Let $x\in\mathbb{R}$. Demonstrate that if the numbers $a = x^3–x$ and $b = x^2 +1$ are rational, then $x$ is rational.
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Any idea, please... – Alexx Apr 21 '18 at 19:06
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Any attempt, please... – amWhy Apr 22 '18 at 00:40
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Any effort, please... – amWhy Apr 22 '18 at 00:41
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amWhy, Why do you say I have to make an effort because I just wrote a solution, did not you see it? – Alexx Apr 22 '18 at 13:29
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I just wanted to check my solution, did you understand? amWhy – Alexx Apr 22 '18 at 13:31
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Hint: $x^3-x=x\bigl((x^2+1)-2\bigr)$
José Carlos Santos
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I'm discussing the 3 cases, x = 1, -1 and x different from 1, -1 then generalize the conclusion? – Alexx Apr 21 '18 at 19:12
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@Alexx If you know how to do it, then post a complete answer. – José Carlos Santos Apr 21 '18 at 19:15
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x^3-x ∈ Q result x(x^2-1)∈ Qresult x(x-1)(x-1)∈Q, case 1:x=1,1∈ Q result x∈ Q;case 2:x=-1, -1∈ Q result x∈ Q:case 3:if x diffrent 1 and x diffrent -1: x different 1result x-1 diffrent 0, xdiffrent -1result x+1diffrent 0: Then (x-1)(x+1) diffrent 0. – Alexx Apr 21 '18 at 19:29
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x(x-1)(x+1)∈ Q and (x-1)(x+1)∈ Q and (x-1)(x+1) diffrent 0 result x(x-1)(x+1)(x-1)(x+1) ∈ Q result x∈ Q. – Alexx Apr 21 '18 at 19:33
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Because in each possible cases, x∈ Q result x∈ Q, in any case. Its complete my solution? – Alexx Apr 21 '18 at 19:35
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@Alexx What about this: $x=\frac{x^3-x}{x^2+1-2}=\frac a{b-2}\in\mathbb Q$. – José Carlos Santos Apr 21 '18 at 19:36
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a and b are rational so a and b-2 are rational so a/(b-2) is rational if b is'nt 2 – Alexx Apr 21 '18 at 19:43
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$x^2+1\in \mathbb Q \implies x^2\in \mathbb Q$
From this, we deduce that $x^3=rx$ for some $r\in \mathbb Q$. It is easy to see that $r=1\implies x=0,\pm 1$ which are all rational. Assume, then, that $r\neq 1$.
But this implies that $x^3-x=x(r-1)\in \mathbb Q\implies x\in \mathbb Q$ as desired.
lulu
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