42 CHAPTER III. 



Replacing x by - and multiplying by x*, we find 



which with the above three sextics give the six existing PIQ\, 2]. 



58. Theorem. The necessary and sufficient conditions that x p x a 

 shall be a PIQ\j), p] are that a be a primitive root modulo p and 

 that a root of y p = y + 1 (modp) belong to the exponent (p p 1)/Q> !) 



If a be an integer not divisible by p, the congruence 



x p = x + cc (mod p) 

 is irreducible by 40. The product of its roots is 



..x~ x p ~ l ==- 

 Setting x = ay, we find that 



y p =y-\- 1 (mod p). 



Hence if # belong to the exponent p p 1, then a is a primitive root 

 modulo p and ?/ belongs to the exponent (p p l)/(p 1). The in- 

 verse is true by 14 ? since p 1 and (# p l)/(jP 1) are relatively 

 prime. 



59. EXERCISES ON CHAPTER III. 



Ex. 1. If 9 be a root of one of the PIQ[2, 5] of 53, then x* g 

 is an I[3, 5 2 ]. Eliminate and derive the following I$[6, 5]: 



x*x 5 + 2, iC 6 2ic 3 2. 



Ex. 2. (Moore). If x be a root of the irreducible congruence 

 x e_ % X 3_ 2 = (mod 5), 



a mark c -f c x + c 2 o; 2 -f- C 3 ic 3 + c 4 # 4 -h c 5 5 of the (rjP[5 6 J will belong 

 to the included field G-F[5 B ] if and only if 



c 3 = 0, c 4 = Scj-f 4c 2 . C 5 = 2c x -f- 3c 2 (mod 5). 



Show that qp = a? + x 2 -}- 2x^ is a primitive root of the 6r.F[5 3 ] and 

 that it satisfies the congruence qo 3 ^ 2qp + 3 (mod 5). 



Ex. 3. (Pellet). If y belong to the G-F[p n ] and m be the least 

 integer for which yP m y, then x p x y is irreducible in the field 

 if neither n/m nor y + y p -\- y p2 H ----- hy pm " be divisible by p; in the 

 contrary case it decomposes in the field into linear factors. Prove this 

 theorem equivalent to that of 40 for I = 1. 



Ex. 4. (Pellet). If p be a prime number which is a primitive 

 root of the prime number , - ~ is irreducible modulo p. 



XV X 1 



Ex. 5. Show that the theorems of 34 and 36 may be combined 

 into the theorem stated without proof by Pellet: 



