How many non-rational complex numbers $x$ have the property that $x^n$ and $(x+1)^n$ are rational?

In this answer, I proved the following statement:

For each positive integer $n$, there are finitely many non-rational complex numbers $x$ such that $x^n$ and $(x+1)^n$ are rational. These complex numbers all have the form $$x(a,b)=\frac{\zeta^b-\zeta^a}{1- \zeta^b},$$ where $\zeta$ is a primitive $n$th root of unity and $a,b$ are different nonzero elements of $\mathbb Z/n\mathbb Z$.

However, a search with a computer suggests that there are even finitely many such complex numbers over all $n$. I have counted $26$ of them.

Let $S_n$ be the set of non-rational complex numbers such that $x^n$ and $(x+1)^n$ are rational. Since $x^{nm}$ and $(x+1)^{mn}$ are rational if $x^n$ and $(1+x)^n$ are, it follows that $S_n \subseteq S_{nm}$. Let $S= \varinjlim S_{n!}$. This is the set of all non-rational complex numbers $x$ for which there exists an $n$ such that $x^n$ and $(x+1)^n$ are rational.

Question: Is $S$ finite?

I have checked with a computer that

$$S_{12}=S_{4!} = S_{5!}=S_{6!}$$ and that $S_n \subseteq S_{12}$ for $n<100$.

Question Is $S=S_{12}$? In other words, if $x^n$ and $(x+1)^n$ are rational, is it true that $x^{12}$ and $(x+1)^{12}$ are rational?

Micah has remarked in a comment to David's answer that any $s \in S$ must be a quadratic irrational. So, the finiteness of $S$ would follow if one could prove that finitely many numbers of the form $x(a,b)$ are quadratic irrationals. In order for this to be true, it must be that $x(a,b)^{\sigma^2} = x(a,b)$ for any $\sigma \in \text{Gal}(\mathbb Q(\zeta)/\mathbb Q) = (\mathbb Z/n\mathbb Z)^\times$. This means that we must have

$$\frac{\zeta^{c^2b}-\zeta^{c^2a}}{1- \zeta^{c^2b}} = \frac{\zeta^b-\zeta^a}{1- \zeta^b}$$ for any $c \in (\mathbb Z/n\mathbb Z)^\times$. Perhaps someone can extract an explicit condition on $(a,b)$ from this?...

The set $S_{12}$ consists of $26$ complex numbers. Here is a list of them (here written in terms of a primitive $24$th root of unity $\zeta$):

$$-1/3\zeta^4 - 1/3$$

$$-1/2\zeta^4 - 1/2$$

$$-2/3\zeta^4 - 2/3$$

$$-\zeta^4 - 1$$

$$-1/2\zeta^6 - 1/2$$

$$-\zeta^6 - 1$$

$$\zeta^4$$

$$-1/2\zeta^4$$

$$-\zeta^4$$

$$-2\zeta^4$$

$$2\zeta^4 - 1$$

$$\zeta^6$$

$$2/3\zeta^4 - 1/3$$

$$-2/3\zeta^4 + 1/3$$

$$-\zeta^6$$

$$-2\zeta^4 + 1$$

$$2\zeta^4 - 2$$

$$z^4 - 1$$

$$1/2\zeta^4 - 1/2$$

$$-\zeta^4 + 1$$

$$\zeta^6 - 1$$

$$1/2\zeta^6 - 1/2$$

$$\zeta^4 - 2$$

$$2/3\zeta^4 - 4/3$$

$$1/2\zeta^4 -1$$

$$1/3\zeta^4 - 2/3$$

Here is where they lie in the complex plane:

Here is a brief preview of the behavior of the function $f(n):=\#S_n$, as a plot of $f(n)$ for $n=2$ to $n=100$:

Solution 1:

Note that $$1 + x = \frac{1 - \zeta^{a}}{1 - \zeta^{b}}.$$ We may (and do) assume that $\zeta$ is a primitive $n$th root of unity and that $(a,b) = 1$. We may also assume that $\zeta^a \ne \zeta^b$.

Claim: If $x$ has degree at most $2$ over $\mathbf{Q}$, then $n = 1$, $2$, $3$, $4$, or $6$.

Proof: Let $E = \mathbf{Q}(x)$, and assume that $[E:\mathbf{Q}] \le 2$. Since $E$ is abelian (it is contained in a cyclotomic field), the complex conjugate of $x$ also lies in $E$. The complex conjugate is obtained by replacing $\zeta$ by $\zeta^{-1}$. Hence the ratio $$\frac{1+x}{ 1+\overline{x}} = \zeta^{a - b}$$ also lies in $E$. Let $\alpha = \zeta^{a-b} \ne 1$. Then, solving for $\zeta^b$ in the equation $$1 + x = \frac{1 - \alpha \cdot \zeta^b}{1 - \zeta^b},$$ we find that $$\zeta^b = \frac{x}{1 + x - \alpha} \in E.$$ It follows that $\zeta^b $ and $\zeta^a = \alpha \cdot \zeta^b$ both lie in $E$, and hence that $\zeta \in E$. In particular, the degree of the field $\mathbf{Q}(\zeta)$ is at most $2$, from which it follows that $n$ is one of $1$, $2$, $3$, $4$, or $6$.

More generally, if one assumes that $x$ generates a field of degree $m$, then one can deduce in the same way that $\varphi(n) \le m$, where $\varphi(n) = [\mathbf{Q}(\zeta):\mathbf{Q}]$ is the Euler $\varphi$ function.