24 CHAPTER III. 



to the exponent k(p n l}/d, d being the greatest common divisor oft 

 and p n 1, are the binomials x 1 Q*, the case p n =4:l l, A = 4^ 

 being excluded. 



Inversely, we obtain by this theorem every binomial irreducible in 

 the GF[p n ]. In the first place, I and t must have no factor in 

 common, since otherwise x l tf would he algebraically reducible. 

 On the other hand, if A contains a prime factor 0, not a factor of p n 1, 

 we can determine ( 7, Note) an integer O v such that 



Since (> ae =p, it follows that Q e **= a is a root of 



x<> ^= 0. 

 Hence x a is a factor of x 9 Q', so that x^ e K divides #' Q*. 



Example. For jp n=s l, we may take $ = 5. Then for A = 2 

 and t= 1, 3, 5, we obtain the irreducible binomials x 2 5, a? 2 -h 1, 

 # 2 3 belonging to the exponents 12, 4, 12 respectively. For I = 3 

 and = 1, 2, 4, 5 respectively, we obtain the binomials 



t// -/ * ^(y TL . JO * 00 O 



irreducible modulo 7 and belonging to the respective exponents 18, 

 9, 9, 18. 



36. Theorem. - - Let p n =2 { t l, i^>2, t odd-, I = 2>'s, j > 2 7 

 s odd, let Jc be the smaller of the integers i and j] finally, let m be 

 odd. Then if, in the N quantics IQ[m, p n ~\ belonging to the exponent 



e = (p nm l)/d, 



we replace x by x 1 , where I == 2->s is prime to d and contains no prime 

 factors other than those occurring in p n<m 1, we obtain N quantics 

 of degree m I each decomposing into 2 k ~ 1 quantics irreducible in the 

 GF[p n ~\, so that we obtain all of the 2 k l N quantics 



belonging to the exponent el. 



If v denote the least integer such that p nv 1 is divisible by el, 

 we find as in 34 that v = qm. In the present case, q is even; for, 

 if q be odd, v would be odd and p nv 1 the double of an odd 

 number, whereas I is divisible by 4. By the restrictions on p n 

 and m, 



13) #= 2 l r - 1 (r odd). 



Raising this identity to the power g, we find 



