TITLE: Merry-go-round problem
QUESTION [4 upvotes]: The problem sounds like this:

There are $n$ seats on a merry-
  go-round. A boy takes $n$ rides. Between each ride, 
  he moves clockwise a certain number of places to a 
  new horse. Each time he moves a different number of 
  places. Find all $n$ for which the boy ends up riding 
  each horse.

So if there are $n$ horses, first the boy could move by one place then he could move by $n + 1$ places then by $2n + 1$ so on and so forth, until he moves $(n-2)n + 1$ places, in which case he'd would have been ridden each horse only one time and taken unique number of steps, which implies that all $n$'s satisfy given condition.
Is there something that I'm missing?
Thank you for your answers

REPLY [3 votes]: Christian Blatter has shown that it cannot be done if $n$ is odd. It remains to show that it can be done if $n$ is even. 
One way to do this is to alternate the odd numbers $1,3,\dots,n-1$ in increasing order with the even numbers $2,4,\dots,n-2$ in decreasing order. The number of places the boy will move each time is given by the sequence
$$1,n-2,3,n-4,\dots,n-(n-2),n-1$$
Note that this is the same as moving one place forward, two places backward, three places forward, four places backward, etc. The horses that the boy has used will always form an unbroken chain (i.e., no gaps), and the boy will always move from one end of the chain to the other end of the chain, extending the length of the chain by one horse.
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