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As a result of the answer I got for this question - Irrational solutions to some equations in two variables - I was wondering if the next statement is always true:

Let $x,y$ be real, irrational numbers such that $x+y\ne0$. And let $n_1,n_2,n_3$ be some positive integers (different from each other) such that $\gcd(n_1,n_2,n_3)=1$.

Prove (or find a counter example) that if: $$x^{n_1}+y^{n_1}$$ $$x^{n_2}+y^{n_2}$$ $$x^{n_3}+y^{n_3}$$ are all rational numbers, then also both: $$x+y$$ $$xy$$ have to be rational numbers.

Community
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M.Wulich
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    You at least need $d= \gcd(n_1, n_2, n_3)$ to be $1$. Otherwise something like $x=\sqrt[d]{2},y=\sqrt[d]{3}$ would be a counterexample. – Arthur Oct 03 '14 at 16:46
  • You are right, I'll edit it in a minute. – M.Wulich Oct 03 '14 at 16:52
  • This holds for the triple $(2,3,4)$. – Andrew Dudzik Oct 04 '14 at 00:02
  • And for $(2,3,5)$... well, let's just say that this is of similar difficulty to your last problem. – Andrew Dudzik Oct 04 '14 at 00:32
  • I think this is true – mercio Oct 05 '14 at 12:27
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    The condition that $x+y$ and $xy$ are rational is equivalent to saying that they are the roots of a quadratic with rational coefficients ($z^2 - (x+y)z + xy$ in particular) and hence that, for some rational $p$ and $q$ that $x=p+\sqrt{q}$ and $y=p-\sqrt{q}$. – Milo Brandt Oct 05 '14 at 16:47
  • @mercio Why? I am finding it very hard to have a good intuition about this problem. – Andrew Dudzik Oct 05 '14 at 17:40
  • I suspect that a necessary stepping stone is, if we let $f(n)=x^n+y^n$, there must be some algebraic relation between $f(n_1),f(n_2)$, and $f(n_3)$, since knowing two, the other can only take finitely many values - i.e. a polynomial $P(f(n_1),f(n_2),f(n_3))=0$. It's not obvious to me what this relation must be, since the only relation I know of is $f(a+b)=f(a)f(b)-(xy)^af(b-a)$, but that introduces $(xy)^a$ as a variable, which is troublesome. – Milo Brandt Oct 05 '14 at 17:44
  • @Slade : my anecdotary evidence about all the $n1<n2<n3<10$ cases. Also the regularity of $[K(x^n+y^n,x^m+y^m) : K(x+y,xy)]$ which is $nm$ when one of them is even. – mercio Oct 05 '14 at 17:49
  • I meant $nm/2$ ... – mercio Oct 05 '14 at 17:58

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Sadly, the statement doesn't hold in general. For instance for $(n_1, n_2, n_3) = (2, 3, 8)$ we can choose $$ x = \frac{1}{2}\left(\sqrt{3} - 1 +\sqrt{2}\sqrt[4]{3}\right) $$ and $$ y = \frac{1}{2}\left(\sqrt{3} - 1 -\sqrt{2}\sqrt[4]{3}\right). $$ Now it's easy to see that $\gcd(2,3,8) = 1$, $x^2+y^2 = 2$, $x^3+y^3 = 2$ and $x^8+y^8 = 8$, and both numbers are irrational, but still $$ x+y = \sqrt{3} - 1 $$ and $$ xy = 1- \sqrt{3} $$ are not rational.

gtrrebel
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  • Nicely constructed example. –  Oct 17 '14 at 15:20
  • I agree that this is easy to check (though, I confess I didn't), but how did you discover it to begin with? – Jason DeVito - on hiatus Oct 17 '14 at 15:43
  • I first chose, n1 and n2 to be as simple as possible, then sum of squares and cubes to be also easy to handle. Then I checked for other solutions for resulting system, and found those with Wolfram Alpha. Then I just checked the sum for other small powers and noticed that the 8th powers add up to integer. – gtrrebel Oct 17 '14 at 15:50
  • Actually the OP has pushed his luck: the natural generalization of his previous question would have to consider three consecutive exponents. – user26857 Oct 17 '14 at 15:59
  • @user26857 how do you figure that "the natural" generalization must do so? – Dustan Levenstein Oct 17 '14 at 18:16
  • @DustanLevenstein Maybe because 4,5, and 6 are consecutive? – user26857 Oct 17 '14 at 18:25
  • @user26857 I mean why that particular choice of generalization? Both your version and OP's version are generalizations of the previous question. One is obviously stronger than the other, but I don't see what criterion you're using to say that one is "more natural" than the other. – Dustan Levenstein Oct 17 '14 at 18:48
  • $\Bbb C(x^2+y^2,x^3+y^3,x^8+y^8) = \Bbb C(x+y,xy)$, but when you express $x+y$ (or $xy$) in terms of those $3$ quantities and evaluate with your example, you get $0/0$. I didn't think about that one happening lol. – mercio Oct 17 '14 at 18:54