Sequence of powers of Gaussian integers capturing all positive integers?

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,

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

2. 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.

Solution 1:

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.

Solution 2:

Update: This proof no longer works as is for the new question about positive integers -- I suspect it could be fixed, but I won't bother since the other one is simpler and not affected by the change.

To generate $1$, we need to have $z^k=a+\mathrm{i}b$, and hence $(\lvert z \rvert^2)^k=a^2+b^2$, with $a=1$ or $b=1$. Since $\lvert z \rvert^2$ is an integer, $k$ cannot be even, for otherwise we would have two squares that differ by $1$, which is only possible if $\lvert z \rvert=1$, which is obviously not a solution.

So $k$ must be odd. Now consider $(x + \mathrm{i}y)^k$ for $k$ odd (and still $a=1$ or $b=1$). Then all terms in the real part contain a factor $x$ and all terms in the imaginary part contain a factor $y$. Since one of these parts is $1$, it follows that one of $x$ and $y$ is $\pm 1$. They cannot both be $\pm 1$, since that is obviously not a solution. Denote the one that isn't $\pm1$ by $w$. (Also clearly $w\neq0$.)

Now consider $(x + \mathrm{i}y)^k$ for arbitrary $k$. All terms in the real and imaginary parts are either $\pm 1$ or multiples of $w$. Thus the only residues modulo $w$ that occur are $\pm1$ and $0$. It follows that $\lvert w \rvert \le 3$, i.e. $w=\pm3$ or $w=\pm2$. Both cases are easily excluded: With $w=\pm3$, both parts of $z^2$ are even, and hence we cannot generate any odd integers other than $\pm1$ and $\pm3$. With $w=\pm2$, the residues modulo 5 of the real and imaginary parts are periodic in $k$ with period $4$ (period $2$ if we drop their sign) and none of the residues in the period is $0$, and hence we cannot generate any multiples of $5$. $\Box$

P.S.: I just realized that I killed this proof when I proposed to save the question by including $z^0$ :-). Fortunately the proof can be saved just as easily; just replace $1$ by $-1$.

P.P.S.: This avenue has now also been closed by the restriction to positive integers.