CLASSIFICATION AND DETERMINATION, etc. 41 



To determine the PIQ[6, 5] satisfied by the primitive root 

 = x 2 2#, we form the powers, 



-x 2 - x-l; Q*=x*-2x\ 



* + 2x + 3, Q= 2x- x*+ x*+ x 2 + x-1. 

 We derive at once the required congruence 



P 6 - <? 5 -f (> 4 - ? 3 -h 20 + 2 ~0 (mod 5). 



57. We can set up the PIQ[2, 2 3 ] and PIQ[6, 2] by means of 

 the theorem: 



34) l*x*+lx + p 



is a PIQ\2, 2 3 ] if, and only if, ft is a root of 



35) J 3 = j 2 -fl (mod 2) 



and A is (my mark except zero and 4 . 



By 40, the quadratic 34) is an IQ[2, 2 s ] for every mark 

 A =)= in the 6r-F[2 3 ] and for every root /3 of the congruence 



/S 4 + /3 2 + /3 + 1 = (ft -f l)(/3 3 -f P 2 + 1) = (mod 2). 



Defining the 6r.F[2 3 ] by means of the irreducible congruence 35), 

 we may take /3 = 1, j, j 2 or j 4 . We first find the exponent ep to 

 which belongs a root | of the congruence 



| 2 -? + ^ (mod 2). 

 Since jj belongs to the GF[2 2 ' 3 ], e^ is a divisor of 2 6 1 ^ 3 2 - 7. But 



Hence for = 1, ^=3; for a root /5 of 35), we find 



so that ep= 2 6 1. The theorem is therefore proven for the case A = 1. 



Setting = A#, it follows that, for =j= 1, # belongs to the 



exponent 2 6 1 unless x 9 = 1 ? which occurs only when A 2 =/J, 



i.e., >l = /3 4 . We therefore reach all |O(2 6 - 1) = ISPIQ[2, 2 s ]. 



Half of them are given in the left members of the identities below. 

 To pick out a set of three whose product gives a PIQ[6, 2], we 

 select three which are like functions of respectively j, j 2 , /, the 

 latter being the roots of 35). We thus find 



x + j 4 ) = 



5 ^ + j 4 ) - 



