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I did not make any progress. The problem is from RMC 2008.

The only idea that I have is:

Try to find sets of irrational numbers such that every number in the set multiplied by another number in the set yields a rational number.

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Suppose $q=x^2+2x$ is rational.

Then, by the quadratic formula, $x=\frac{-2 \pm \sqrt{4+4q}}{2}=-1 \pm \sqrt{\alpha}$, where $\alpha=q+1$ can be any rational number that is not a perfect square (since $x$ is irrational).

Now, $$x^3-6x=(-1 \pm \sqrt{\alpha})^3-6(-1\pm\sqrt{\alpha}) = \pm \alpha^{3/2} - 3\alpha \mp 3 \sqrt{\alpha}+5$$ will be rational if and only if $\alpha^{3/2}-3\sqrt{\alpha}$ is (since the other terms are always rational). But this is equal to $\sqrt{\alpha}(\alpha-3)$. Since $\alpha-3$ is rational and $\sqrt{\alpha}$ irrational, the entire quantity will be rational iff it is zero: that is, iff $\alpha=3$.

Thus the numbers $x$ that satisfy the condition are $-1 \pm \sqrt{3}$.

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Suppose $x^2 + 2x = q_1$ and $x^3 - 6x = q_2$ with $q_1, q_2\in \mathbb{Q}$. Clearly $[\mathbb{Q}(x):\mathbb{Q}] = 2$. Thus the polynomial $t^2 + 2t - q_1$ divides $t^3 - 6t - q_2$; that is, there exists some $\alpha$ such that $$t^3 - 6t - q_2 = (t^2 +2t + q_1)(t - \alpha) = t^3 + (2 - \alpha)t^2 + (q_1 - 2\alpha)t - \alpha q_1.$$ Equating corresponding coefficients of $t^n$ gives $\alpha = 2$, $q_1 = -2,$ and $q_2 = 4$. Thus both polynomials are rational iff $x^2 + 2x + 2 = 0$; that is, iff $x = -1 \pm \sqrt{3}$.

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  • $\begingroup$ What does $[\mathbb{Q}(x):\mathbb{Q}] = 2$ mean ? $\endgroup$ Sep 3, 2014 at 2:45
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    $\begingroup$ The degree of the extension $\mathbb{Q}(x)/\mathbb{Q}$. The point is that any algebraic number $x$ has a minimal polynomial $f$ (minimal here being in terms of degree, and we can assume that $f$ is monic to make it unique) such that any other nonzero polynomial $g$ with $g(x) = 0$ has $f| g$. Since $x$ is not rational (i.e., is not the zero of a linear polynomial over $\mathbb{Q}$), $f(t) = t^2 + tx - q_1$ must be that minimal polynomial. Thus it divides the polynomial $g(t) = t^3 -6t - q_2$ (as it also has $g(x) = 0$), and the computation described above gives you the $q_i$. $\endgroup$
    – anomaly
    Sep 3, 2014 at 2:59
  • $\begingroup$ ...and (since I ran out of characters above) the degree of that minimal polynomial $f$ is $[\mathbb{Q}(x):\mathbb{Q}]$, more or less by definition. $\endgroup$
    – anomaly
    Sep 3, 2014 at 3:00
  • $\begingroup$ Alternatively, one can compute cubic mod quadratic, which must be $0,\,$ see my answer. $\endgroup$ Sep 3, 2014 at 3:12
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I add yet another solution hopefully a simple one.

Since $x^2+2x$ is rational so too is $x^2+2x+1=(x+1)^2$.

Let $y=x+1$, so $y^2$ is rational, then $x^3-6x=y(y^2-3)-3y^2+5$ is rational, which means that $y(y^2-3)$ is rational.

Now since $y$ is irrational and $y^2-3$ rational we must have $y^2-3=0$, thus $y=\pm \sqrt{3}$ and $x=-1\pm\sqrt{3}$

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$q,r\in\Bbb Q,\,\ x^2 = \color{#0a0}{q\!-\!2x},\,\ r = x(x^2\!-\!6) = x(\color{#0a0}{q\!-\!2x}\!-6) = (q\!-\!6)x - 2(\color{#0a0}{q\!-\!2x}) = (\color{#c00}{q\!-\!2})x - 2q$

Thus $\, x\not\in\Bbb Q\,\Rightarrow\, \color{#c00}{q=2}\,\Rightarrow\,r=-2q = -4,$ so the polynomials are $\,f = x^2\!+2x-2,\,$ and $x^3-6x+4 = (x\!-\!2)f,\,$ with irrational roots being the roots of $\,f,\,$ i.e. $\,-1\pm \sqrt 3$

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Let $x$ be an irrational number with $$x^{2} + 2x = \frac{p_{1}}{q_{1}},\ x^{3} - 6x = \frac{p_{2}}{q_{2}}$$ for some integers $p_{1}, q_{1}, p_{2}, q_{2}$ such that $q_{1}, q_{2} \neq 0$ and $(p_{1}, q_{1}) = (p_{2}, q_{2}) = 1.$ Then we have $$2x^{2} - (6 + \frac{p_{1}}{q_{1}})x + \frac{p_{2}}{q_{2}} = 0,$$ so that $$x = \frac{6 + \dfrac{p_{1}}{q_{1}} \pm \sqrt{(6 + \dfrac{p_{1}}{q_{1}})^{2} - \dfrac{8p_{2}}{q_{2}}}}{4},$$ whence if $$(6 + \dfrac{p_{1}}{q_{1}})^{2} - \dfrac{8p_{2}}{q_{2}} > 0$$ and is not a perfect square, then $x$ is irrational.

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  • $\begingroup$ haha, i also came to this...but I was thinking of a more detailed set... anyway, thanks. $\endgroup$ Sep 3, 2014 at 2:37
  • $\begingroup$ If that condition applies then $x$ is complex, not irrational! $\endgroup$
    – Winther
    Sep 3, 2014 at 3:32
  • $\begingroup$ @Winther: Yeah thank you so much. I have corrected it. $\endgroup$
    – Yes
    Sep 3, 2014 at 3:36
  • $\begingroup$ This problem has only two solutions (see other answers), your answers seems to suggest there are many?! This method is not very useful in my opinion. $\endgroup$
    – Winther
    Sep 3, 2014 at 3:36
  • $\begingroup$ @Winther: Thank you so much for your patience and attention. Will you please help me check this proof math.stackexchange.com/questions/917701/… $\endgroup$
    – Yes
    Sep 3, 2014 at 4:14

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