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The next statement is a conjecture of mine, so I dont know if it's true (though quite sure):

Let $x,y$ be irrational numbers such that $x^4+y^4=1$. Prove (or disprove) that $x^5+y^5$ is irrational or $x^6+y^6$ is irrational (or both of them are irrational).

Edit: just to remove any doubt, my meaning is to prove that at least one of the numbers $x^5+y^5$ and $x^6+y^6$ has to be irrational.

M.Wulich
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  • I hope I'm overlooking something, but this problem seems deviously difficult. In principle, it comes down to finding rational points on a planar curve. If there are counterexamples, there should be only finitely many. – Andrew Dudzik Oct 02 '14 at 21:14
  • well there is a relation between $x^4+y^4, x^5+y^5, x^6+y^6$ whose degree (in $x$ and $y$) is at most $222$. So yeah this doesn't look easy. Also I'm curious about the degree of $\Bbb C(x^4+y^4,x^5+y^5,x^6+y^6)$ over $\Bbb C(x+y,xy)$ – mercio Oct 02 '14 at 21:26
  • @mercio I've undeleted my partial answer below, since it might have some calculations of interest in this regard. – Andrew Dudzik Oct 02 '14 at 21:35
  • Actually the smallest relation only has degree $60$ in $x,y$. This gives a big polynomial relation of degree $10$ in $x^6+y^6$ and degree $12$ in $x^5+y^5$, of which we want a rational point. – mercio Oct 02 '14 at 22:11
  • though there always is the trivial solution $x^n+y^n = 1$ for $n=4,5,6$. Maybe this one can give a counter-examples ? I'm not sure (I have no idea of the degree of the field extension) – mercio Oct 02 '14 at 22:27
  • @mercio Just by dogged elimination, it looks like $x+y$ and $xy$ live in a cubic extension of $\mathbb{Q}$. So given the values of $x^5+y^5$ and $x^6+y^6$, we're looking at a degree $6$ extension at worst. Just setting them to $1$ doesn't work—it just gives the trivial solution—but maybe I can get a counterexample with some guessing and crunching. – Andrew Dudzik Oct 03 '14 at 06:08

1 Answers1

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It turns out that $K(x^4+y^4,x^5+y^5,x^6+y^6) = K(x+y,xy)$.

Letting $a = x^4+y^4, b = x^5+y^5, c = x^6+y^6$, we have the identities

$$x+y = - \frac{4c^{6}b-10c^{4}ba^{3}+80c^{3}b^{3}a^{2}-100c^{2}b^{5}a+14c^{2}ba^{6}+16cb^{7}-68cb^{3}a^{5}+70b^{5}a^{4}-6ba^{9}}{2c^{6}a-4c^{5}b^{2}-5c^{4}a^{4}+70c^{3}b^{2}a^{3}-170c^{2}b^{4}a^{2}+c^{2}a^{7}+128cb^{6}a+28cb^{2}a^{6}-16b^{8}-37b^{4}a^{5}+3a^{10}} \\ xy = - \frac{-6c^{6}a+12c^{5}b^{2}+5c^{4}a^{4}-50c^{3}b^{2}a^{3}+90c^{2}b^{4}a^{2}-5c^{2}a^{7}-64cb^{6}a+8b^{8}+9b^{4}a^{5}+a^{10}}{-8c^{5}a^{2}+40c^{4}b^{2}a-40c^{3}b^{4}+8c^{3}a^{5}-60c^{2}b^{2}a^{4}+100cb^{4}a^{3}-4ca^{8}-32b^{6}a^{2}-4b^{2}a^{7}} $$

So $a,b,c$ are rational if and only if $x+y$ and $xy$ are rational. Then $x,y$ are irrational whenever $(x-y)^2 = (x+y)^2-4xy$ is not a rational square (and $x,y$ will be in a quadratic extension of $\Bbb Q$)

Since $x^4+y^4 = (x+y)^4 - 4(x+y)^2xy + 2x^2y^2$, to get a counter-example we want to find rational points on the curve $1 = s^4-4s^2p+2p^2$, which is an elliptic curve.

We have the obvious points $s= \pm 1,p=0$, which gives rational $x,y$.

We also have the solutions $s = \pm 1, p =2$. This gives counter examples $x,y = \frac{1\pm \sqrt{-7}}2$ giving $a=1,b=11,c=9$ ; and $x,y = \frac{-1\pm \sqrt{-7}}2$ giving $a=1,b=-11,c=9$

There is a rational parametrization sending the curve to the curve $E : y^2 = x^3- x$, and it is known that $E(\Bbb Q) = (\Bbb Z/2\Bbb Z)^2$. The $4$ points I've given are thus the only $4$ rational points, so there isn't any real counter-example.

However, if you allow to replace $\Bbb Q$ with some quadratic extension $K$ of $\Bbb Q$, you can easily find some real counter-example to that more general statement, simply by picking some small $s$ close to $\pm 1$ and then solving for $p$ and choosing the solution close to $0$ (actually the region of the curve where $s^2 > 4p$ is quite small). Most of those $(s,p )$ should correspond to couples $(x,y)$ lying in turn in a quadratic extension of $K$.

mercio
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  • Well done! (though I have obviously not checked your equations) Doesn't that curve have genus $3$, though? – Andrew Dudzik Oct 03 '14 at 10:16
  • Nevermind, I think I see a singularity at $\infty$ and I think it has delta invariant $2$. Was there a more immediate way to see it? – Andrew Dudzik Oct 03 '14 at 10:28
  • @Slade well the curve can be put (over $\Bbb C$) in the form $y^2 = 1 + x^4$, and this kind of hyperelliptic curve is an elliptic curve. The big polynomial of degree $15$ in $a,b,c$ (which I didn't write down) also describes the same elliptic curve, though it has quite a lot of singularities to make up for the high degree. – mercio Oct 03 '14 at 13:29
  • and if I'm not mistaken the curve has complex multiplication by $i$. I have no idea if we have the full group of rational points though – mercio Oct 03 '14 at 13:41
  • @mercio First I highly appreciate your effort in trying to solve this question, thank you ! But the counter examples you gave are complex numbers, which are not considered irrational, aren't they ? However, can it be derived from your proof that if we seek only real irrational solutions, then there are none ? Or on the other hand, is it possible to manipulate the $x,y$ you gave to be real numbers ? – M.Wulich Oct 03 '14 at 16:09
  • Well irrational means not rational. Clearly those nonreal complex numbers are not rationals. So they are irrational. To get real counterexamples I would have to go and find more points on the curve. I might do that in a few days if noone else wants to do it. – mercio Oct 03 '14 at 16:13
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    @M.Wulich Requiring that the answers be real, given that $x+y$ and $xy$ are, amounts to the further condition $(x+y)^2 \geq 4xy$. Semialgebraic conditions are much more difficult to work with than algebraic ones, so mercio's approach is almost certainly the right one: start figuring out the rational points on the elliptic curve mentioned, then sort out which ones give real solutions later. I agree that a real solution would be interesting, but having found a complex one is definitely an achievement (especially because, in principle, more can be generated with the group law). – Andrew Dudzik Oct 04 '14 at 00:22
  • @mercio: This is a my conjecture which is some what similar to this question. Can you check it. I think my current knowledge is not enough for this. Thank you. http://math.stackexchange.com/questions/956415/is-x1-frac1n-1-x1-frac1n-always-irrational – Bumblebee Oct 05 '14 at 09:41