# 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.

• Hi. Do you have any idea whether this is actually true? Dec 15, 2014 at 23:27
• If this is true, the fact that $x$ is real must be crucial. Note that if $x$ is a primitive cube root of unity, then $x^6=(x+1)^6=1$. This makes me skeptical that there's any simple algebraic proof... Dec 15, 2014 at 23:40
• It is very strange that someone wants to close a question with 16 upvotes and 7 stars. Dec 16, 2014 at 0:02
• @RobertLewis, "PSQ" stands for "Problem Statement Question," which is generally discouraged at MSE (and probably explains the votes to close). The OP, however, has edited the question to show that he's given the problem some thought. Dec 16, 2014 at 0:18
• What is the source and context of this problem? Number theory textbook? Algebra textbook? Contest math?
– mrf
Dec 16, 2014 at 8:32

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.

• Very nice David! This is probably the simplest proof. Dec 17, 2014 at 0:47
• This is great. Note that it generalizes to show that $x$ can be at worst a quadratic irrational even if it's not real... Dec 17, 2014 at 2:09
• Splendid argument! Thank you very much. Dec 17, 2014 at 15:39

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 $n$.

• Fantastic proof. This is just a basic statement about real and rational numbers, and it can be solved with Galois theory. On the other hand, I am pretty confident that we can do essentially the same argument in more elementary language, without using any abstract algebra. We "just" have to show that every complex number $x$ such that $(x+1)^n$ and $x^n$ are real numbers already is also such a fraction containing roots of unity. Dec 16, 2014 at 10:45
• Thanks @Martin! I, too, would love to see a proof using only high-school mathematics. I'm sure there is one, but I don't see it at the moment. Dec 16, 2014 at 10:48
• Lovely.${{{}}}$ Dec 16, 2014 at 21:48
• @BrunoJoyal For a proof using not much more than high-school mathematics, see my answer. Dec 17, 2014 at 0:41
• I love it. This post should go to your list "Some fun answers of mine" in your profile page :) Dec 17, 2014 at 0:54

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.

• I've seen this Conjecture (or something very similar), including a proof, on www.artofproblemsolving.com some years ago. Dec 16, 2014 at 9:23
• Hi Alex, where do you use the assumption that $x$ is real? Dec 16, 2014 at 9:47
• @MartinBrandenburg Thanks. Dec 16, 2014 at 9:55
• @BrunoJoyal Hi. I did not use this assumption. So in Micah's comment should be essential that 6 is not a power of a prime. Dec 16, 2014 at 10:00
• Actually, if $x$ is a cube root of unity then $(x+1)^3=-1$, so you'd better be assuming that $x$ is real somewhere. And indeed you are -- your "without loss of generality we can take $p$th roots" claim only works if $p$th roots are well-defined... Dec 17, 2014 at 2:08

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 $$x \in \mathbb{Q}$$.