Is $x^{1-\frac{1}{n}}+ (1-x)^{1-\frac{1}{n}}$ always irrational if $x$ is rational?

Question

Let $x$ be rational with $0<x<1$ and let $y$ be the rational defined by $y = 1 - x.$ Let $n$ be any natural number with $n>2.$ Then I want to prove that $$x^{(1-1/n)}+ y^{(1-1/n)}$$ will never be a rational.

Here is my attempt:
Let $x=\dfrac{a}{c}$ and $y=\dfrac{b}{c}$ where $a,b$ are positive integers such that $a+b=c.$
Then it is equivalent to show that the following number is irrational, $$a\Big(1+\frac{b}{a}\Big)^{1/n}+b\Big(1+\frac{a}{b}\Big)^{1/n},∀n\in\mathbb N\setminus\{1\}.$$ After this I was stuck. How can I continue? Hints are also welcome.

Answer 1

Let $u=x^{n-1}$ and $v=y^{n-1}$ By Boreico’s theorem (see section "Higher powers" on pages 91-92 at http://www.thehcmr.org/issue2_1/mfp.pdf), $u^{\frac{1}{n}}+v^{\frac{1}{n}}$ is rational iff $u^{\frac{1}{n}}$ and $v^{\frac{1}{n}}$ are both rational. Since $n-1$ and $n$ are coprime, this is equivalent to $x^{\frac{1}{n}}$ and $y^{\frac{1}{n}}$ being both rational. As Jack M remarked, your conjecture is then equivalent to Fermat’s last theorem.

Answer 2

Your conjectured statement isn't true. Take $x=\frac{36}{100}$ and $n=2$. Then

$$x^{1-\frac1n}+y^{1-\frac1n} =\left(\tfrac{36}{100}\right)^{\tfrac12}+\left(\tfrac{64}{100}\right)^{\tfrac12}=\tfrac75 $$

Answer 3

This is not an answer, but some insight and too long for a comment. I would be inclined to say yes, for the following reason. The equation above is equivalent to:

$x^{n-1} + (1-x)^{n-1} = z^n$.

Let $z = \frac{p}{q}$ with $p, q \in \mathbb{Z}$. Therefore,

$f(x) = q^nx^{n-1} + q^n(1-x)^{n-1} - p^n$

is a polynomial with integer coefficients. By the Rational root theorem, this means that there exists a rational solution for $f(x) = 0$.

However, since the domain is restricted to $(0, 1)$, this may not be correct. But there exists a rational solution for general choice of x and n.

Answer 4

If $m$ and $n$ are relatively prime positive integers, there are an infinite number of solutions to $x^n + y^n = z^m$ in positive integers.

This can be done using the fact that there are an infinite number of integer solutions to $an-bm = 1$.

In this case, since $n = m-1$, they are obviously relatively prime.

Answer 5

Your conjecture that this has no solutions (when $n>2$) is actually equivalent to Fermat's last theorem. You just need to find an $x$ such that both $x$ and $1-x$ are rational $n$-th powers. If $x=\frac {a^n} {b^n}$, then $1-x=\frac {b^n - a^n} {b^n}$, so we're trying to solve $b^n-a^n=c^n\iff a^n+c^n=b^n$. By Fermat's last theorem this has no solutions when $n>2$ other than the ones in which $x$ is either $0$ or $1$. When $n=2$ it does, however, and the usual method for generating pythagorean triples will get you all of them.