A number being both Gaussian integer and Eisenstein integer If a complex number is Gaussian integer and Eisenstei integer at the same time, is it true that the number can only be ordinary integer? Why?

Such a number would lie in $\Bbb Q(i)\cap\Bbb Q(i\sqrt3)$, but that is $\Bbb Q$.
Otherwise, $\Bbb Q(i)$ would equal $\Bbb Q(i\sqrt3)$, since their intersection
must then be a $\Bbb Q$-vector space of dimension $\ge2$. Then $-3$ would be
a square in $\Bbb Q(i)$: $-3=(a+bi)^2=a^2-b^2+2abi$, $a$, $b\in\Bbb Q$.
Either $a=0$ or $b=0$ so that $-3=a^2$ or $-3=b^2$. Both are impossible.
The same argument shows that $\Bbb Q(\sqrt r)\cap\Bbb Q(\sqrt s)=\Bbb Q$
for non-square $r$, $s\in\Bbb Q$ unless $rs$ is a square in $\Bbb Q$.
