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.

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Nice question! It seems one should first figure out whether the liminf of the absolute values of the real or imaginary parts is finite or infinite. – Qiaochu Yuan Mar 13 '11 at 0:01
@Qiaochu Yuan: I was thinking more along the lines of "density", i.e. since the absolute value grows exponentially the range of angles available to produce "small" integers shrinks exponentially, so it might not be possible to produce them all without violating equidistribution of the angles. – joriki Mar 13 '11 at 0:03
@joriki: yes, that's what I mean. I don't know how small the relevant cosines can get relative to the powers of the absolute value and this already seems like an interesting question. – Qiaochu Yuan Mar 13 '11 at 0:06
@Qiaochu Yuan: It seems to me these are slightly different things. It might be possible to get arbitrarily small integers arbitrarily late in the sequence (this must have something to do with irrationality measures) and yet it might still not be possible to hit them all because that would violate the equidistribution of angles. – joriki Mar 13 '11 at 0:15
@joriki: yes, but it's a closely related question. If one could show a strong enough negative version (say, it escapes exponentially fast to infinity) then the problem would be resolved. – Qiaochu Yuan Mar 13 '11 at 0:19

2 Answers

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.

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I wish I could say more than thank you. – Chulumba Mar 13 '11 at 10:40
@Chulumba: You're quite welcome :-) – joriki Mar 13 '11 at 10:52
@joriki Very Nice Solution. I would like to ask you about what happens if we want any even positive integer be either the real part or the imaginary part of that sequence of the powers ?! – Nemes Aug 7 '13 at 13:17
@Nemes: That isn't possible, either. Consider the residues mod $2$. If $z=0+0\mathrm i\bmod2$ or $z=1+1\mathrm i\bmod2$, then $z^2=0\bmod2$, and the exponentially accumulating factors of $2$ prevent you from hitting all even integers. On the other hand, if $z=0+1\mathrm i\bmod2$ or $z=1+0\mathrm i\bmod2$, then each power has one even part and one odd part, so due to the period $4$ in the residues mod $10$ you miss at least one of the five even residues mod $10$. – joriki Aug 8 '13 at 7:41
@joriki Thats right, thanks you. Fix a Gaussian integer $z$. I am wondering if there is any way to estimate the density of the set $\lbrace |Im(z^n)|, |Re(z^n)| \; | \; n = 1,2,...\rbrace$. I mean how large this density could be !? – Nemes Aug 8 '13 at 12:12

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.

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thanks very much for the answer. Do you think it would be fair if we edited the question to require only whether is is possible to get only positive integers. In that case, $1$ will be done away with as you suggested by including $z^0$ into the sequence. Let's me try. – Chulumba Mar 13 '11 at 9:47
@Chulumba: Feel free to do that, as I was about to post a simpler proof that will still solve that new question :-) – joriki Mar 13 '11 at 9:57