(1) The points of the polygon are inscribed on a circle. [given]
(2) Construct a triangle on each side of the polygon with the centre of the circle from (1) as the third point.
(3) Each triangle from (2) is isosceles. [2 of the sides are radii of the circle from (1)]
(4) The area of an isosceles triangle can be given as 1/2a^2sqrt((b^2)/(a^2)-1/4) where a is the length base of the triangle and b is the length of each of the other two sides. [proof left to the reader!]
(5) The square of each side of the polygon is divisible by n. [given]
(6) For each triangle from (2), double its area is divisible by the square of the side it is built on. [from (4), (6.1)]
(6.1) sqrt((b^2)/(a^2)-1/4) is an integer. [proof missing]
(7) For each triangle from (2), double its area is divisible by n. [from (5) and (6)]
(8) The area S of the polygon is equal to the sum of the triangles from (2). [by construction]
(9) If n divides a and b then it divides a+b
(10) 2S is divisible by n, QED. [from (7), (8), and (9)]
Of course, this is not complete.
For example, we also have to show that 2S is an integer, which really comes down to showing that sqrt((b^2)/(a^2)-1/4) is an integer. I should add, this is also needed for (6) as if that statement is not an integer than (6) is false.
[edit] I've added the missing step (6.1) though it is incomplete for now.
[edit 2] Actually, the missing step is different! The sum of the (6.1) pieces from each triangle needs to be an integer, not each individually (though if you could prove individually than you would get together, obviously).
[edit 3] I'm silly! We know that 2S is an integer because of the general form of the area of a polygon, given vertices (x_1, y_1), (x_2, y_2) etc. That is:
