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Say we are given the norm of an Eisenstein integer $N(f)=7$. How do we actually find the integer?

The norm for any Eisenstein integer is defined as $$N(f)=(a+b\epsilon)(a+b\bar\epsilon)=a^2-ab+b^2$$ where $\epsilon=-\frac{1}{2}+\frac{\sqrt{3}i}{2}$

Is there any algorithm for going ''backwards''?

Thanks

user26857
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H.E
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  • Except if the norm is $0$, there are several Eisenstein integers with that norm. Would any of these suffice, or are you looking for a specific one? – Daniel Fischer Feb 28 '14 at 20:28
  • I am trying to find out how to determine these: $f=2+3\epsilon$, $f=3+2\epsilon$ and $f=3+\epsilon$. All of them have norm 7. – H.E Feb 28 '14 at 20:34
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    Well, you can look at $$(2a-b)^2 + 3b^2 = 28,$$ which gives you the options $b = \pm 1,,\pm 2,,\pm 3$. The corresponding values for $2a-b$ are $\pm 5,, \pm 4, , \pm 1$. Then combine. – Daniel Fischer Feb 28 '14 at 20:38
  • @DanielFischer where did you get $(2a-b)^2+3b^2=28$ from? – H.E Feb 28 '14 at 20:45
  • You start with $a^2 - ab + b^2 = 7$. Multiply by $4$, get $4a^2 - 4ab + b^2 + 3b^2 = 28$. I find it easier to work with it in that form. – Daniel Fischer Feb 28 '14 at 20:47
  • Ah yes, brilliant. Thank you – H.E Feb 28 '14 at 20:54

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