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Question - Math 324 Assignment 6 (due October 29)

1. (7) If m ≥ 1 is an odd integer with exactly s di stinct prime factors, determine the
integer N of Problem 4.7. Hint: Theorem 4.13 and Problem 4.4.

2. ( 6) If r is a primitive root for a prime p ≡ 1 mod 4, show that − r is also a primitive root
mod p.

3. ( 3 + 6) Let q be an odd prime. Show that
a) there are exactly two real-value d Dirichlet characters mod q, and
b) since the principal character χ0 mod q is one of the two above, let χ1 be the other one
and show that, for all integers b with (b, q) = 1, we have
(i) χ1(b) = 1, if x2 ≡ b mod q has a solution x, and
(ii) χ1(b) = − 1, if x2 ≡ b mod q has no solution mod q.

...Read More 4. ( 3 + 3 + 3 )(4.4 #2) Let f(x) = x3 + 8x2 − x − 1. Find all solutions of
a) f(x) ≡ 0 mod 11, b) f(x) ≡ 0 mod 112, and c) f(x) ≡ 0 mod 113.

5. ( 3 + 6) For each integer m ≥ 1, set ω = ωm := em(1) (note that ω m = 1) and consider the
set R of all complex numbers of the form ∑
0 ≤ i < m a iω i with integer coefficients a i.
Show that a) R is a subring of C, and
b) for each µ ≠ 0 in R, define congruences mod µ R, by α ≡ β mod µR ⇔ α − β = µρ for
some ρ ∈ R ( ⇔ α − β ∈ µ R with µR:= { µρ: ρ ∈ R}). Show that
i) this defines an equivalence re lation on R, and (for brevity),
ii) that, if α ≡ β mod µ R and χ ≡ δ mod µR, then αχ ≡ βδ mod µR. ...Read Less

Solution Preview - lutions x mod m of x2 ≡ 1 mod m. Let m = p1a(1) … psa(s) be the prime power factorization of m, with a(i) ≥ 1 for all i. Each solution x of x2 ≡ 1 mod m is a simultaneous solution of the equations x2 ≡ 1 mod pia(i) for 1 ≤ i ≤ s. For each i, this equation mod pia(i) has exactly 2 solutions x ≡ ± 1 pia(i), by Problem 4.4, so each

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