265 



Proof I. Ic is a prime number 



a), k is a divisor of r — 1 and n; consequently D =z k. 



(I [D) <f {D) Xv Tn, '^ = (—1 ) (^ — 1) Xi {n, 1) 



= — (A — 1) , according to 4. IV, 



and 



k _ 



m=l "'=1 



, 2nim , 1 inim 



k _ * — 1 



2 XvK k)e '' z= 2 e ^ = — I, 



7/1 = 1 »i=l 



consequently, according to definition 6, 



a, (n, k) = - (k—1) = {I {D) <f (D) Xv U^]- 



i^). If k be a divisor of n and not of r — 1, we shall have 



Xv ( n, J j = Xv (", k) z= 0, f accord ing to 4, III), 



, 2nimn , 



2 Xv (m, A') ^"^ == ^ Xv (^, '^) 



»«=! '"=1 



= , (according lo 4, VII), 



consequently 



a-j («, k) := 0, 



80 that now both members of the sought relation ai-e equal to zero. 

 y). r.et k be no divisor of n, so that we may make 



nn' ^ 1 {mod. k), mn ^ m' {mod. k), k'^ m' ^ I . 

 According to proposition 5 



Xv {n, k) = Xv {n\ k), 

 consequently 



Xv {n, k) Xv (m', l^ = Xv (w', k) Xv (m', ^). = Xv ('«'"'' '^O = "/v (m, A), 



because 



m'n' ^ m7in' ^e m (moii. ^). 



Hence it follows 



2 Xv {m, k)e ^ = Xv 0', ^') ^ h (^'^ '^O ^ ^^ » 



7/i=l ''*'=' 



consequently 



a, (,^ /') = Xv ('', /•) = f' {i>>) <t m Xv ( n, -^J , 



because I) = 1. 



II. y(: is the product of two different prime factors. Take k =^ i\k^ 



