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Suppose we have the domain D={a+b$\sqrt{10}$|a,b $\in \mathbb{Z}$}. I want to show this is not a UFD and I am given the hint to find two factorization of 6 using the norm N(a+b$\sqrt{10}$)=$a^2-10b^2$. I know the norm of 6 is 36. But to find the factors of 6 do I just proceed by brute force?

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  • $\begingroup$ In this case wouldn't that identity just be the norm? $\endgroup$ – user227630 Mar 31 '15 at 0:47
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Hint 1 :Note that $N(xy)=N(x)N(y)$. So $N(x)$ divides $N(xy)$. Try to find a solution to $N(x)=6$.

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