$ r_n\equiv u_n (\mod 1986)$ which $ 0\leq r_n\leq 1985$ Because $ \{r_n\}$ take finite value so it has $ k,l\in N$ such that : $ k > l$ $ r_k = r_l,r_{k + 1} = r_{l + 1}$ (1) We prove that $ r_{n} = r_{n + k - l},\forall n$ Because (1) we has :$ r_{k - 1}\equiv r_{l - 1} (\mod 1986)$ so $ r_{l - 1} = r_{k - 1}$ $ r_{k + 2} = \equiv r_{l + 2}(\mod 1986)$ so $ r_{k + 2} = r_{l + 2}$ Contine this progress we has $ r_{n} = r_{n + k - l},\forall n\in N$ Because $ r_2\equiv 0(\mod 1986)$ so $ r_{h(k - l) + 2}\equiv 0(\mod 1986)$ So has infinite $ u_n$ such that $ 1986|u_n$

this is not a problem in China 1991. In fact, it is in Canada 1986. maybe that is the reason why the sourse of M27 was shown as "Canada 1986":http://www.mathlinks.ro/viewtopic.php?p=849911#849911

Prove TTsphn is not complete. May be period $ r_{k+T}\equiv r_k \ \forall k>C$. In these case we can continue sequense $ n\to n-1$: $ u_n=u_{n+1}^2-u_{n+2}$, therefore $ C=0$.