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Factor $3+\sqrt{3}$ into irreducibles in $\mathbb{Z}[\sqrt{3}]$.

I have done the norm which is $6= 2 \cdot 3$. And i have tried two find $a$ and $b$ such that the norm of $a+b\sqrt{3}$ is $2$ and similarly with $3$, but then i get that there are no solutions (and that means that it is irreducible?) so I think I may be doing something wrong. I am not quite sure how to solve this.

Thank you Sorry for the format i am not used to it-

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  • $\begingroup$ No problem, the \$ should enclose the expressions, so you would use $\sqrt{3}$ for $\sqrt{3}$ instead of just $\sqrt{3} $\endgroup$ – RGS Jan 27 '18 at 15:22
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$$3 + \sqrt{3} = (1 - \sqrt{3})(-3 - 2\sqrt{3})$$

This is one possible factorization of $3 + \sqrt{3}$ into irreducibles. Noting that the first factor has norm $2$ and the second one $3$, proves that the factors on the RHS are indeed irreducibles.

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  • $\begingroup$ The norms are $-2$ and $-3$. $\endgroup$ – lhf Jan 27 '18 at 15:32
  • $\begingroup$ @lhf I tend to use $N(a + b\sqrt{3}) = |a^2 - 3b^2|$. Considering the absolute value changes nothing, right? $\endgroup$ – Stefan4024 Jan 27 '18 at 15:34
  • $\begingroup$ thank you, but how do you find such numbers , for example, for a^2-3b^2=2 ,how do you get to the solution ? trying which one fits? $\endgroup$ – angela Jan 27 '18 at 15:39
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    $\begingroup$ @AngelaIg More or less yes. I tried to solve the equation $a^2 - 3b^2 = \pm 2$ and that's how I found the factorization. $\endgroup$ – Stefan4024 Jan 27 '18 at 15:49
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You need to consider factors with norm $\pm 2, \pm 3$.

Anyway, it is easy to find a factorization of $3+\sqrt{3}$ by inspection: $$ 3+\sqrt{3} = \sqrt{3}(\sqrt{3}+1)=(0+\sqrt{3})(1+\sqrt{3}) $$ The first factor has norm $-3$ and the second factor has norm $-2$. Since the norms are prime, the factors are irreducible.

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