Irrational solutions to some equations in two variables 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.

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 $K$.
