<s>[INST] Prove that x is rational Let $x$ be a real number with the properties that $x^3+x$ and $x^5+x$ are rational.
Prove that $x$ is rational. Denote $a=x^3+x$; $b=x^5+x$. We can multiply and add them together until we get the desired result. I also know some non-elementary proofs of this, but have you some nice elementary proofs?
Thank you.
 [/INST] $$x^5+x^3=ax^2$$
$$x^5+x=b$$
$$x^3-ax^2-x-b=0$$
If $a=0$ then the only option is $b=0$ and $x=0$. Assume $a\neq0$.
$$x^3+x=a$$
$$a-x-ax^2-x-b=0$$
$$ax^2+2x+b-a=0$$
So we get that $$x=\frac{-2\pm2\sqrt{1-a(b-a)}}{2a}$$
If $\sqrt{1-a(b-a)}$ is rational we are done. Assume it is not. Plug in $x^3+x=a$.
We get $$\frac{-4}{a^3}-\frac{4}{a}-a+\frac{3 b}{a^2}+\frac{4\sqrt{1-a (-a+b)}}{a^3}+\frac{2\sqrt{1-a (-a+b)}}{a}-\frac{b\sqrt{1-a (-a+b)}}{a^2}$$
So, $0=\frac{4}{a^3}+\frac{2}{a}-\frac{b}{a^2}=\frac{4+2a^2-ba}{a^3}$, i.e. $4+2a^2-ba=0$.
We should also have $0=\frac{-4}{a^3}-\frac{4}{a}-a+\frac{3 b}{a^2}=\frac{-4-4a^2-a^3+3ab}{a^3}$, i.e. $-4-4a^2-a^3+3ab=0$
I am lazy. Solve for $a$ and $b$ from these two equations (cubic in $a$ so it is easy). Then solve for $x$ (the cubic so it is easy). Check if the irrational solutions for $x$ satisfy the quintic with the value of $b$. Done.
</s>