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I'm trying to solve this problem from my abstract algebra course:

Being $z_1=39-8i$ and $z_2=7+i$ elements from the ring of Gauss integers, $\mathbb{Z}[i]=\{a+bi \mid a,b\in\mathbb{Z}\}$. Find a greater common divisor of both terms, $d=\text{gdc}(z_1,z_2)$, and find elements $x_1,x_2\in\mathbb{Z}[i]$ such that $d=x_1z_1 + x_2z_2$.

I have no idea how to proceed with this kind of exercises, I've seen other ones similar to this one in my book and want to find a method to approach these kind of exercises. I decided to post this one in particular because the ring $\mathbb{Z}[i]$ seemed especially weird but interenting, so I assumed this must be one of the most difficult ones of this kind of problems.

How can I solve this? Any help will be appreciated, thanks in advance.

Alejandro Bergasa Alonso
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    Euclid's algorithm works in the ring $\Bbb{Z}[i]$ as well. You need a little bit extra care in selecting the appropriate "integer approximation to the quotients". Instead of bluntly rounding the quotients towards zero, which is what we do in $\Bbb{Z}$, we need to round to the closest Gaussian integer. Here $$ \frac{z_1}{z_2}=\frac{53}{10}-\frac{19}{10}i.$$ We see that the Gaussian integer closest to it is $q_1=5-2i$. Therefore the first remainder will be $$r_1=z_1-q_1z_2=2+i.$$ Then in the next step we calculate $$ \frac{7+i}{2+i}=3-i. $$ – Jyrki Lahtonen Jan 23 '21 at 16:13
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    (cont'd) We see that $q_2=3-i$, and the next remainder is $r_2=z_2-q_2r_2=0$. As usual, the last non-zero remainder is the gcd, so $d=\gcd(z_1,z_2)=2+i$. To write $d$ as a linear combination of $z_1$ and $z_2$ you can follow the usual procedure. Alternatively you can combine the two stages like (inserting a link soon). – Jyrki Lahtonen Jan 23 '21 at 16:15
  • Cool! I think I understand it, although it's a bit confusing considering the ring is $\mathbb{Z}[i]$. Thank you for your help! – Alejandro Bergasa Alonso Jan 23 '21 at 16:17
  • See here for another answer of mine on this theme. I think I have examples of calculating $x_1,x_2$ as well. Stay tuned. – Jyrki Lahtonen Jan 23 '21 at 16:26
  • In this thread we have more comprehensive answers. – Jyrki Lahtonen Jan 23 '21 at 16:27
  • Here I show a method for calculating the gcd as well as $x_1,x_2$ in one go by organizing the data into a table. The context there is different because we expected the gcd to be $1$, and were interested in the modular inverse. There are likely better explanations of that use of row operations on augmented matrices. IIRC Robjohn and Bill Dubuque have posted some, but I cannot find them right now. – Jyrki Lahtonen Jan 23 '21 at 16:31
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    More examples 1, 2. Look up the first for Bill Dubuque covers the details rather well. There's also Wikipedia. The links in this comment are about $\Bbb{Z}$ rather than $\Bbb{Z}[i]$. As long as you round the quotients to nearest Gaussian integer rather than towards zero, the same ideas will work. – Jyrki Lahtonen Jan 27 '21 at 20:11
  • @JyrkiLahtonen Thanks for your help! I'll take a look to them. – Alejandro Bergasa Alonso Jan 27 '21 at 20:16

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I'd work in the ring $R = \Bbb Z[x]/\langle x^2+1\rangle = \{ax+b\mid a,b\in\Bbb Z\}$, where $x^2+1 = 0$ in $R$.

Then division with rest gives

$-8x+39 = (-8)(x+7) + 95$, i.e., $95 = 1\cdot (-8x+39) + 8\cdot (x+7)$.

Then over the rationals: $1 = \frac{1}{95}(-8x+39) + \frac{8}{95}(x+7)$.

Wuestenfux
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  • The last argument only holds over the rationals. – Wuestenfux Jan 23 '21 at 14:53
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    $(7+x)(7-x)=49-x^2=50-(x^2+1)\equiv50$. And $2\cdot50-95=5$, so you get at least $5$. Then $7+x-5=2+x$. As $(2+x)(2-x)=4-x^2\equiv5$, we see that $5$ is divisible by $2+x$. So $2+x$ is the gcd. – Jyrki Lahtonen Jan 23 '21 at 18:15