106 



and also 



fl"! + f2<'2 + hfr, + Ui = 4. 

 From these two relations we see, remembering that A is divisible by the 

 square of a prime ideal (^), that the required factorization of p is either 



(p)=Ai2.A2.A3 

 where Ai, A2, A3 are prime ideals of first degree, or 



(p) = A1-A2 

 where Ai is of first degree and A2 of second degree. 

 Hence the factors of F(x) are either 



F(x) = I Pi(x) } ' P,(x) . F,{x) (mod. p), 



where Pi(x), P2(x), PsCx) are prime functions of first degree, or 



F(x) = {Pi(x)|' P,(x) 



where Pi(x) is of first degree and P,(x) of second degree. 



In order to find the prime ideal factors of p we have thus to resolve F(x) 

 into its prime factors with re.-pect to the modulus p. To do this we set 

 X* + ax2 +bx + c = (x + l)2 (x2 + mx + n) (mod. p) 



= x^ + (m + 21) x* + (n + P + 2ml) x^ + (ml^ + 

 21n)x + nl2. (mod. p.) 

 Hence, for determining 1, m, n we have the congruences 

 m + 21 = o "I 



n + 2ml+P = a 

 mP + 21n = b 

 nl2=c J 



Eliminating m and n, we get 



4P + 21a = b ) 



31* + 3.I2 ^c j 



which give 



2al2 ^3bl — 4c (mod. p). 



Having thus obtained the values of 1, m, and n, we set 



X2 +mx + n = (x + r) (x + s) (mod. p) 



= X2 +(r + s) x + rs. (mod. p). 



Hence, 



r-|-8 = m ) 



_ > (mod. p). 



rs^n J 



or 



(r — 8)2 = — 4 (a 4-212) (mod. p). 



} (mod. p). 



(1^ Hilbert, p. 201. (2) Hilbert, p. 195. 



