OF ELLIPTIC FUNCTIONS. 
425 
1 . 2 U 3 
to verify which observe that, substituting herein for M its value 1 + — , the equation 
becomes 
(2y 3 + 3wV — u)(v -f- 2 u 3 ) — 3 vu(v 2 u 2 + 2 u 5 v + 1 ) = 0 ; 
that is, 
2v 4 -f iv 3 u 3 — £vu— 2w 3 =0, as it should do. 
33. Any expression whatever of M in terms of u, v is in fact one of a system of four 
expressions ; viz. we may simultaneously change 
w=7(mod.8) 
+ + ~ 
+ + + 
+ + — 
that is, 
signs are 
u 
V 
1 
M 
n 
= 1 
n= 3 
n = 5 
into 
v, 
(- 
n 2_ i 
■)“ 
u, 
(-K 1 
vM 
+ 
+ 
+ 
+ 
+ - 
- 
or 
l 
u 
1 
©’ 
V 4 
+ 
+ 
+ 
+ + + 
+ + 
+ 
or 
1 
~ 5 
V 
(- 
n 2 — 1 
.) s 
1 
m’ 
( -p 1 
V 4 ’ 
+ 
+ 
+ 
+ 
+ - 
+ 
1 
Thus n= 3, starting from — =1+ — , we have 
° M v 
~ =1+-, — 3M=1 
M v 
j1=i+?h, 
m 4 M m 3 ’ 
3M=1-- 
V 3 
and of course if from any two of these we eliminate M, we have either an identity or 
the modular equation ; thus we have the modular equation under the six different forms : 
(1, 2) (v+2u 3 )(u-2v 3 ) + 3uv=0, 
(1, 3) v s (v+2u 3 )— u{u 3 -\-2v) =0, 
(1,4) (v+2u 3 )(v 3 -2u)+3u i =0, 
(2.3) (u- 2?; 3 )(% 3 +2y)+3y 4 =0, 
(2, 4) v(v 3 -2u)-u 3 (u-2v 3 ) =0, 
(3.4) (u 3 +2v)(v 3 -2u)+3u 3 v 3 = 0. 
34. Nextw=5. Here, starting from u the changes give 
° M v(l — wry ° ° 
1 v — U S U + V b V 4 lP(v — U b ) U 4 r -w- U 3 (ll + V 5 ) 
M v(l— uvpy u(l+u 3 v)’ m 4 M u 4 (l—uv 3 y v 4 ' v 4 (l+u 3 v)’ 
viz. the third and fourth forms agree with the first and second forms respectively ; that 
is, there are only two independent forms, and the elimination of M from these gives 
5uv(l — uv% 1 -f u 3 v ) — (v — u 5 )(u + v 5 )= 0, 
which is a form of the modular equation. 
35. In the case n= 7, starting from u yyy 7 - (as to this see post. 
No. 43), the forms are 
