Concyclic points on a regular polygon So if you have two distinct concyclic points, what is the criterion that they are vertices (I find that this is a more interesting problem) of a regular n-gon? Three points? Four points? 5 points? k points?
This was all the information I got on the problem, I can try to clarify if needed.

Lets take the case for 2 distinct concyclic points, say A and B, lying on a circle centered at O. Now, as the above answer says quite correctly, if A and B have to be vertices of a regular n-gon (whose circumcircle should be the same circle as the one given to us, of course), then that n-gon's center must be nothing but O itself. 
For any regular n-gon, the angle subtended by any side of the n-gon at its center is $2\pi/n$. This means that any two vertices (not necessarily consecutive) will subtend at the center, an angle which will always be an integral multiple of $2\pi/n$ (again, as given in the answer above). There you have it! 
So, to check whether the points A and B are the vertices of a regular n-gon, just join OA and OB, and check whether the angle made by these two lines is an integral multiple of $2\pi/n$ or not (for integral n, n>2). Lets do a little bit of math and see where does this lead.
Let the angle between OA and OB be $x$ degrees. For A and B to be the vertices of a regular n-gon, we must have -
$\pi$$x/180$ = $2\pi/n$ times any positive integer $k$; (for integral $n$ > 2 and of course, $k$ cannot be greater than $n$ itself), which further means that
$k/n$ = $x/360$; (for integral n > 2) ................. (1)
$x$ will always be less than 360 degrees. So $k$ (if found) will never be greater than $n$, by the above equation. So that's not a worry. 
Also, since $k$ and $n$ are both integers, the left hand side is always a rational quantity. Thus, right hand side must also be rational. For this to happen, one cannot take $x$ to be an irrational number. This gives us the criterion, which ultimately says- 
"Points A and B will be the vertices of a regular n-gon only when the angle between OA and OB, measured in degrees, is a rational number"
Additional Note 1: The condition "integral n>2" is also not of any worry, as whatever fraction we may get on the right side of equation 1 (which will be less than 1, remember?!), we can always multiply its numerator and denominator by a suitable integer to get a denominator > 2. 
Additional Note 2: The case for more than 2 concyclic points now becomes easy. Lets say we have k-points. Take two of those consecutive points in this group which make the smallest possible angle at the center, out of all possible consecutive pairs of points. Check the above condition for this angle, and then check whether all other angles made by consecutive points are integral multiples of this angle or not. If yes, then they are concyclic!
