OPERATORS AND CYCLIC SUBGROUPS etc. 257 



the number of distinct odd prime factors of p n 1, then the con- 

 gruence 246) has exactly 2 x +<" solutions m. 1 ) The solution m E^ 1 is to 

 be excluded. Consider the 2 x +<" 1 substitutions with the multipliers 

 a, a m , a. 1 , m > 1. They generate as many cyclic groups. In fact, 

 (a m ) x =K requires x = m (mod p n 1); while (a~ m ~~ l )y = a is im- 

 possible since m + 1 has a factor > 1 in common with p n 1. 

 Moreover, the sets of multipliers of the substitutions of period p n 1 

 in each cyclic group are the same in pairs. Hence these special cyclic 



groups contain altogether yO(^ TC 1) (2 X +/* 1) distinct sets of 

 unequal multipliers. 



(iii) If a mr =a', a ms =a t ) a mt =a r , we find that 



= (mod ^> w -l). 



Hence Jf = m 2 -f m -f 1 must be divisible by j9 n 1. Since m (m -f 1) 

 is even, M is an odd number. Hence p n 1 must be odd and there- 

 fore p n = 2 W . Since d = 1, 3 is not a factor of p n 1. Hence each 

 prime factor g of ^) w 1 is of one of the forms 6&-J-5, 6& + 1. 

 Now .M" and hence also w 3 1 must be divisible by q. If # = 6&-f 5, 

 Fermat's theorem gives w 6 *+ 4 =l (mod #). Since w 3 =l, we have 

 m = 1 (mod g') and therefore Jf =3^0 (mod g), which is impossible. 

 Hence must q = 6fc -f 1. Inversely, if ^ = 67^ + 1, m 6 * 1 = (mod q) 

 has 6& distinct integral solutions. But the left member is divisible 

 by m 3 1 and therefore by M . Hence M = (mod #) has two 

 distinct solutions. Each of these solutions leads to one, and but one, 

 solution of M = (mod (f). To give a proof by induction from 

 T = e to r = e-fl, let w 3 -l = ^. Then 



(m + x(f 1 EE Qq e + 3m 2 xq e (mod 2 2e ) 

 and will therefore be divisible by g e+1 if, and only if, 



::^Q (mod q). 



Since 3 and m are prime to q y x is uniquely determined mod q. 

 Hence each m determines one solution y^.m-\- x<f of 



if 1 = (mod 2 e + 1 ). 



Hence, if m 2 + m -f 1 be divisible by # e , </ 1 will be prime to q 

 and hence y 2 + / + 1 w ^ ^ e divisible by #*+ 1 . Supposing that the 

 prime factors of 2 n 1 are all of the form 6& -f- 1 and that the 

 number of distinct ones is y, it follows that M=Q (mod 2 n 1) 

 has 2 y solutions m. But, if m be a solution, then m 1 will be 



1) Dirichlet, Zahlentheorie, 37. 



DlCKSON, Linear Groups. 17 



