For all integers $m$ such that $0≤m<n$ find $a,b,c\in D_n$ such that
- $a(rR^m ) a^{-1}=R^2$
- $b(rR^m ) b^{-1}=r$
- $c(rR^m ) c^{-1}=rR$
$D_n$ is dihedral group of an $n$-gon represented by
$$D_n=\{I,R,R^2, R^3, ...R^{n-1},r,rR,rR^2,..,rR^{n-1}\},$$
$r$ is the basic reflection and $R$ is the basic rotation in $D_n$
Consider the following group presentation. $$D_{n}=\langle R,r | R^n=r^2=1, r^{-1}Rr=R^{-1} \rangle$$ In particular, note that $Rr=rR^{-1}$ (call this identity $\star$). We can apply this identity to any word in $R$ and $r$ to move all the $r$'s to the left of all the $R$'s. Thus we may conclude that any $z\in D_{n}$ has the form $z=r^xR^y$ for some $0\leq x \leq 1$ and $0 \leq y \leq {n-1}$.
Of course, you already knew this, as evidenced by your list of the elements of $D_{n}$, but the technique by which this is achieved is important to understand. In fact, a similar use of $\star$ is the key to solving this problem.
Let's figure out the forms of $a(rR^m)a^{-1}$. We know we may write $a=r^xR^y$, so this becomes $$r^xR^y(rR^m)(r^xR^y)^{-1}=r^xR^yrR^mR^{-y}r^{-x}$$ From here we divide into two cases: $x=0$ and $x=1$. If $x=0$, $$\begin{eqnarray*} \require{cancel} \cancel{r^x}R^yrR^mR^{-y}\cancel{r^{-x}}&=&\color{blue}{R^yr}R^{m-y}\\ &=&\color{blue}{rR^{-y}}R^{m-y}\\ &=&rR^{m-2y}\\ \end{eqnarray*}$$ Let's keep this result in mind and compute the $x=1$ case. $$\begin{eqnarray*} \require{cancel} r\color{blue}{R^yr}R^m\color{blue}{R^{-y}r}&=&r\color{blue}{rR^{-y}}R^m\color{blue}{rR^{y}}\\ &=&\cancel{r^2}\color{red}{R^{m-y}r}R^{y}\\ &=&\color{red}{rR^{y-m}}R^{y}\\ &=&rR^{2y-m}\\ \end{eqnarray*}$$ (Note, by the way, that we could have performed this second calculation more quickly by applying $r^{-1}R^sr=R^{-s}$ to the first calculation, followed by $\star$.)
Now let's think about your questions.
In both of the resulting equations, you can see that no matter what $m$ and $y$ are, we'll still be left with an $r$, so there's no way that either can come out to be equal to $R^2$.
We see that this amounts to solving the equations $m-2y\equiv 0 \pmod n$ and $2y-m\equiv 0 \pmod n$.
We see that this amounts to solving the equations $m-2y\equiv 1 \pmod n$ and $2y-m\equiv 1 \pmod n$.
With these observations, can you complete the proof? (Be careful: the existence of solutions may depend on the parity of $n$!)