Existence of elements of order $2$ in $S_n$ whose Product has order $n$ 
Show that all for $n>2$, $S_n$ contains two elements $x,y$, both of order $2$, such that their product is of order $n$.

I can find such elements for $n$ small, but I don't know how to make up an algorithm to produce these $x,y$ for large $n$. Any hints are appreciated.
 A: Consider the dihedral group as a subgroup of the symmetric group with $s$ the reflection and $r$ the rotation generators. Then the elements $sr$ and $sr^2$ each have order $2$ and their product is $r$ which has order $n$. Note that we've tacitly used that $n > 2$.
A: One pattern that works for $n=2k+1$ is $$(2\,3)(4\,5)\ldots(2k\,2k+1)\circ(1\,2)(3\,4)(5\,6)\cdots(2k-1\,2k) 
=(1\,3\,5\,7\,\ldots\, 2k-1\,2k+1\,2k\,2k-2\,\ldots\,6\,4\,2)$$
and a similar pattren for $n=2k$.
A: Two reflections (generally) make a rotation.

A: For $n= 3$, we note that 
$$ (1\ 3 ) (2\ 3) = (1\  2\ 3), $$
and 
$$ \big| (1 \ 2) \big| = 2 = \big| (2 \ 3) \big|, $$
whereas 
$$ \big| (1 \ 2 \ 3) \big| = 3. $$
For $n=4$, we find that 
$$ \big[ (1 \ 2) ( 3\ 4) \big] \big[ ( 1 \ 3)  \big] = ( 1 \ 2 \ 3 \ 4 ),   $$
and 
$$ \big|  (1 \ 2) ( 3\ 4)  \big| = 2 = \big| (1 \ 3) \big|, $$
whereas 
$$ \big| (1 \ 2\ 3 \ 4) \big| = 4. $$
For $n= 5$, we find that 
$$ \big[ (1 \ 2) ( 3 \ 5) \big] \big[ (1 \ 3) ( 4\ 5)  \big] = (1 \ 2 \ 3 \ 4 \ 5), $$
and 
$$ \big|  (1 \ 2) ( 3 \ 5) \big| = 2 = \big| (1 \ 3) ( 4\ 5)  \big|, $$
whereas 
$$ \big| (1 \ 2\ 3\ 4\ 5) \big| = 5. $$
Hope this helps. 
