186 
ON BICURSAL CURVES. 
[510 
gives, when rationalised, an equation in sn 2 u of the order 2n + 2k ; the roots of this 
equation are sn 2 cq, sn 2 a 2 ... sn 2 ct 2n+nJc . Considering the functions (1, sn 2 u) n+k and 
(1, sn 2 u) n+k ~ 2 as indeterminate, the coefficients can be found so that all but one of 
the roots of the equation in sn 2 u shall have any given values whatever, sn 2 <x 1 , 
sn 2 a 2 ,... sn 2 a 2 n+2*-i 5 the theorem then shows that the remaining root is sn 2 a 2?l+2 fc, where 
®2n+2& Ci x Cl. 2 (X 2n +2k—l » 
which is, in fact, Abel’s theorem.} 
Now, supposing that the three functions of 0 all vanish for 2k common values 
of 6, each of the functions of u will contain the same 2k H functions, say these are 
H (u — a 2n+1 )... H (u — a 2n+2 k). Omitting these and also the denominator factor © 2A: (u), 
we have the set of equations 
H(u — H(u — a 2 ) ... H(u — a m ) 
%- n (u) 
{x, y, z) = G 
where, however, 
a a + a 2 ... + a 2Jl 0, 
(the values a 1} a 2 ... a. 2n are of course different for the three coordinates x, a, z respec 
tively} ; viz. we have 
a i + a 2 • • • + — (®2?l+l ••• tt-zn+nk) 
we have 
.V» 
or, changing the common denominator, 
(«> y, 
where 
a i + a 2 ... + a. 2n — 0 j 
or, what is the same thing, 
(x, y, z) = (1, sn 2 u') n + (1, sn 2 A) 71-2 sn v! sn'. u'; 
viz. writing sn u = O', and ©' = 0' (1 — 0') (1 — №0'), this is 
O, y, z) = ( 1, 0') n + (1, 6>') w “ 2 V©', 
a normal representation of the curve of the order 2n.