TITLE: Sequence of powers of Gaussian integers capturing all positive integers?
QUESTION [17 upvotes]: Fix a complex number $z=x+iy$ where $x,y\in \mathbf{Z}$ Consider the sequence generated by the powers $$z^0, z^1, z^2, z^3,z^4 \ldots$$ The question is whether it is possible to capture any positive integer as either the real part or the imaginary part of this sequence of the powers of a fixed complex number. From some calculation and reasoning,

it is very easy to see that in order to generate all positive integers, $x$ and $y$  have to be relatively prime;

it appears that the sequence of real parts and the sequence of imaginary parts blows up very early and hence it would not be unwise to conjecture that such a complex number does not exist.
Is there any way to get hold of this problem? Thanks in advance.


My curiosity partly grew out of observations on various polynomial bijection questions e.g., I saw on MathOverflow. I hope that at least the simple statement of the problem appeals to one's thoughts.

REPLY [21 votes]: Here's another, simpler proof:
$$z^5=(x^5-10x^3y^2+5xy^4) + \mathrm{i}(y^5-10y^3x^2+5yx^4)
\stackrel{\mathrm{A}}{=}
x^5+\mathrm{i}y^5
\stackrel{\mathrm{B}}{=}
x+\mathrm{i}y=z\mod 10\;.$$
A is because either $x$ or $y$ must be even, else $z^2$ has only even parts. B is because as a ring $\mathbb{Z}/10\mathbb{Z}$ is isomorphic to $(\mathbb{Z}/5\mathbb{Z})\oplus(\mathbb{Z}/2\mathbb{Z})$ and taking the fifth power is the identity in each component.
Thus the residues modulo $10$ have period $4$, so there are at most $8$ different ones and we can't produce numbers with the remaining two.