A real number $x$ such that $x^n$ and $(x+1)^n$ are rational is itself rational

Let $x$ be a real number and let $n$ be a positive integer. It is known that both $x^n$ and $(x+1)^n$ are rational. Prove that $x$ is rational.

What I have tried:

Denote $x^n=r$ and $(x+1)^n=s$ with $r$, $s$ rationals. For each $k=0,1,\ldots, n−2$ expand $x^k\cdot(x+1)^n$ and replace $x^n$ by $r$. One thus obtains a linear system with $n−1$ equations with variables $x$, $x^2$, $x^3,\ldots x^{n−1}$. The matrix associated to this system has rational entries, and therefore if the solution is unique it must be rational (via Cramer’s rule). This approach works fine if $n$ is small. The difficulty is to figure out what exactly happens in general.

Answer

Here is a proof which does not require Galois theory.

Write
$$f(z)=z^n-x^n\quad\hbox{and}\quad g(z)=(z+1)^n-(x+1)^n\ .$$
It is clear that these polynomials have rational coefficients and that $x$ is a root of each; therefore each is a multiple of the minimal polynomial of $x$, and every algebraic conjugate of $x$ is a root of both $f$ and $g$. However, if $f(z)=g(z)=0$ then we have
$$|z|=|x|\quad\hbox{and}\quad |z+1|=|x+1|\ ;$$
this can be written as
$$\def\c{\overline} z\c z=x\c x\ ,\quad z\c z+z+\c z+1=x\c x+x+\c x+1$$
which implies that
$${\rm Re}(z)={\rm Re}(x)\ ,\quad {\rm Im}(z)=\pm{\rm Im}(x)=0\tag{*}$$
and so $z=x$. In other words, $f$ and $g$ have no common root except for $x$; so $x$ has no conjugates except for itself, and $x$ must be rational.

As an alternative, the last part of the argument can be seen visually: the roots of $f$ lie on the circle with centre $0$ and radius $|x|$; the roots of $g$ lie on the circle with centre $-1$ and radius $|x+1|$; and from a diagram, these circles intersect only at $x$. Thus, again, $f$ and $g$ have no common root except for $x$.

Observe that the deduction in $(*)$ relies on the fact that $x$ is real: as mentioned in Micah’s comment on the original question, the result need not be true if $x$ is not real.

Comment. A virtually identical argument proves the following: if $n$ is a positive integer, if $a$ is a non-zero rational and if $x^n$ and $(x+a)^n$ are both rational, then $x$ is rational.

Attribution
Source : Link , Question Author : Dan Ismailescu , Answer Author : David