(III.) 
of Edinhurgli, Session 1883-84. 393 
Here also there are two cases, viz,, 
(1) q odd, M even, =2A:4- 1, 2n say, 
(2) q even, p even, m even, = 2r, 2s, 2n say, 
the corresponding identities being 
+ 2^ + 1 )N + + 4jiA: + + 2^ + + 2A: + 1 ) } ‘ 
+ (4^7c2 + ipk + 2p + 2k+l )2N + (47:2 + 47: + 2f] 
_( I 1 111 
“i / + ^ +27:+l + 27:+l+^? +2{ }+... 
^[ { (1 6r2s2 + 8rs+4r2 + 1 )N + (4rs2 + r + 5 ) (4s2 + 1 ) j 2 + (4rs2 + r + 5)4N + (4s2 + 1 )2] 
= 1 Ui- 1- ^ i- 
I ) 2r + 2s+ 2s+ 2r+2| |+... 
Putting in the first of these k = 0, we have 
J{2p^ + 2p + 1 )n 4 - 2p+ 1}^ + {2p + 1)2n + 4 
f ] 1 i_ 1 I_ 1 
j’^i9+l + l-'rJ9+2{ }+... 
Specialising still further by taking n = 0, we have 
sl{^P^^f + ^ = ifP+^)^J^T+T+J+2{2p + l)+ ... 
* 
which is M. de Jonquieres’ theorem V., and which is to be found 
given as an example at page 31 of my pamphlet on The Expression 
of a Quadratic Surd as a Continued Fraction, Glasgow, 1874. 
9. Considerable interest attaches to the above identity deduced 
from (III.) by putting A; = 0. Written in the unexpanded form of 
the general theorem, it is 
J I K(j9,l,l,i^N + iK(p,l,l)K(l,l) |2 + K(^,1,1)2 n + K^^ 
_ i I Jl i_ i_ 1 1 
\ ]'^p+l^l+p+2{ )+... 
How 
K(p,l,l,^’)N + JK(^D,l,l)K(l,l) 
= K(p,1,1^)n + K(p,1,1) 
= K(p,1,1,^,n); 
