122 
is a sufficient solution of the correspondence problem last enun- 
ciated. But it is not a solution of the original problem: for the 
series will go on cbaabede...and not repeat itself, so that the 
points cbaab do not form a proper in-and-circumscribed pentagon. 
Thus the problem is in general impossible. If however there 
is any proper solution, the equation of the fourth degree 
(which determines the improper solutions) will have more than 
four roots, and will therefore be identically satisfied by any 
number whatever; so that whatever point x we start with, the 
point w will come to coincide with it. 
Precisely similar reasoning is applicable to the cycles of an 
even order. Thus e.g. for a quadrilateral we get the four im- 
proper solutions y828 got by starting from the points y. I pass 
to the consideration of correspondences of higher orders. 
In an (7, 7) correspondence there are 
2r united points a ; 
their remaining correspondents form 
2r (r — 1) points 0 ; 
to these again correspond 
2r (r — 1) points ¢, and so on. 
Similarly there are 
2r (r — 1) points a, 
each of which is such that two of its correspondents coincide ; 
viz. these are 
2r (r — 1) points £, 
_ to which also correspond 
2r (r —1)’ points y, and so on. 
Now if we attempt to form a closed cycle of the n™ order, 
we shall be led to a correspondence 
Para De aly 
which has 27.(7—1)"" united points. From this number we 
shall have to subtract the number of improper solutions as given 
