15
$\begingroup$

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.

$\endgroup$
21
$\begingroup$

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 $\pmod 5$.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Thomas Andrews Mar 4 '13 at 21:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.