Question

9. Given any nonconstant polynomial f(x) with integral coefficients, prove that there are infinitely many primes p such that f(x) = 0 (mod p) is solvable. (H)

Answer 1

f(a)0 (mod p) By Taylor's series expansion, f(a+tp) = f(a) + tp° f'(a)+ tp2i " (a)/2+ "pif(n)(a)/n!, (3.2)

where n is the degree of f(x). Observe that if fe)-Σ.', T-0 then Σ. (Ε) k! rk Therefore, f() (a) k! is an integer for 0<k<n. We now suppose that f(a) 0 (mod p) Since p a) ! 0 (mod p+) for k1 (2j> j for j 1), we conclude from (3.2) that (mod p) f(a+tp) = f(a)+ tp'f'(a) (3.3) We now return to (3.3). We see that we need to choose f(a) (mod p) tf'(a) (3.4) We now split our investigation into cases: Case 1. pt f'(a) In this case, there is a unique solution t for (3.4) and hence we obtain THEOREM 3.2 (Hensel's Lemma) Suppose that f(x) is a polynomial with in- tegral coefficient. If f(a) 0 (mod p) and f(a)# 0 (mod p), then there is a unique t (mod p) such that f(a+ tp)0 (mod p+) Case 2. f(a) then (3.4) is not solvable. If plf'(a) but pt р Case 3 If plf'(a) and plthen there are p solutions for t in (3.4).