Question

39. Suppose that the polynomial congruence f(x)0 (mod 7) has two distinct so- 0 lutions, what are the possible number of solutions of the congruence f(x) (mod 49)?

Answer 1

Let us recall the taylor expansion of a polynomial(f(x) is the derivative of f, f"() is the second derivative and so on) f" (r) 2! Now say 0 < a, b < 6 are the solution for the congruence f(z) Ξ 0(mo(17) Also let 0 < c < 48 be a solution for the congruence f(z) 0(110)(149). In this case /(c) 0(m()(17), hence c Ξ a, bmo(17. Therefore any s(lution of f(x) 0(mod49) will be in a sense a lift of a solution of f(x) 01n0d7. Let us discuss the lifts of a i.e. ca7t,0 t 6. Now we have two cases: . Case 1: f,(a) 0(mo(17). Using the taylor expansion, we have f(a+ 7t)f(a) f(a)7t.Taking congruence modulo 49 we have /(a + 7t) f(a) (mod49). Therefore if a is the solution of f(z) Ξ 0(mod49), then we have 7 distinct lifts of a. If not then a can't be lifted to any solution . Case 2: f(a) f 0(mod7). Proceeding similar to case before we have f (at 7t) (a) +/(a) 7t (mo(149) (we have second term because "(a)诺 0(mod7)). Let f(a)7z. If a 7t is the solution, then we have 7(z + t/(a)) 0(mod49) → z + tf"(a) Ξ 0(In()(17). Now since we have f,(a) 0(111()(17), we get thatョg, f,(a)q 1(mod7). Now if t mod7), we see that a tp is indeed the solution. Therefore a can be lifted uniquely in this case. (This is the result of Hensel's Lemma Therefore the given congruence can have 0(no solution could be lifted), 1,2 (unique lift), 7(one didn't lift, other didn't ft uiquely), 8, 14 solutions