108 



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

 p we set 



X* -|- ax^ -}-'jx + c= (x + 1) (x^ — lx--\-nix-]-xx) (mod p). 



= x^-)-(m — P) X- -I- (n -j- Im) X -]- In (mod p) 

 hence, 



m — P =a 1 



n-[-lna = b }-(modp). 

 In^c j 



from which we get 



(1) l* + aF — b] = — c. (mod. p). ' 



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



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

 of p is 



(p) = (p, + 1) (P, ©^-le^ + me + n). 



B) If (1) 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 + F) (x^^ + sx + t) (mod. p). 



where 



s = -(l + K 



(mod. p;. 



t~a — P— 1'-— IV 

 and hence, 



(p) = (p, e + 1) (p, e + n (p, 62-1-89 + 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 + 1') (x + \") (X + V") (mod. p). 

 and hence, 



(P) = (p, e -f 1) (p, B + Y) (p, e 4- n (p, e + r^O- 



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



set 



F(x) ^ (x^ -f i^x + n) (x^ — mx + n') (mod. p). 



^^x^ -)- (n -{-n' — m^) x^ -|- m (n^ — n) x -f- nn^ (mod. p). 



Hence, 



n-f-n^ — m^ii^a T 



(2) m(n^— n)=b |- (mod. p). 



I 



nn'^== c J 



If the system (2j is soluble we have 



F(x)=x2 4-mx + n) (x^— mx + nO (mod. p). 



