Don't know where to use the hypothesis Let a be a real number such that $|a| > 2 $. Prove that if $a^{4}-4a^{2}+2$ and $a^5-5a^3+5a$ are rational numbers, then $a$ is a rational number as well.
My attempt is the following.
$a^5-5a^3+5a = a(a^{4}-4a^{2}+2 -a^2 +3)$. Which can be written as $a(\frac{c}{d}-a^2+3)$ for some $c,d$ in the integers. Note, $a(\frac{c}{d}-a^2+3)= \frac{g}{h}$ by our hypothesis. Now we have $\frac{ac-da^3+3ad}{d}=\frac{g}{h}$. I have played around with expression but cannot get the result.
 A: By doing polynomial division, you can prove that $a^3 = \lambda a + \kappa$, where $\lambda, \kappa \in \mathbb{Q}$. From this, it also follows that $a^2 = \mu a + \nu$ for some $\mu, \nu \in \mathbb{Q}$. The rest follows readily from those observations.

In a little more detail: note that $a^5-5a^3+5a = a(a^4-4a^2+2)-(a^3-3a)$. Let $q_1 := a^5-5a^3+5a \in \mathbb{Q}$ and $q_2 := a^4-4a^2+2 \in \mathbb{Q}$. Then $a^3-3a = aq_2-q_1$, so $a^3 = aq_3-q_1$ with $q_3:=q_2+3 \in \mathbb{Q}$. From this relation we get $$a^4=a^2q_3-aq_1$$ and $$a^5=a^3q_3-a^2q_1 = (aq_3-q_1)q_3-a^2q_1 = -q_1a^2+q_3^2a-q_1q_3.$$
We know that $a^4-4a^2+2 = q_2 \in \mathbb{Q}$, so $a^2q_3-aq_1-4a^2+2 = q_2$, thus $a^2(q_3-4) = aq_1+q_2-2$. We may assume that $q_3 \neq 4$, since $a>2$ implies $q_2>2$ hence $q_3 >5$. Then 
$$a^2 = a\frac{q_1}{q_3-4}+\frac{q_2-2}{q_3-4} = \mu a + \nu.$$
Combine the expressions for $a^5$ and $a^2$ we have found to get
$$a^5 = -q_1(\mu a + \nu)+q_3^2a-q_1q_3 = (q_3^2-q_1\mu)a - q_1(\nu+q_3)$$
Finally, take $a^5-5a^3+5a = q_1$ and substitute the expressions we have:
$$(q_3^2-5q_3-q_1\mu+5)a - q_1(\nu+q_3)+5q_1=q_1,$$
which implies that
$$(q_3^2-5q_3-q_1\mu+5)a = q_1(\nu+q_3-4) \in \mathbb{Q}.$$
There's still a little more work involved to decide whether the coefficient for $a$ is $0$ or not, but you should be able to finish off the problem.
