[806 
lO 
XtH 
1—1 
o 
00 
807. 
tion of the 
A PROOF OF WILSON’S THEOREM. 
= 0, 
[From the Messenger of Mathematics, vol. xii. (1883), p. 41.] 
Let n be a prime number; and imagine n points, the vertices of a regular 
polygon; any polygon which can be formed with these n points as vertices is either 
0...(2), 
regular or else it is one of a set of n equal and similar polygons. For instance, 
n — 5, the polygon as shown in the figure is one of a set of 5 equal and similar 
>, c, f, g, h) 
3 
(3). 
2 </ 
aates (a,...) 
X \/ .4 
(4), 
(5), 
/ 
(6), 
1 5 
b, c, f, g, h. 
p.q.r = 2pqr 
efficients of 
f the other 
y, z), and 
i) is 
polygons; in fact, if the points taken in their cyclical order, but beginning at pleasure 
with any one of the 5 points are called 1, 2, 3, 4, 5, then we have 5 such polygons 
13254; and so in general. The whole number of polygons is ^ . 1.2.3 ... (n — 1); and 
the number of the regular polygons is \{n— 1); hence the number of the remaining 
polygons is = \ (n — 1) {1 • 2 ... (n — 2) — 1}; and this number must therefore be divisible 
by n; that is, 1.2 ... (n — 1) — n +1 is divisible by n; or, what is the same thing, 
1.2... (n — 1) + 1 is divisible by n, which is the theorem in question. 
us is = 4n ;