Claim 1: Exists a polynomial $Q(x) \in \mathbb{Q}\left [ x \right ]$ such that $\textrm{lim}\left ( \sqrt[3]{P(n)}-\sqrt[3]{Q(n)^3+3Q(n)+1} \right )=0.$ Proof: $\, \,$ The result is equivalent to: $$\textrm {lim}\left (\frac{P(x)-Q(x)^3-3Q(x)-1}{\left ( \sqrt[3]{P(x)} \right )^2+\sqrt[3]{P(x).\left [ Q(x)^3+3Q(x)+1 \right ]}+\left ( \sqrt[3]{Q(x)^3+3Q(x)+1} \right )^2} \right )=0$$$\, \,$ Or we need $deg(P(x)-Q(x)^3-3Q(x)-1) \leq \frac{2}{3}.2016-1=1343.$ $\, \,$ In all $Q(x) \in \mathbb{Q}\left [ x \right ],$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $ choose $Q_1(x)$ such that $deg(P(x)-Q_1(x)^3-3Q_1(x)-1)$ takes minimum value. $\, \,$ Clearly, $degQ=672$ and $Q$ is a monic polynomial. $\, \,$ Assume $deg(P(x)-Q_1(x)^3-3Q_1(x)-1)=d \geq 1344.$ $\, \,$ Or $P(x)-Q_1(x)^3-3Q_1(x)-1=ax^d+f(x),$ where $ 1344 \leq d < 2016$ and $degf \leq d-1.$ $\, \,$ Choose $Q_2(x)=Q_1(x)+\frac{a}{3}.x^{d-1344}.$ We must have $deg(P(x)-Q_2(x)^3-3Q_2(x)-1) \geq d.$ $\, \,$ But, \begin{align*} P(x)-Q_2(x)^3-3Q_2(x)-1&= P(x)-\left [ Q_1(x)+\frac{a}{3}.x^{d-1344} \right ]^3-3.\left [Q_1(x)+\frac{a}{3}.x^{d-1344} \right ]-1 \\ &= ax^d-aQ_1(x)^2.x^{d-1344}+f(x)-\frac{a^2}{3}.Q(x).x^{2d-2688}-\frac{a^3}{27}.x^{3d-4032} \end{align*}$\, \,$ Notice that: $degQ=672,$ $Q(x)$ is monic, $degf <d,$ and $d<2016,$ so $$\left\{\begin{matrix}deg\left( ax^d-aQ_1(x)^2.x^{d-1344}\right) <d & & & \\deg\left(\frac{a^2}{3}.Q(x).x^{2d-2688} \right) =2d-2016 < d & & & \\deg\left( x^{3d-4032}\right)=3d-4032 < d & & & \\ \end{matrix}\right. \Rightarrow deg\left(P(x)-Q_2(x)^3-3Q_2(x)-1 \right) < d ,$$$\, \,$ adsurb. q.e.d Turn back to the problem, Assume there are infinitely positive integers $(a_i): a_1<a_2<...<a_n$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $ such that $P(a_i)$ is nice, which are in the form $b_i^3+3b_i+1,$ respectively. Now, we have: $\textrm{lim}\left ( \sqrt[3]{P(n)}-\sqrt[3]{Q(n)^3+3Q(n)+1} \right )=0.$ $$\Rightarrow 0=\textrm{lim}\left ( \sqrt[3]{Q(a_i)^3+3Q(a_i)+1}-\sqrt[3]{P(a_i)} \right )=\textrm{lim}\left ( \sqrt[3]{Q(a_i)^3+3Q(a_i)+1}-\sqrt[3]{b_i^3+3b_i+1} \right )$$So for all suffice large $n,$ $\mid \sqrt[3]{Q(a_n)^3+3Q(a_n)+1}-\sqrt[3]{b_n^3+3b_n+1} \mid \leq 1.$ $\bullet$ $\textrm{lim}\left ( a_n \right )=+ \infty \rightarrow \textrm{lim}\left ( b_n \right )=+ \infty.$ $\, \,$ For all suffice large $n,$ if $Q(a_n) \geq b_n+2,$ then \begin{align*} \sqrt[3]{Q(a_n)^3+3Q(a_n)+1}-\sqrt[3]{b_n^3+3b_n+1} &\geq \sqrt[3]{(b_n+2)^3+3(b_n+2)+1}-\sqrt[3]{b_n^3+3b_n+1} \\ &=\frac{6b_n^2+12b_n+14}{\left ( \sqrt[3]{(b_n+2)^3+3(b_n+2)+1} \right )^2+ \sqrt[3]{(b_n+2)^3+3(b_n+2)+1}.\sqrt[3]{b_n^3+3b_n+1}+\left ( \sqrt[3]{b_n^3+3b_n+1} \right )^2} \end{align*}$\, \,$ but $$\underset{x \rightarrow + \infty} {\textrm{lim}}\left(\frac{6x^2+12x+14}{\left ( \sqrt[3]{(x+2)^3+3(x+2)+1} \right )^2+ \sqrt[3]{(x+2)^3+3(x+2)+1}.\sqrt[3]{x^3+3x+1}+\left ( \sqrt[3]{x^3+3x+1} \right )^2} \right )=2,$$$\, \,$ adsurb. $\, \,$ Similarly, if $Q(a_n) \leq b_n-2,$ we have a contradiction.[/i] So for all suffice large $n,$ we always have: $|Q(a_n)-b_n| \leq 1.$ Thus, $\exists M,N \in \mathbb{N}^*: M.Q(x) \in \mathbb{Z}\left [ x \right ] \rightarrow |M.Q(a_n)-M.b_n| \leq M,$ but $M.Q(a_n)-M.b_n \in Z.$ $$\Rightarrow M.Q(a_n)-M.b_n \in \left\{-M;-M+1;...;M-1;M \right\}, \forall \, n \geq N.$$Consider $2M+1$ sets $A_k=\left\{n \mid n \geq N, \, M.Q(a_n)-M.b_n =k \right\}, \forall \, k=\overline{-M;M}.$ $$\left|\bigcup_{i=-M}^{M}A_i \right|= + \infty \rightarrow \exists \, j: \left| S_t \right|= + \infty.$$Thus, there are infinitely $n: b_n=Q(a_n)- \frac{t}{M}=H(a_n),$ where $H(x)=Q(x)- \frac{t}{M} \in \mathbb{Q}\left [ x \right ].$ Or there are infinitely positive integers $x$ such that: $P(x)=H(x)^3+H(x)+1.$ $$\Rightarrow P(x)=H(x)^3+3H(x)+1, \forall \, x \in R.$$$\bullet $ If $H(x) \not\in \mathbb{Z}\left [ x \right ],$ let $M>1$ be the smallest positive integer such that $M.H(x) \in \mathbb{Z}\left [ x \right ], (1).$ $\, \,$ From $(1),$ we know that $\exists$ a prime $p \mid M$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $ and a biggest degree $k$ of $M.H(x)$ such that $p$ doesnt divide its coefficient. $\, \,$ We have: $M^3.P(x)-3M^2.M.H(x)-M^3=(M.H(x))^3.$ $\, \,$ Thus, all coefficients of $(M.H(x))^3$ is divisibled by $p.$ $\, \,$ Assume $M.H(x)=T(x)+a_k.x^k+...+a_1x+a_0,$ where $degT \geq k+1$ and $a_k \not\vdots p.$ $\, \,$ So the coefficient of $x^{3k}$ is: $$X=a_k^3+\sum_{i,j \neq k}^{0 \leq i < j; i+2j=3k}a_ia_j^2+\sum_{m,n,p \neq k}^{0 \leq m < n<p; m+n+p=3k}a_ma_na_p$$$\, \,$ Note that: $ i+2j=3k; \, m+n+p=3k \rightarrow max\left\{i,j \right\}; \, max\left\{m;n;p \right\}>k \rightarrow p \mid a_ia_j^2, \, a_ma_na_p.$ $$\Rightarrow p \not | X,$$$\, \,$ adsurb. So $H(x) \in \mathbb{Z}\left [ x \right ].$ Thus, for any $x,t \in \mathbb{Z},$ all the numbers $P(x),P(x+t), P(x+2t)$ are nice. Q.E.D