Hmm... It's strange nobody didn't post solution

Hint

drkim wrote: A infinite sequence $\{ a_n \}_{n \ge 0}$ of real numbers satisfy $a_n \ge n^2$. Suppose that for each $i, j \ge 0$ there exist $k, l$ with $(i,j) \neq (k,l)$, $l - k = j - i$, and $a_l - a_k = a_j - a_i$. Prove that $a_n \ge (n + 2016)^2$ for some $n$. Does it mean that there exists some $n$ which satisfies $a_n \ge (n + 2016)^2$?

Suppose to the contrary that $(n+2016)^2>a_n>n^2$ for all $n\in \mathbb{Z}^+.$ For positive integer $n>2016$. Choose $(i,j)=(0,n)$, we get that there exist positive integer $k$ that $a_{k+n}-a_k=a_n-a_0.$ We have $(n+2016)^2-1\geq a_n-a_0=a_{k+n}-a_k\geq (k+n)^2-(k+2016)^2+1.$ So, $k\leq \frac{2016n+O(1)}{n-2016}<C$ for positive constant $C$ for all $n>2016.$ This give us for all positive integer $n>2016$, at least one of the following $C$ equations must holds: $$a_{n+i}-a_i=a_n-a_0 \qquad \text{for each} \quad i=1,2,...,C.$$Note that each equation is equivalent to $a_{n+i}-a_n=a_i-a_0.$ For an integer $i_0>2016$, there exists positive integer $i_1<C$ that $a_{i_1+i_0}-a_{i_0}=a_{i_1}-a_0$. Then, there exists positive integer $i_2<C$ that $a_{i_2+i_1+i_0}-a_{i_1+i_0}=a_{i_2}-a_0$, and so on. For positive integer $\ell$ large enough, summing up these first $\ell$ equations give us $$a_{i_{\ell}+i_{\ell-1}+...+i_0}-a_{i_0}=-\ell a_0+\sum_{j=1}^{\ell}{a_{i_j}}.$$We have $LHS\geqslant \left( \sum_{j=0}^{\ell}{i_j}\right)^2-(i_0+2016)^2$ and $RHS\leqslant -\ell a_0+D\ell$ where $D=\max_{1\leq j\leq C} \{a_j\}$ is a constant. So, $-\ell a_0+D\ell \geqslant \left( \sum_{j=0}^{\ell}{i_j}\right)^2-(i_0+2016)^2\geqslant \ell^2-O(1)$, which is clearly false for large enough $\ell$, done.