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Show that the only rational solution to the title curve is $x=0$.

My attempt: Squaring both sides we have $e^{2x} - e^x + \frac{1}{4} = x^3 + \frac{1}{4}$, which yields $e^{2x} - e^x = x^3$. I suspect that the left hand side is irrational for any rational $x\neq 0$, and if so, we reach a contradiction and conclude that the only possibility is $x=0$.

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    $\begingroup$ I think that you need to explain why "the left hand side is irrational for any rational $x\neq0$". $\endgroup$ – barak manos Jan 10 '16 at 11:54
  • $\begingroup$ I think that's the heart of the question @Barack, which is why i didn't post my attempt as a solution. $\endgroup$ – Isaac. Jan 10 '16 at 11:58
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    $\begingroup$ After work of Baker (Fields Medal) and others it is known (in particular) that for all $x$ non zero rational $e^x$ is trascendental and you'd have $trascendental =algebraic$, absurde $\endgroup$ – Piquito Jan 10 '16 at 12:41
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You haven't sufficiently explained why the left-hand side is irrational for any rational $x \ne 0$.

Note that if $x = \frac{p}{q}$, then the equation would imply $q^3 (e^x)^2 - q^3 e^x -p^3 = 0$, in contradition to the fact that $e^x$ is transcendental for any rational $x \ne 0$.

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  • $\begingroup$ I think that you need to explain why "$e^x$ is transcendental for any rational $x\neq0$" (although it is far more obvious than why "$e^{2x}-e^x$ is irrational for any rational $x\neq0$"). $\endgroup$ – barak manos Jan 10 '16 at 12:03
  • $\begingroup$ I thought the transcendence of $e^x$ for any rational $x\neq 0$ was established by Hilbert in the 19th century. $\endgroup$ – Isaac. Jan 10 '16 at 12:05
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    $\begingroup$ @barakmanos Well, it depends on the tools you have to solve this exercise. The claim would e.g. follow from the Lindemann-Weierstrass theorem. $\endgroup$ – Dominik Jan 10 '16 at 12:12
  • $\begingroup$ Yep, I agree (or OP's comment above yours). +1 in any case. $\endgroup$ – barak manos Jan 10 '16 at 12:14
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We can write equation as $$e^x(e^x-1)=x^3$$ therefore $e^x-1=\frac{x^3}{e^x}$ now we know that $e$ is irrational(2.781...) so any value of $x$ else than $0$ makes the the equation irrational so $0$ is the only rational solution.

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  • $\begingroup$ The LHS would also be irrational, so what exactly have you proved here? $\endgroup$ – barak manos Jan 10 '16 at 12:05
  • $\begingroup$ @barak manos i have edited it $\endgroup$ – Archis Welankar Jan 10 '16 at 12:16
  • $\begingroup$ What do you mean "makes the the expression irrational"? It's an equation, if both sides are irrational then they can be equal to each other. $\endgroup$ – barak manos Jan 10 '16 at 12:18

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