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7) we have t h a t s ' = s t l . 7) and being m = sp ) we have 2 R t h a t k/p must be an i n t e g e r , t h a t i s k = c p R , w i t h c p o s i t i v e i n t e g e r . Hence Moreover by v i r t u e o f ( 4 . 8) R m=sp ,n = (st1)p' ,k = cp R (s,c p o s i t i v e i n t e g e r s , 1 G II G h-1). 9) R P m

We o b t a i n t h e r e f o r e wn # 0. If i t were wf = q, by v i r t u e we should have k =q(pktm) = qn. As kl and k2 are 2 s o l u t i o n s o f ( 4 . 1). We o b t a i n t h e r e f o r e wn # q. 6) 1 < wn1 s: wn2 < q-1. By previous arguments, a p o s i t i v e i n t e g e r u and an i n t e g e r A , w i t h 1 e x i s t , such t h a t : < X < h-1, M. 7) 1 X w 2 = ( 0 t l ) pX , n wn = u p , 1 G X G h-1. 8) we have (as k 2 - k l k 2 = mq a,where A + o$tll t +'I1. i s the discriminant of ( 4 .

7) we have t h a t s ' = s t l . 7) and being m = sp ) we have 2 R t h a t k/p must be an i n t e g e r , t h a t i s k = c p R , w i t h c p o s i t i v e i n t e g e r . Hence Moreover by v i r t u e o f ( 4 . 8) R m=sp ,n = (st1)p' ,k = cp R (s,c p o s i t i v e i n t e g e r s , 1 G II G h-1). 9) R P m

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