Given $\lambda^3 - 2\lambda^2- 1 = 0$ for some real $\lambda$ prove that $[\lambda[\lambda[\lambda n]]] - n$ is odd for any positive integer $n$ .
(I Voronovich)
Nice problem! Maybe proving such $\lambda$ exist is harder.
Since
\begin{align*}[\lambda[\lambda[\lambda n]]] &=[\lambda[\lambda(\lambda n-\{\lambda n\})]] =[\lambda[\lambda^2n-\lambda\{\lambda n\}]]\\&=[\lambda(\lambda^2n-\lambda\{\lambda n\})-\lambda\{\lambda^2n-\lambda\{\lambda n\}\}]\\&=[\lambda^2(\lambda-2)n+2\lambda^2 n-\lambda ^2\{\lambda n\}-\lambda\{\lambda^2n-\lambda\{\lambda n\}\}]\\&=n+2[\lambda^2n]+[2\{\lambda^2n\}-\lambda^2\{\lambda n\}-\lambda\{\lambda^2n-\lambda\{\lambda n\}\}]\end{align*}Observe that $[2\{\lambda^2n\}-\lambda^2\{\lambda n\}-\lambda\{\lambda^2n-\lambda\{\lambda n\}\}]=0$ so we are done.$\Box$