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I'm trying to find the GCD of $29-3i$ and $11+10i$ in $Z[i]$

I know that one way I can do this, is to factorize each of the numbers and find the GCD through that. But the norms in these two numbers are fairly large so it seems like a tedious way to do it. I was wondering if there's a trick that I'm not seeing to get the GCD.

I tried subtracting the numbers and got $18 - 13i$, so we know our GCD divides that number. Here the norms are still too large, and this is definitely not a gaussian prime. Is there a way to reduce these numbers to something more manageable?

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    18^2 + 13^2 = 493 = 17 * 29 – 2017-02-18
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    That's the one I know: If both a and b are nonzero then, a+bi is a Gaussian prime iff a^2+b^2 is an ordinary prime. Is there another one..? – 2017-02-18
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    ohhh sorry lol I meant gaussian PRIME – 2017-02-18
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    I was trying to say that the difference is reducible so we basically got nowhere – 2017-02-18
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    The Euclidean algorithm works, you just have to be a bit more clever about your coefficients. For instance, $29-3i - (1-i)(11+10i)$ looks a bit more manageable. – 2017-02-18
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    ok.. where did you get (i-1) from though? – 2017-02-18
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    I just wonder. Is there an agreement on what is "the" gcd of two Gaussian integers? There are four possible gcd's, but which one is "the" gcd. – 2017-02-18
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    $\frac{29-3i}{11+10i}\approx 1.3-1.5i$. This is not a Gaussian integer, but $1-i$ is pretty close. Close enough to give a noticeably smaller norm, at least. – 2017-02-18
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    Hm.. well it seems so wishy washy. I mean, why not $1-2i$? I guess I'm looking for a systematic way, rather than just look at it and know the factorization. But I guess that's the only way? – 2017-02-18
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    no cleverness needed. typing answer – 2017-02-18
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    I recommend you to look up how to do [Euclid's algorithm in the Gaussian integers](http://www.math.ucsd.edu/~jverstra/Gaussian1.pdf). – 2017-02-18
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    you want me to divide the two numbers? well I already know how to do that. – 2017-02-18

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Hint $\ $ The gcd divides the gcd of their norms $= (850,221) = 17 = (4-i)(4+i) = \frak p_1 \frak p_2$

$\quad {\rm mod}\,\ 17,\,\color{#c00}{i\!-\!4}\!:\ \ 29-3\color{#c00}i \equiv 29-3(\color{#c00}4)\equiv 0,\ \ 11+10\color{#c00}i \equiv 11+10(\color{#c00}4)\equiv 0$

Thus the prime $\,{\frak p_1} = i - 4\,$ is a common divisor, necessarily greatest, since $17\nmid 11+10i$

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$$\frac{29 - 3i}{11+10i} = \frac{289 - 313i}{221} = 1-i + \frac{68 - 92i}{221} $$ confirm $$ 68^2 + 92^2 = 13088 < 48841 = 221^2 $$ $$ \color{blue}{ (29 - 31) = (1-i)(11+10 i) + (8-2i)} $$

$$\frac{11+10i}{8-2i} = \frac{68 +102i}{68} = 1+i + \frac{34i}{68} $$ confirm $$ 34 < 68 $$ $$ \color{blue}{ (11+10i) = (1+i)(8-2i) + (1+4i)} $$

$$\frac{8-2i}{1+4i} = -2i $$ confirm $$ \color{blue}{ (8-2i) = -2i (1+4i) + 0} $$ $$ $$ $$ \color{red}{ \gcd(29 - 3i, 11+10i) = 1 + 4i} $$ $$ $$ $$\frac{29 - 3i}{1+4i} = \frac{17 - 119i}{17} = 1-7i $$ $$ (1+4i)(1-7i) = 1 + 28 + 4i - 7i = 29 - 3i $$ $$ $$ $$\frac{11 +10i}{1+4i} = \frac{51 - 34i}{17} = 3-2i $$ $$ (1+4i)(3-2i) = 3 + 8 + 12i - 2i = 11 + 10i $$

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    Let me ask you the following question: When we dividing one gaussian integer into another gaussian integer why do we find the nearest gaussian integer and consider it as the "quotient"? I can't get this point. Would be very grateful for clarification! – 2018-06-01
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One more trick that you can use is the fact that $gcd(a,b)=gcd(a,b+ra)$ for any element $r$, which is helpful to reduce the number you work with. For example, in your case you can use $$gcd(29-3i,11+10i)=gcd(29-3i+i(11+10i),11+10i)=gcd(19+8i,11+10i) \\ =gcd(19+8i-(11+10i),11+10i)=gcd(8-2i,11+10i)$$ You can stop here and use the prime decomposition $8-2i=2(4-i)=i(1-i)^2(4-i)$ and check which of these divide $11+10i$. You can also continue with this method until you get "smaller" numbers, or use the norms method as in Bill Dubuque's answer.