Steve is piling $m \geq 1$ indistinguishable stones on the squares of an $n\times n$ grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions $(i, k), (i, l), (j, k), (j, l)$ for some $1\leq i, j, k, l \leq n$, such that $i < j$ and $k < l$. A stone move consists of either removing one stone from each of $(i, k)$ and $(j, l)$ and moving them to $(i, l)$ and $(j, k)$ respectively,j or removing one stone from each of $(i, l)$ and $(j, k)$ and moving them to $(i, k)$ and $(j, l)$ respectively. Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves. How many different non-equivalent ways can Steve pile the stones on the grid?

Let sequence $\{a_n\}$ satisfy $a_0=0, a_1=1$, and $a_n = 2a_{n-1}+a_{n-2}$. Show that $2^k\mid n$ if and only if $2^k\mid a_n$.

Let $\{a_n\}$ be a sequence defined as $a_n=\lfloor{n\sqrt{2}}\rfloor$ where $\lfloor{x}\rfloor$ indicates the largest integer not exceeding $x$. Show that this sequence has infinitely many square numbers.

Show that for any positive integer $n$, the value of $\displaystyle\sum_{k=0}^{n}2^{3k}\binom{2n+1}{2k+1}$ is not a multiple of $5$.

Let $a_0$, $a_1$, $a_2$, $\cdots$ be an increasing sequence of non-negative integers such that every non-negative integer can be expressed uniquely in the form of $(a_i + 2a_j+4a_k)$ where $i$, $j$, and $k$ are not necessarily distinct. Determine $a_{1998}$.

Let $p$ be an odd prime number. Find the number of subsets $\mathbb{A}$ of the set $\{1, 2, \cdots, 2p\}$ such that $\mathbb{A}$ has exactly $p$ elements and the sum of all elements in $\mathbb{A}$ is divisible by $p$.

A permutation $\{x_1,\ x_2,\ \cdots,\ x_{2n}\}$ of the set $\{1,\ 2,\ \cdots,\ 2n\}$, where $n$ is a positive
integer, is said to have property $P$ if $\mid x_i − x_{i+1}\mid = n$ for at least one $i$ in $\{1,\ 2,\ \cdots,\ 2n − 1\}$. Show that, for each $n$, there are more permutations with
property $P$ than without.