Use the hint. Essentially, you have that and with . Do you see why this is? Therefore, for every such that there is a unique element which also has order 3 and is non-trivial. Thus, there are an even number of elements of order 3. Do you understand why?

Now, can you think of the only element in your group which does not have order 3 but which the equation still holds? This element does not have a pair, and so you have 2n+1 elements which are solutions to this equation.