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.
 A: We may take $x>0$. $\frac{x^5+x}{x^3+x}$ is rational, so $bx^4+b = ax^2 +a$ for some integers $a,b$ and $x = \sqrt{k}$, where $k=c+\sqrt{d}$ is the root of a quadratic equation, with $c,d\in\mathbb{Q}$, $d=0$ or $d>0$ not a square.
If $k$ is rational ($d=0$) but not a square, $x^2-1$ is rational and $x^5+x = (x^3+x)(x^2-1)+2x$ is irrational, a contradiction.
If $k$ is irrational, ($d\neq 0$), then expanding $x^3+x$ we get
$$x^3+x = c^3+3cd+c + (3c^2+d+1)\sqrt{d}$$
is irrational, a contradiction.
A: $$x^5+x^3=ax^2$$
$$x^5+x=b$$
$$x^3-ax^2-x-b=0$$
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.
A: Suppose $a=x(x^2+1)$ and $b=x(x^4+1)$ are rational. 
We may disregard the trivial case when $x=0$, so suppose $x \neq 0$.
Then
$a \neq 0$.
By polynomial long division, we get
$$
b = a(x^2 - 1) +2x.
$$
Thus
$$
a x^2 + 2 x - (a + b) = 0.
$$
Using the quadratic formula, we see $x$ is rational if and only if
$
4+4a(a+b)
$
is a perfect square
as a rational number.
Factoring out the perfect square 4, we see
$x$ is rational if and only if
$1+ab+a^2$ is a perfect square .
Then I noticed
$$
1+ ab + a^2 = 
\left( \frac{1}{x}(a+b)-1 \right)^2,
$$
but I don't know if that's helpful.
Can this really be proved using "elementary" methods? The question is tagged "algebra-precalculus" but can it really be solved using only the tools available to a precalculus student?
