Let
![\alpha](/media/m/f/c/3/fc35d340e96ae7906bf381cae06e4d59.png)
be the positive root of the equation
![x^{2} = 1991x + 1](/media/m/6/9/5/6959ea150c54ada1a92c2f45b871a8d2.png)
. For natural numbers
![m](/media/m/1/3/6/1361d4850444c055a8a322281f279b39.png)
and
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
define
Prove that for all natural numbers
![p](/media/m/1/c/8/1c85c88d10b11745150467bf9935f7de.png)
,
![q](/media/m/c/1/d/c1db9b1124cc69b01f9a33595637de69.png)
, and
![r](/media/m/3/d/f/3df7cc5bbfb7b3948b16db0d40571068.png)
,
%V0
Let $\alpha$ be the positive root of the equation $x^{2} = 1991x + 1$. For natural numbers $m$ and $n$ define
$$m*n = mn + \lfloor\alpha m \rfloor \lfloor \alpha n\rfloor.$$
Prove that for all natural numbers $p$, $q$, and $r$,
$$(p*q)*r = p*(q*r).$$