Classes of Congruent Integers. 27 



is also sufficient can easily be seen; for, then, the congruence 

 becomes 2 + z=0 (mod. f), which has evidently a solution x = 2. 

 Thus we get a case 



d = l, n>2, p>2. (45) 



Next, putting /==! in (ii), we get d=l, n=2, no restriction 

 being laid upon p, in other words p^2. Hence, combining this 

 case with (45), we get 



d=l, n>\, #»2, 



and d=], n=% p=2. \ (46) 



Lastly from (iii), we get «— I — J =1. Putting I — = 



— - + 1 — e, 0<e^], we can transform it as follows: 



1 2 , s 

 — +— =1 + — , 

 p ii n 



1 2 

 or, since we suppose n>l, 1< 1 — <2. 



Solving this inequality, we obtain 



f P>% 

 1 n=2, 



and it will be found on verification that these values all satisfy 

 n—\ — = 1- If P = 2, ?i=2, 3, then fc=d and consequently, from 

 the condition n<zd + k ; we obtain <i>l. If p>2, n=2, the condi- 

 tion n<zd + k becomes 2<.d+ l _ . , from which d>l follows. 

 Thus we get the two cases, 



p=2, «=2,3, d>], 



p=2, 

 n = 2, 3, 



(47) 

 p>2, n=2, ä>\. J 



Combining n=l with each of (46) and (47), we arrive at the 

 following conclusion : 



Let p be a prime ideal of the first degree and suppose that its norm p 

 is divisible by p' but not by p i+ \ then there exist primitive roots of p" when 



