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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.

Bumblebee
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5 Answers5

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

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

MPW
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    Oooh. How about $n>2.$ – Bumblebee Oct 03 '14 at 04:03
  • Shall I edit the question? – Bumblebee Oct 03 '14 at 04:04
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    Normally I don't like question edits that invalidate good answers, but the existence of Pythagorean triples does trivialize the problem for $n=2$, and it's a somewhat deep problem for $n\geq 3$. – Andrew Dudzik Oct 03 '14 at 09:48
  • @Nilan: I don't know if it's true for $n>2$ or not. Do you have some reason to think it is true? Are you asking for help proving this, or for help discovering whether or not it's true? – MPW Oct 03 '14 at 15:23
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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.

Ewan Delanoy
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  • It would be nice to see a simpler proof of this fact, but this deals with the problem quite nicely. – Andrew Dudzik Oct 13 '14 at 08:18
  • @Slade in fact Boreico’s argument is not very complicated, and an ad-hoc reformulation could be made here, though I don’t have the time to write that out now. – Ewan Delanoy Oct 13 '14 at 08:23
  • Perhaps I should have said "more digestible". The ideas do look simple, but I need some time to process exactly what is going on. – Andrew Dudzik Oct 13 '14 at 08:26
  • @EwanDelanoy: I get the paper that you gave. It will take a little time to me understand that:) Thanks for your help. – Bumblebee Oct 13 '14 at 10:39
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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.

Jack M
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  • If $x^{1/n},(1-x)^{1/n}\in\mathbb{Q}$, then this is surely correct. But is it at all obvious that this follows from $x,y,x^{1-1/n}+y^{1-1/n}\in\mathbb{Q}$? – Andrew Dudzik Oct 03 '14 at 08:45
  • @Slade You're right - this started out as a proof that there are no counterexamples for $n>2$ of that specific type, and then I forgot what I was trying to do. There may be counterexamples of a more general form. – Jack M Oct 03 '14 at 09:00
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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.

Ryan Dougherty
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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.

marty cohen
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  • How does this answer the question? Note that the OP is considering rational $x$, and $y = 1 - x$. Furthermore $x^{\frac{n-1}{n}} + y^{\frac{n-1}{n}} = z$ does not imply $x^{n-1} + y^{n-1} = z^n$ (although this seems to be the $3^\text{rd}$ time this mistake has been made on this page) – zcn Oct 03 '14 at 06:17