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I've found the statement on the internet that the polynomial $x^3-5x$ is injective on the rational numbers, but without any comments on how to prove it. I think it means it must be easy, but I don't see how I can prove it. I'm getting a complicated expression, in which I don't see anything.

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1 Answer 1

up vote 20 down vote accepted

If $x^3 - 5x = y^3 - 5y$ with $x\neq y$ then $x^3-y^3 = 5(x-y)$ or $x^2+xy+y^2=5$.

Solving for $x$ in terms of $y$:

$$x = \frac{-y\pm \sqrt{y^2-4(y^2-5)}}2$$

So $20-3y^2$ has to be a square of a rational number. This is equivalent to finding integer solutions to $ 20q^2 - 3p^2= n^2$ with $(p,q)=1$. Show that this isn't possible $\pmod 5$.

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+1 clever approach. –  Rustyn Mar 4 '13 at 21:42
    
It's the same as the "complete the square" approach in the other answer, I'm just too lazy to complete the square myself :) @RustynYazdanpour –  Thomas Andrews Mar 4 '13 at 21:52

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