VMO

We will prove the following lemma

Lemma. Consider positive integer $m=2k+1$ with $k\ge 2$, and the square table of size $m\times m$ in which each cell is filled by one of $n$ colors. Then the number of different ways to color (not duplicated by the rotation) is equal to $$\frac{n(a^4+a^2+2a)}{4}\text{with }a=n^{k^2+k}.$$ Proof. Note that the middle cell, with symbol C as shown in the image, is not affected by the rotation so there is always $n$ ways to fill it.



Consider the collection of 4 squares $k\times k$ in the corners as set $A=A_1\cup A_2\cup A_3\cup A_4$ and 4 rectangles $1\times k$ as set $B=B_1\cup B_2\cup B_3\cup B_4$ in such a way that $B_t$ is immediately followed by $A_t$ for $1\le t\le 4$. We consider 4 pairs consists of two subsets $(A_t,B_t)$ of $A,B$. The number of cells on each pair $(A_t,B_t)$ are $k^2+k$, so the number of ways to fill in each pair of subsets is $a=n^{k^2+k}$. Here, we consider each pair like that as a vertex of some square $XYZT$. We will count the number of coloring for the vertices $X,Y,Z,T$ so that they do not duplicate by the rotation. We have the following cases

(1) All vertices are filled with the same color, there are $a$ ways.

(2) The coloring is alternative, so we just concern how to fill in 2 adjacent vertices with the number of ways is $\frac{a^2-a}{2}$.

(3) Otherwise, there is some identical pairs that is not cyclical of 2, then the number is $a^4-(a^2-a)-a$. This way of coloring has a circular permutations and will generates $\frac{a^4-a^2}{4}$ different ways.

In total, the number of coloring is $$\frac{a^4-a^2}{4}+\frac{a^2-a}{2}+a=\frac{a^4+a^2+2a}{4}.$$

Back to the problem, Using the above lemma when $2k+1=5$, we have a number of ways to fill in the paper $5\times 5$ that not duplicate each other by the rotation is $$n\left(\frac{n^{24}+n^{12}+2n^6}{4}\right)=\frac{n^{25}+n^{13}+2n^7}{4}.$$ Denote $S$ as a set of these coloring ways. In this problem, we also need to consider the reflection transformation. We separate $S$ to two types of papers, namely $A$, $B$: papers can and cannot create itself through the vertical and horizontal reflections.

Notice that each paper in $A$ generates 2 different papers in $S$ (should only be counted as 1 in the original problem); meanwhile, each paper in $B$ generates exactly 1 paper in $S$. Hence, $$2A+B=\frac{n^{25}+n^{13}+2n^7}{4}.$$ Also, it is easy to count $B=n^{15}$ (fill the left half of the piece of paper, also the middle line as well). From there we can calculate $$2A+2B=(2A+B)+B=\frac{1}{4}(n^{25}+n^{13}+n^7)+n^{15}$$ so $$A+B=\frac{n^{25}+4n^{15}+n^{13}+n^7}{8}.$$ This is the number of different pieces of paper we need to count. The problem is completely solved. □