of Edinburgh, Session 1883 - 84 . 581 
Kow as {§ 2) is even, -wliieh is prime to it, is odd 3 hence 
(-1VA + P,.,Q,_, 
a-l 
= an integer = m say. 
A = (-1)'P,.,M-(-1)T,.,Q,_, .... (a). 
Substituting this in the expression for P^.^D we have 
V-=D = ( - lr2P,_,Q,.,M - ( - l)*2P,.,Ql.i + 2 m 
= ( - ir2M{P._,Q,_, + ( - 1)^} - ( - 1) 2P,_,Q^, 
= (-1)^2mP,_,Q._,-(-1)^2P,_,QL, 
and.-. D = (-l)^2MQ,.,~(-ir2QL, (P). 
As H = A® + D it is seen that (a) and (/ 5 ) give the desired expression. 
5 . As an example of the use of the preceding result, let us try to 
find all the numbers whose square roots have culminate cycles of 
eight elements. 
If we denote the symmetric portion of the cycle by 
a, b, c, d, c, b, a, 
the initial condition which we have to satisfy is 
(J,c) = 2(a, h); 
i.e., bc +1 = ‘lab + 2 ; ■ 
i.e., b{c - 2a) = 1 j 
the solution of which in integers is clearly 
6=1, and c = 2a+ 1 . 
We thus have 
l^a-i = (^5 I5 2a + 1) = 2a^ + 4a + 1 , 
Pa-2 = («, 1) =a+l, 
Qz-2 = 1 p 
SO that the general result desired is 
x/{(2a2 + 4a -M)m » (a + 1 }2 + 4(a + 1)M - 2 
_ f ) 1 1 1 1 1 111 
( / ■*'S+ 1 +2a + l + { } + 2a+l+ l + a + 2{ }+•••’ 
* * 
where for shortness there is put 
{ } for {(2a2 + 4a+ 1 )m - (a + 1)} . 
