- #36

archaic

- 688

- 210

Incomplete attempt:

Let ##p/q## and ##r/s## be in ##\mathbb{Q}## such that ##f(p/q)=f(r/s)##.

$$\begin{align*}

f\left(\frac{p}{q}\right)=f\left(\frac{r}{s}\right)&\Leftrightarrow\left(\frac{p}{q}\right)^3-2\frac{p}{q}=\left(\frac{r}{s}\right)^3-2\frac{r}{s}\\

&\Leftrightarrow \left(\frac{p}{q}\right)^3-\left(\frac{r}{s}\right)^3-2\left(\frac{p}{q}-\frac{r}{s}\right)=0\\

&\Leftrightarrow\left(\frac{p}{q}-\frac{r}{s}\right)\left(\left(\frac{p}{q}\right)^2+\frac{p}{q}\frac{r}{s}+\left(\frac{r}{s}\right)^2\right)-2\left(\frac{p}{q}-\frac{r}{s}\right)=0\\

&\Leftrightarrow\left(\frac{p}{q}-\frac{r}{s}\right)\left(\left(\frac{p}{q}\right)^2+\frac{p}{q}\frac{r}{s}+\left(\frac{r}{s}\right)^2-2\right)=0\\

\end{align*}$$

$$\frac{p}{q}=\frac{r}{s}\text{ or }\left(\frac{p}{q}\right)^2+\frac{p}{q}\frac{r}{s}+\left(\frac{r}{s}\right)^2=2$$

$$\begin{align*}

\left(\frac{p}{q}\right)^2+\frac{p}{q}\frac{r}{s}+\left(\frac{r}{s}\right)^2=2&\Leftrightarrow\left(\frac{p}{q}+\frac{r}{s}\right)^2-\frac{p}{q}\frac{r}{s}=2

\end{align*}$$

We will analyse ##(ps)^2+pqrs+(qr)^2=2(qs)^2##:

For the LHS to be even, I need to have 2 odd terms and 1 even term, or 3 even terms.

**1)**If both ##(ps)^2## and ##(qr)^2## are odd then ##p,\,s,\,q## and ##r## are all odd, and so ##pqrs## cannot be even. Thus, this configuration doesn't work.

**2)**That all terms are even: //to be finished.