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 September 28th, 2013, 08:14 AM #1 Member   Joined: Jun 2013 Posts: 31 Thanks: 0 Rational number Let x be a real number with the property: x^3+x and x^5+x are rational. Prove, that x is rational.
 October 7th, 2013, 02:38 AM #2 Member   Joined: Mar 2013 Posts: 90 Thanks: 0 Re: Rational number If $x^3+x$ is rational, $x$ must be of the form either$x\=\ a\,+\,b\sqrt m$ where $a,\,b$ are rational and $m$ is a positive non-square rational, or$x\=\ a\,+\,b\sqrt[3]n\,+\,c\left(\sqrt[3]n\right)^2$ where $a,\,b,\,c$ are rational and $n$ is a positive non-cube rational. Let us look at each of these cases. Case 1: $x\=\ a\,+\,b\sqrt m$ We have $x^3+x\=\ a(1+a^2+3mb)\,+\,b(1+3a^2+mb^2)\sqrt m$ For this to be rational, the coefficient of $\sqrt m$ must vanish. Hence $b=0$ (since $1+3a^2+mb^2\,\ne\,0$) and so $x=a$ is rational. Case 2: $x\=\ a\,+\,b\sqrt[3]n\,+\,c\left(\sqrt[3]n\right)^2$ We have $x^3+x\=$ $(a+a^3+nb^3+n^2c+6nabc)\,+\,(b+3a^2b+3nb^2c+3nc^2a )\sqrt[3]n\,+\,(c+3ab^2+3nbc^2+3ca^2)\left(\sqrt[3]n\right)^2$. Then $b\,+\,3a^2b\,+\,3nb^2c\,+\,3nc^2a\=\ 0\ \ldots\,\fbox1$ $c\,+\,3ab^2\,+\,3nbc^2\,+\,3ca^2\=\ 0\ \ldots\,\fbox2$ $\fbox1\times c\,-\,\fbox2\times b$ yields $3nc^3a\,-\,3ab^3=0$ $\Rightarrow$ $a\,=\,0$ or $nc^3\,=\,b^3$. (i) $a=0$ This leads to $b\,+\,3nb^2c\=\ 0$ and $c\,+\,3nbc^2\=\ 0$. If $b,\,c$ are not both zero, this means $x$ must take one of the following forms: $x\=\ b\sqrt[3]n\,-\,\frac1{3nb}\left(\sqrt[3]n\right)^2$ if $b\,\ne\,0$ $x\=\ -\frac{\sqrt[3]n}{3nc}+c\left(\sqrt[3]n\right)^2$ if $c\,\ne\,0$ Take the first one. $x^2\=\ -\frac23\,+\,\frac{\sqrt[3]n}{9nb^2}\,+\,b^2\left(\sqrt[3]n\right)^2$ $x^3\=\ \left(nb^3-\frac1{27b^3}\right)\,-\,b\sqrt[3]n\,+\,\frac{\left(\sqrt[3]n\right)^2}{3nb}$ Now examine the coefficient of $\sqrt[3]n$ in $x^5\,=\,x^2\,x^3$. It works out as $\frac10b\,-\,\frac1{243nb^5}$. Then the coefficient of $\sqrt[3]n$ in $x^5-x^3$ is $\frac{19b}9\,-\,\frac1{243nb^5}$. Since this cannot be zero, $x^5-x^3$ is irrational. But if $x^5+x$ and $x^3+x$ are both rational then their difference must be rational as well. This contradiction shows that $x\=\ b\sqrt[3]n\,-\,\frac1{3nb}\left(\sqrt[3]n\right)^2$ is impossible. The other one $x\=\ -\frac{\sqrt[3]n}{3nc}+c\left(\sqrt[3]n\right)^2$ leads to a similar impasse. Thus if $a\,=\,0$, we must have $b\,=\,c\,=\,0$ and so $x\,=\,0$ which is rational. (ii) $nc^3\,=\,b^3$ Since $n$ is not a perfect cubic rational, this has no non-zero rational solution in $b,\,c$ so again $b\,=\,c\,=\,0$ and $x\,=\,a$ is rational.

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