Eisenstein integers with norm value of 5

Question

I am trying to find out the Eisenstein integers that have norm values of 3,5,7.... I want to see if there is some pattern.

For the norm value of 3, I was able to find six Eisenstein integers. For the norm value of 7, I was able to find twelve Eisenstein integers.

I am unable to find even a single Eisenstein integer with a norm value of 5. Is this correct? Do Eisenstein integers with a norm value of 5 not exist or am I missing something? Does it have to do anything with the fact that 5 is a prime Eisenstein integer?

If we had $5=(a+b\omega)(a+b\omega^2)$, then $5$ wouldn’t be a prime Eisenstein integer — J. W. Tanner, Feb 18 '24 at 09:08
$(2a-b)^2+3b^2=20$ has no integer solution. Brute force shows it very quickly. — Anne Bauval, Feb 18 '24 at 09:20
@AnneBauval yes, i was not able to find any by brute force either! — Disha, Feb 19 '24 at 15:38
And it is the easiest way: $20$, $20-3$, $20-3\cdot2^2$ are not squares, that's all. — Anne Bauval, Feb 19 '24 at 16:29
@AnneBauval, yes, I see that. I tried it with $(a-b)^2+ab=5$ and then took the possible cases. $(a-b)^2 \neq 5, 3, 2$ so those three cases get rejected. And in the case of $(a-b)^2=1, 4$. I could not find any a and b which satisfy that and $ab=4,1$ consecutively. — Disha, Feb 19 '24 at 16:38

score 0 · Accepted Answer · answered Feb 22 '24 at 16:14

You are not missing anything. There are no Eisenstein integers

$(a+b\omega$ with $a,b\in\mathbb Z$ and $\omega$ a primitive cube root of $1)$ with a norm of $5.$

One way to see this is to note that $N(a+b\omega)=(a+b\omega)(a+b\overline\omega)$ $=a^2-ab+b^2$

$\equiv a^2+2ab+b^2$ $=(a+b)^2\not\equiv2\pmod3,$ as shown here.

Another way to see this, as suggested in comments by Anne Bauval, is to note that

$N(a+b\omega)=a^2-ab+b^2=\left(a-\frac12b\right)^2+\frac34b^2=5\iff(2a-b)^2+3b^2=20,$

which has no solutions in integers, because $20, 20-3\cdot1^2,$ and $20-3\cdot2^2$ are not squares

and $20-3\cdot b^2<0$ for integers $b>2$.

Finally, if we had $N(a+b\omega)=(a+b\omega)(a+b\overline\omega)=5$,

then $5$ would not be a prime Eisenstein integer.

If $a+b\omega$ were a unit, then we'd have $N(a+b\omega)=1$ – J. W. Tanner Feb 23 '24 at 14:53 — J. W. Tanner, Feb 23 '24 at 14:53

Eisenstein integers with norm value of 5

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