<s>[INST] Prove that $x\in\mathbb Q$ Let $a\in\mathbb Q$ and $a>\dfrac43$. Let $x\in\mathbb R$ and $x^2-ax,x^3-ax\in\mathbb Q$. Prove that $x\in\mathbb Q$.
EDIT:
Thsi is my attempt:
Let $x^2-ax=b$ and $x^3-ax=q$ for some $b,q\in\mathbb{Q}$. Then I tried to write $x^2$ and $x^3$ in linear terms. I got $x^2=ax+b$ and substituting in second equation I got
$$
x\cdot x^2-ax=q\\
x(ax+b)-ax=q\\
ax^2+bx-ax=q\\
a(ax+b)+bx-ax=q\\
a^2x+ab+bx-ax=q
$$
But, what next? Since it is known that $a>\dfrac43$ I know that I must use discriminant of some quadratic equation, but how to get quadratic equation from here?
 [/INST] Put $x^2-ax=b$ and $x^3-ax=c$ with $b,c\in \mathbb{Q}$. As $x\in \mathbb{R}$ is a real solution of $y^2-ay-b=0$, we have $D=a^2+4b\geq 0$. Now:
$$x^3=xx^2=x(ax+b)=ax^2+bx=a^2x+ab+bx=(a^2+b)x+ab$$
This is also $ax+c$. If we suppose that $x\not \in \mathbb{Q}$, we get that $a=a^2+b$. Hence $D=a^2+4b=4a-3a^2<0$ as $\displaystyle a>\frac{4}{3}$, a contradiction.
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