This was not difficult. – Dietrich Burde Jun 08 '24 at 10:50
  • In this duplicate you can see, how the Euclidean Algorithm (EA) gives you $x$ and $y$ via "back substitution". Bill recommends the "forward version". – Dietrich Burde Jun 08 '24 at 11:11
  • Re: your edit after closure: The extended Euclidean algorithm works the same in any Euclidean domain - see the newly added dupe targets. – Bill Dubuque Jun 08 '24 at 11:22
  • Also, as in this answer, a Bezout equation for their coprime integer norms, say $, j(\bar a a) + k (\bar b b) = 1,$ is already a Bezzout equation for $,\color{#c00}a,\color{#0af}b,,$ when simply rewritten in the form $, (j \bar a) \color{#c00}a + (k \bar b)\color{#0af} b = 1. \ \ $ – Bill Dubuque Jun 08 '24 at 13:54
  • Here we have; $\bmod \color{#0af}{53}!:\ \dfrac{1}{\color{#c00}{125}}\equiv \dfrac{1}{19}\equiv \dfrac{3}{57}\equiv \dfrac{56}4\equiv\color{#0a0}{14},,$ by Gauss's algorithm, so $$\begin{align}1 ,&=, \color{#0a0}{14}(\color{#c00}{125})\ -\ !33(\color{#0af}{53})\[.4em] &=, 14(2!-!11i)\ \color{#c00}{(2+11i)} -33(2!+!7i)\ \color{#0af}{(2-7i)}\end{align}\qquad\qquad$$ – Bill Dubuque Jun 08 '24 at 13:54
  • I added a proof here that shows more generally how adjoining the norms of the gcd arguments simplifies the (Euclidean) gcd computation. – Bill Dubuque Jun 08 '24 at 19:27

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