400 
AN EXAMPLE OF THE HIGHER TRANSFORMATION OF A BINARY FORM. [525 
As a verification, suppose (a, b, c, d, e) (x, y) 4 = x* + y 4 (whence 1=1, J = 0). And take 
x x (x + y) 2 — y x (x — y) 2 = 0 for the transforming equation, that is, (a, /3, 7) = (1, 1, 1) and 
(a', /37') = (—1, 1, —1). We have P=R = x 1 —y l and Q = x 1 + y 1 , and thence 
that is, 
whence 
also 
Det. = (5 2 + A 2 ) 2 - 16 PQ 2 5 + 16 Q 4 
= (2 P- — 4Q 2 ) 2 = (— 2^'j 2 — 12 x x y x — 2y x 2 ) 2 , 
(da, b lt Cj, d x , e 1 )(x x , 2/j) 4 = 4 + 6^ + y*) 2 , 
4096 _ 2^ _ _ 262144 _ _ 2 18 
ll ~ 3 5 “ 3 ’ 27 ’ 27 ; 
A =-16, 5 = 8, 
and the equations for / 1( Ji become 
4096 
3 
262144 
27 
= 4(4.64 + 1256), 
= 8(|. -16.64 + ¥ y 4096), 
which are true. More generally, assuming 
(a, 6, c, d, e) (#, yY = x* + 6©# 2 y 2 + y 4 , 
(whence I = 1 + 3® 2 , </=© — © 3 ), and the same transforming equation, we have 
(<h, b x , c 1} d x , e x ){x 1} ytf = 4 {(1 + 3@) x? + (3 - 3®) 2x x y x + (1 + 3®)y?} 2 , 
whence 
012 018 
7, = T (l-30)=, /, = -^(1-30)»; 
also 
A =-16, 5 = 8(1-®). 
Substituting these different values in the equations for I x , J lf we obtain 
16 (1 - 3®) 2 = 12(1+ 3® 2 ) (1 - ®) 2 - 72 (® - ® 3 ) (1 _ ®) + 4 (1 + 3® 2 ) 2 , 
and 
- 8 (1 - 3©) 3 = 27 (® - @ 3 ) (1 - ©) 3 - 9 (1 + 3© 2 ) 2 (1 - ®) 2 
+ 27 (1 + 3© 2 ) (© - © 3 ) (1 - ©) - 54 (© - ® 3 ) 2 + (1 + 3© 2 ) 3 , 
which are in fact satisfied identically. 
Cambridge, 26 July, 1871.