Homework #10 due November 16
Math 365
Fall 2007
book problems
9.1 #7 If p = 2k + 1 is prime, verify that every quadratic nonresidue
of p is a primitive root of p. Let a be a quadratic nonresidue of p.
Then by Eulers Criterion, a(p1)/2 1 (mod p). Since p i
Study guide for exam 2 on November 16
Math 365
Fall 2007
I. Be able to state:
1. Theorem 6.7: Mbius Inversion formula
o
2. How to calculate , , and . In particular, Theorems 6.2 & 7.3
3. Theorem 7.5: Eulers Theorem
4. Theorem 8.3 (on the order of ah (mod
Math 365
Homework 2 Select solutions
Fall 2007
book problems
2.3
#18 Let k, k + 1, k + 2 be three consecutive integers. By the division
algorithm, one of them is divisible by 3, hence 3 | k (k + 1)(k + 2).
By a similar argument, 2 | k (k + 1)(k + 2). Henc
Selected solutions, Homework 3
Math 365
Fall 2007
book problems
3.2 #13(c) If gcd(m, n) = 1 then gcd(Rm , Rn ) = 1.
Let x, y Z be such that 1 = mx + ny . Then,
Z
10 1
9
10mx 10ny 1
=
9
10mx 10ny 10mx + 10mx 1
=
9
mx
= 10 Ryn + Rmx
1=
By 13(a), Ry | Ryn an
Math 365
Select solutions, Homework 4
Fall 2007
book problems
3.3 #20 p is equal to one of 3k , 3k + 1 or 3k + 2. When p = 3 we can check
directly what happens, so we need only consider the case p = 3k for
k > 1. In this case p is not prime so we neednt c
Homework # due #
Math 365
Fall 2007
book problems
4.4 #11 Let d = gcd(m, n). Then, d|n and d|m. If x a (mod n) and
x b (mod m), then d|x a and d|x b. Therefore, d|(x b) (x a),
implying that d|a b. Conversely, if d|a b then there is a solution to the
linea
Math 365
Homework #6 due Oct 19
Fall 2007
book problems
6.1 12 a). According to the formula, if n = r=1 pki , then (n) = r1 (ki +
i
i
i
1). Therefore, if (n) = 10 we have only a limited number of possibilities. 10 can be written as a product in only two w
Math 365
Homework #7 due Friday, Oct 26
Fall 2007
book problems
7.3 #9 Since gcd(2, 77) = 1 and (77) = 60, by Eulers
Theorem 260 1 (mod 77). Therefore,
5
210 = (260 )1666 240 240 23 (mod 77)
#11
Let p be a prime. a). Show (p!) = 2 (p 1)!). Since p is
prim
Math 365
Homework #8 due Nov. 2
Fall 2007
book problems
8.1 # 4. Let r be the order of ab (mod n). By Thm 8.1 to show r | hk
it is enough to show that (ab)hk 1 (mod n). But this is clear since
(ab)hk = (ah )k (bk )h 1 (mod n). To see that if gcd(h, k ) =
Math 365
Homework #9 due Friday November 9
Fall 2007
book problems
8.2 #1
a). There exist at most 2 incongruent solutions to x2 1 (mod p) by
Lagranges Theorem. Since 1 and p 1 are incongruent and both satisfy
x2 1 (mod p), these are the only two solutions
Math 365
Exam 1 study guide
Fall 2007
I. Be able to state:
1. Well Ordering Principle
2. Binomial Theorem
3. Division Algorithm (2.1)
4. GCD Theorem (2.3)
5. Relatively Prime Theorem (2.4)
6. Fundamental Theorem of Arithmetic & canonical form corollary (3