Why is $x^3-5x$ injective on the rationals?

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 $\pmod 5$.