394 
and 
Proceedings of the Royal Society 
Kfel,l)2N + K(l,l)^ 
= 2[K(p,1,1)n + K(1,1)} 
The identity may therefore he put in the form 
— _ 1 1 1 
s/K(p,1,1,p,n)2 + 2K(1,1,p,n) = 
■^p + l + l+ p+2{} + ... 
* * 
But 
-f K(l, 1 ,p,n) =i'+y + y + y+ 4' 
and thus we have the curious theorem,— 
If the periodic continued fraction for + %i (d prime to A) he 
loanted and the continued fraction equivalent to A-~d he found to he 
1111 
of the form ^ + ^hen the periodic continued 
.,..1111 1 
fraction required ^«A+— — 
Example — 
V338= = + __ 
* * 
f 18 9,4 
for y-=^ + y . 
„ 1 1 1 
= ^ + Y + T+ 3 ’ 
_2 + l 1 1 J . 
1 + 1 + 2 + 1 
10. Taking the case of the general theorem where Z = 6, we have 
^\{\{pP‘h‘^c + 2ahc + ^a% + 2a + c)m - ^{ah^^c + hc + 2ah + \)(fi^c + 2&)}2 
+ {ab^c + 6c + 2a& + 1 )m - {h'^c + 26)2] 
:r.n+Ll.J__LJL _L 
a+6 + c + 6 + a+ 2| | 
* 
Here the only case which is not possible is a, 6, c, all odd : that is 
to say, we may have 
