By inspection we see that the divisors of $6$ are $1,2,3,6$
For $4+2\sqrt{5}$ we have $4+2\sqrt{5}=2(2+\sqrt{5})$
showing that $\gcd(6,4+2\sqrt{5})=2$
Is this method correct; if not, how can I do this successfully?
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It seems right. – Apr 05 '14 at 15:16
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could there be a complex number like $1+i$ or a rational number of the form $a+b\sqrt{c}$ that divides this two,this question was worth 17 points in my number theory exam,i wonder if this is all to be done – Jonas Kgomo Apr 05 '14 at 15:23
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@Jonas12 Have you already studied basic number theory of quadratic number fields? Please give some context. – Bill Dubuque Apr 05 '14 at 15:49
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yes i have studied quadratic number fields,it is in the context of rational numbers $D$ and complex numbers $s+t\sqrt{D}$ – Jonas Kgomo Apr 05 '14 at 15:58
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only basics,i havent went much further – Jonas Kgomo Apr 05 '14 at 16:07