CLASSIFICATION AND DETERMINATION, etc. 37 



Hence 



Since 5-5 = 1 (mod 2 3 ), JF 5 (#) = and F^af) = are equivalent in 

 the GF&. Let 



-iz = x ax . 

 If a? be a primitive root in the 6rjF[3 2 ], # 8 = 1, #*= 1. Hence 



FaF = # 10 a# 5 + & = x 2 - ax + 6. 



Hence 

 giving 



a' =26, & 2 = 1 (mod 3). 



Hence 6 = - 1, o = 1 (mod 3), so that the two PIQ[2, 3] 



are a: 2 a? 1. 



53. For j) w = 5, m = 2, we have 



1) a; 12 -!- 1 



- - ^ - ** L 



The eight integers t less than and prime to 24 are 

 1, 5; 7, 11 = 5-7; 13, 17 = 5-13; 19, 23 = 5-19 (mod 24). 



Each pair of integers furnishes a single F t (x). For each of the 

 eight values of x y we have r 2 = 1 (mod 24). Hence F t (x) = is 

 identical with F^tf*) = in the (jr-F[5*], For a primitive root x, we 

 have # 12 = 1. We have therefore in the field, 



F^x) = x*+ ax + 6, j; (a; 18 ) = # 2 - a^ + ft, 

 a; 8 ^(a 11 ) = bx*- ax -f 1, o; 2 ^^ 23 ) = 6^ 2 -f aa; + 1. 

 The product of these four quadratics is therefore identical modulo 5 

 with V(a*- x*+ 1). 



It follows that 5 2 = 1 (mod 5) and, by subsequent expansion, 



2a 2 =6 (mod 5). 



Hence the four PIQ[2, 5] are x 2 + ax + 2a 2 , viz., 

 30) x 2 x + 2, x* 2x - 2. 



Another method of solving this example is to require that 

 #2-1- ax 4. ft shall divide x 8 x*-\- 1 modulo 5. We reduce the latter 

 function by means of the relation 



#*_ -ax -I [a 4 = 6 4 = 1 (mod 5)], 

 and find, modulo 5, that 



^s_ x + i = (- a b b - ab*)x - a*b 2 - a 2 6 3 + 2. 

 Hence 6=_1, 2a 2 EE& (mod 5). 



