108 



In order to decompose F(x) into its prime factors with respect to the modulus 

 p we set 



x' -|- ax^-j-bx ~[-c= (x + 1) (x^ — Ix^ -|- mx 4- n) (mod p). 



= x' -]- (m — 1^) x^ 4- (n -<- Im) x + In (mod p) 

 hence, 



m — P = a ~| 



I 



n-[-lna = b }• (mod p]. 



ln = c j 



from wliich we get 



(1) l^ + aP — bl = — c. (mod. p). 



A) If (1) has one solution only, then the prime factors of F(x) are 



(x-)-l) and (x^ — Ix'* + mx + n) and the required factorization 

 of p is 



(p) = (p, + 1) (p, 63 — 1 92 -j- me + n). 



B) If (I) has two solutions 1 and V. Then F(x) contains two factors of 



Ist degree and one of 2d degree and we have 

 F(x)~(x + 1) (x + 10 (x* + sx + t) (mod. p). . 

 where 



s = -(l + 10 ) 



and hence, 



(p) = (p, e + 1) (p, fi + V) (p, 02 + 89 + 1). 



C) If (1) has three solutions in which case it evidently must have four 



solutions 1, V V V, then 

 F(x) = (X + 1) (X + 10 (X 4- n (X + K'O (mod. p). 

 and hence, 



(p) = (p, 9 + 1) (p, 9 + 1') (p, 9 4- rO (p, 9 + K-0. 



D) If (1) has no solution, F(x) has no factors of 1st degree. Then we 



set 

 F(x) = (x2 -f mx + n) (x^ — mx + n') (mod. p j. 



= x^ -(- (n -j-n' — ^) x2 -|- m (n' — n) x -(- nn' (mod. p). 

 Hence, 



n-f- n^ — m2 =a ] 



(2) m(n'— n)=b j- (mod. p). 



I 



nn'=:= c J 



If the system (2) is soluble we have 



F(x) =x2 -(-mx + n) (x^ — mx + n^) (mod, p). 



