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 $x\in\mathbb Q$.

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?

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    $\begingroup$ What's you thought on this question? Did you make any progress? $\endgroup$ – Stefan4024 Aug 18 '14 at 15:26
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    $\begingroup$ If you would have shown your work, I would have upvoted your question $\endgroup$ – Koenraad van Duin Aug 18 '14 at 18:07
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    $\begingroup$ More to the point, where did you get the question? $\endgroup$ – Will Jagy Aug 18 '14 at 18:19
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    $\begingroup$ @KoenraadvanDuin: Isn't answering a question more of an endorsement than upvoting? $\endgroup$ – Jonas Meyer Aug 18 '14 at 18:27
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    $\begingroup$ @Jonas Meyer, I find that the question interesting, but the way it is shown does not deserve an upvote. $\endgroup$ – Koenraad van Duin Aug 18 '14 at 18:34

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.

  • $\begingroup$ Very nice (+1). Is this approach tailored to the case at hand or is it part of a more general framework? $\endgroup$ – Did Aug 19 '14 at 15:37
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    $\begingroup$ Thank you for your appreciation. To me it was a natural method, given the hypothesis. $\endgroup$ – Kelenner Aug 19 '14 at 15:56

Let $x^2-ax=c$ and $x^3-ax=d$. Let $P(y)$ be the minimal polynomial of $x$ over $Q$.

Then $P(Y)| Y^2-aY-c$ and $P(y) | Y^3-aY-d$. In particular the degree of $P$ is at most two.

Case 1 $\deg P=1$, it follows immediately that $x \in \mathbb Q$.

Case 2 $\deg P=2$, then $P(Y)=Y^2-aY-c$ and hence $$ Y^2-aY-c| Y^3-aY-d \,.$$

This shows that $$Y^3-aY-d = (Y^2-aY-c)(Y-e)$$

The coefficient of $Y^2$ on the RHS is $-(a+e)$ which shows that $e=a$. This implies

$$Y^3-aY-d = (Y^2-aY-c)(Y+a)=Y^3-Y(a^2+c)-ac$$

Therefore $$a^2+c=a$$ $$d=ac$$

The first relation implies $$P(Y)=Y^2-ay+a^2-a$$

As $P$ has real solutions $\Delta \geq 0$ and hence $a^2-4(a^2-a)>0 \Rightarrow 4a-3a^2 >0$, which contradicts $a>\frac{4}{3}$.

Note For $a=\frac{1}{2}$ we get $$P(Y)=Y^2-\frac{1}{2}Y-\frac{1}{4}$$ and $$c=\frac{1}{4}, d=\frac{1}{8} \,.$$

As $$(Y^2-\frac{1}{2}Y-\frac{1}{4})(Y+\frac{1}{2}) = Y^3-\frac{1}{2}Y-\frac{1}{8}$$ it follows that the roots $$x_{1,2} =\frac{1 \pm \sqrt{5}}{4}$$ of $$Y^2-\frac{1}{2}Y-\frac{1}{4}=0 $$ satisfy $$x^2-\frac{1}{2}x=\frac{1}{4}$$ $$x^3-\frac{1}{2}x=\frac{1}{8}$$

So the condition $a > \frac{4}{3}$ is indeed needed.


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