406 
PROFESSOR CAYLEY ON THE TRANSFORMATION 
the first of these has at the point a=0, y=0 a double point, the second a triple point ; 
or there are at the point in question 6 intersections ; but 4 of these are the points which 
give the foregoing reduction 16 — 4=12 ; we have thus the point a=0, y=0, counting 
twice among the twelve points. Writing in the two equations 3=0, the equations 
become # 2 a 3 y— ay 3 = 0, /cV— y 4 =0, viz. these will be satisfied if #V — y 2 =0, that is, the 
curves pass through each of the two points (3=0, y=+#a), and these values satisfy 
(as in fact they should) the third equation, 
P(2ay + 2 a 3 + 3 2 )a(a+ 23 )-y(y+ 23 )( 2 ay+ 23 + 3 2 ) = 0; 
it is moreover easily shown that the three curves have at each of the points in question 
a common tangent ; viz. taking A, B, C as current coordinates, the tangent at the point 
(a, 3> y) of the second curve has for its equation 
A(2a 3 + 3« 2 3)& 4 + B(& V - y 3 ) - C(2y 3 + 3y 2 3) = 0 ; 
and for 3=0, y=+fca this becomes 2#A+B(/clfl)ip2C=0, viz. this is the line from 
the point (3 = 0, y=+#a) to the point (1, —2, 1). And similarly for the other two 
curves we find the same equation for the tangent. 
Hence among the 12 points are included the point (y=0, a=0) twice, and the points 
(3 = 0, y= +#a) each twice : we have thus a reduction =6. 
15. Writing in the equations y=0, a= 0, the first and third are satisfied identically, 
and the second becomes (3 2 =D.(3 2 , that is the equations give O =1 ; writing (3=0, they 
become 
# 2 a 2 =Oy 2 , ay = Day, y 2 = 0& 2 a 2 , 
viz. putting herein y 2 =& 2 a 2 , the equations again give 0 = 1; hence the factor of the 
order 6 is (O — l) 6 , and the equation of the twelfth order for the determination of O is 
(O— 1) 6 {(0, 1) 6 }=0, 
where (O, 1) 6 =0 is the 0#-modular equation above written down. 
16. Beverting to the equation 
1 -y (l— a?) (P— Qa?) 2 
I + y“(l+«)(P + Q*)* 
it is to be observed that for a=0, y=0, that is P=0, this becomes simply y=x, which 
is the transformation of the order 1 ; the corresponding value of the modulus \ is \=k, 
and the equation X = 0 2 & then gives 0 2 =1, which is replaced by 0—1=0. 
If in the same equation we write (3=0, that is Q=0, then (without any use of the 
equation y 2 =#V) we have y=x, the transformation of the order 1 ; but although this 
is so, the fundamental equation 
(P 2 -I- 2PQr ! + QV)* = 0£ 2 (P 2 + 2PQ+ Q V), 
which, putting therein Q=0, becomes (P 2 )* = 0£ 2 P 2 ; that is (/c 2 fa + y) 2 = OP(a + yx 2 f 
is not satisfied by the single relation 0—1 = 0, but necessitates the further relation 
y 2 =£ 2 a 2 . 
