CLASSIFICATION AND DETERMINATION, etc. 27 



Hence, if |/ = Uj is a root of E(%) = 0, we have 



Then, since p n + 1 = 2% t odd, we have 



^+1+ i=o ; uf - 1/u.. 

 Applying 24, we have modulo p, 



so that every root 1 ) of E(%) = belongs to the GrF[p n ~\. Hence 



Substituting in this identity 



Q* U Q* (JCA/2 



and clearing the equation of fractions, we obtain formula 15). 



38. As a simple example, let p n = 1 2 3 1, I 4. The 



binomials 



# 16 - 5 s (s = 1, 3, 5) 



can be readily decomposed into irreducible quartics. The congruence 



(i) = r+4| 2 +2 = (mod 7) 

 has the roots 1 and + 3. Further 



^i6_ 5, = ^ie + 5, + 8== ^i6 + 52^ ( s + 3 = 22 = 4, 6, 8). 



Since S'-^^S 2 ' (mod 7), equation 15) becomes 



-& 1 (mod 7), 

 holding for Z = 4, 2, 3. Taking each in turn, we have modulo 7: 



X + 4 == (^_ ^2 _ 4) ^4 + ^2 _ 4) ^4_ 2 ^2_ 4) ^4_|_ 2a; 2_ 4^ 



^i6_j_ 2 = (^_ 2^2_ 2) (0*+ 2ic 2 - 2) (a; 4 - 4z 2 - 2) (^+ 4ic 2 - 2), 



1) For another proof see Serret, Cours d'Algebre sup<rieure, II, pp. 160 3. 

 Compare 82 below. 



