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. 
 A: 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.
A: Let $x$ be a real number such that $x^n$ and $(x+1)^n$ are rational. Without loss of generality, I assume that $x \neq 0$ and $n>1$.
Let $F = \mathbb Q(x, \zeta)$ be the subfield of $\mathbb C$ generated by $x$ and a primitive $n$-th root of unity $\zeta$. 
It is not difficult to see that $F/\mathbb Q$ is Galois; indeed, $F$ is the splitting field of the polynomial $$X^n - x^n \in \mathbb Q[X].$$ The conjugates of $x$ in $F$ are all of the form $\zeta^a x$ for some powers $\zeta^a$ of $\zeta$. Indeed, if $x'$ is another root of $X^n - x^n,$ in $F$, then
$$(x/x')^n = x^n/x^n = 1$$
so that $x/x'$ is an $n$-th root of unity, which is necessarily of the form $\zeta^a$ for some $a \in \mathbb Z/n\mathbb Z$.
Assume now that $x$ is not rational. Then, since $F$ is Galois, there exists an automorphism $\sigma$ of $F$ such that $x^\sigma \neq x$. Choose any such automorphism. By the above remark, we can write $x^\sigma = \zeta^a x$, for some $a \not \equiv 0 \pmod n$. 
Since also $(1+x)^\sigma = 1+ x^\sigma \neq 1 + x$, and since $(1+x)^n$ is also rational, the same argument applied to $1+x$ shows that $(1+x)^\sigma = \zeta^b (1+x)$ for some $b \not \equiv 0 \pmod n$. And since $\sigma$ is a field automorphism it follows that
$$\zeta^b (1+x) = (1+x)^\sigma = 1 + x^\sigma = 1 + \zeta^a x$$
and therefore
$$x(\zeta^b - \zeta^a) = 1 - \zeta^b.$$
But $\zeta^b \neq 1$ because $b \not \equiv 0\pmod n$, so we may divide by $1-\zeta^b$ to get
$$x^{-1} = \frac{\zeta^b - \zeta^a}{1-\zeta^b}.$$
Now, since $x$ is real, this complex number is invariant under complex conjugation, hence
$$x^{-1} = \frac{\zeta^b - \zeta^a}{1-\zeta^b} = \frac{\zeta^{-b} - \zeta^{-a}}{1-\zeta^{-b}} = \frac{1 - \zeta^{b-a}}{\zeta^b-1} = \zeta^{-a}\frac{\zeta^a - \zeta^b}{\zeta^b-1} = \zeta^{-a} x^{-1}.$$
But this implies that $\zeta^a = 1$, which contradicts that $x^\sigma \neq x$. So we are done. $\qquad \blacksquare$
The following stronger statement actually follows from the proof:

For each $n$, there are finitely many non-rational complex numbers $x$ such that $x^n$ and $(x+1)^n$ are rational. These complex numbers belong to the cyclotomic field $\mathbb Q(\zeta_n)$, and none of them are real. 

Indeed, there are finitely many choices for $a$ and $b$.
In this related answer, Tenuous Puffin proves that there exist only $26$ real numbers having this property, allowing for any value of $n$.
A: It seems the following.
We can prove that $x$ is rational provided  $n$ is a power of an odd prime $p$. Denote $x^n=r$ and $(x+1)^n=s$ with $r$, $s$ rationals. If both $r$ and $s$ are $p$-th powers of  rational numbers, then we descent from $n$ to its $p$-th root. So without loss of generality we can assume that one of the numbers $r$ and $s$ (for instance, $r$) is not a $p$-th power of a rational number. This answer implies that in this case a polynomial $x^n-r$ is irreducible over a field $\mathbb Q$. But the polynomial $x^n-r$ has a common root $x$ with a polynomial $(x+1)^n-s$, so $0<\deg GCD(x^n-r, (x+1)^n-s) <n$, a contradiction.
Concerning the general case, we have $\sqrt[n]s-\sqrt[n]r -1=0$. So the positive answer immediately follows from a general 
Conjecture. Let $m,n>1$ be integers and $a_1,\dots,a_m>1$ be mutually different integers such that no $a_i$ is divisible by an $n$-th power. Then the numbers  $1, \sqrt[n]{a_1},\dots, \sqrt[n]{a_m}$ are linearly independent over $\mathbb Q$. 
Some years ago I proved a similar conjecture for $n=2$ by means of Galois theory. A general conjecture may be a separate question for MSE or  MO. 
A: Here's a hybrid approach which don't require much insight (possibly exact same as the first solution ?).
Assume $x \not \in \mathbb{Q}$, and let $F = \mathbb{Q}(x, \zeta_n)$. Then $F$ is a galois extension of $\mathbb{Q}$ (because it's the splitting field of $y^n - x^n$), and $F$ is a nontrivial group. As $x \not \in \mathbb{Q}$, there exists a nontrivial permutation $\sigma \in \operatorname{Gal}(F/\mathbb{Q})$, such that $\sigma(x) \neq x$. As $\sigma$ permutes the roots of $y^n - x^n$, $\sigma(x)$ must be of the form $\zeta_n^a x$ for some constant $a$. Let $\tau = \zeta_n^a \neq 1$. 
Since $(x+1)^n$ is also rational, it's invariant under $\sigma$, so we get 
\begin{align}
(x+1)^n &= \sigma((x+1)^n) = (\sigma(x)+1)^n = (\tau x +1)^n 
\\ \Rightarrow (x+1)^2 
 =||x+1||^2 &= || \tau x + 1 ||^2 = (\tau x + 1) \overline{(\tau x + 1)} = (\tau x + 1)(1 + \frac{x}{\tau})  
\\ \Rightarrow x^2+2x+1 &= x^2+1+x(\tau+\frac{1}{\tau})
\\ \Rightarrow \tau^2 - 2 \tau + 1 &= 0 \Rightarrow \tau = 1
\end{align}. 
A contradiction ! So $x \in \mathbb{Q}$. 
