How prove this $[r_{31}]^{k}$ is divided by 2012 then the remainder is $30$ 
Define polynomial $f_{n}(x)=x^n-2012\prod_{i=0}^{n-2}((n-1)x-i)+(-1)^n(n\ge 1)$,and let $r_{n}$ is $f_{n}(x)=0$ lagest real root,
show that: the sequence $\{[r_{31}]^k\}_{k\ge 0}$ have infinite many items,  such
$[r_{31}]^{k}$ is  divided by 2012 then the remainder is $30$

where $[x]$ s the largest integer not greater than $x$

My try:
$$f_{n}(x)=x^n-2012(n-1)x[(n-1)x-1][(n-1)x-2]\cdots[(n-1)x-(n-2)]+(-1)^n(n\ge 1)$$

and the following  How we find the lagest real root? Thank you

$f_{31}(x)$ has 31 real roots, 30 lying in the interval $(0,1)$ and one giant root around $x\approx 2012\cdot 30^{30}$. Let $A_n$ be the sum of the $n$-th powers of the roots of $f_{31}(x)$: $\{A_n\}_{n\in\mathbb{N}}$ is a linear recurring sequence of integers with characteristic polynomial $f_{31}(x)$. 
Note that $A_0=31, A_1=2012\cdot 30^{30}$ and that
$$f_{31}(x)\equiv x^{31}-1 \pmod{2012},$$
so $A_{62}\equiv A_{31}\equiv A_{0}\equiv 31\pmod{2012}$. Moreover, $A_{62k}$ is just a bit more that $\lfloor r_{31}\rfloor^{62k}$, so if you prove that, for any $k$ big enough, $A_{62k}-\lfloor r_{31}\rfloor^{62k}<2$, then $A_{62k}=1+\lfloor r_{31}\rfloor^{62k}$ follows and
$$\lfloor r_{31}\rfloor^{62k} \equiv 30\pmod{2012}$$
follows too.
