Rational solutions to $e^x - \frac{1}{2} = \sqrt{ x^3 + 1/4}$

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 $x=0$.

• I think that you need to explain why "the left hand side is irrational for any rational $x\neq0$". – barak manos Jan 10 '16 at 11:54
• I think that's the heart of the question @Barack, which is why i didn't post my attempt as a solution. – Isaac. Jan 10 '16 at 11:58
• 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 – Piquito Jan 10 '16 at 12:41

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 $x \ne 0$.

• 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 $x\neq0$"). – barak manos Jan 10 '16 at 12:03
• I thought the transcendence of $e^x$ for any rational $x\neq 0$ was established by Hilbert in the 19th century. – Isaac. Jan 10 '16 at 12:05
• @barakmanos Well, it depends on the tools you have to solve this exercise. The claim would e.g. follow from the Lindemann-Weierstrass theorem. – Dominik Jan 10 '16 at 12:12
• Yep, I agree (or OP's comment above yours). +1 in any case. – barak manos Jan 10 '16 at 12:14

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.

• The LHS would also be irrational, so what exactly have you proved here? – barak manos Jan 10 '16 at 12:05
• @barak manos i have edited it – Archis Welankar Jan 10 '16 at 12:16
• 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. – barak manos Jan 10 '16 at 12:18