MAT 115B HOMEWORK 6 SOLUTION
14.1.4: (a) No, since | |. (b) Yes, since (2 + i)(6 - 3i) = 15.
14.1.10: Solve the equation a + bi = u(a - bi) for u cfw_1, i, -1, -i and we get is a multiple of 1, 1 + i, i or 1 - i. 14.1.16: = 79-8i . Take = 6 - i, then = -
Math 115b: Number theory Homework 3
This problem set is due Wednesday, January 30. GK3.1. Show that if x Z p is a p-adic integer for a prime p, and if x has square roots in the p-adic numbers Q p , then actually its square roots lie in the p-adic integers
MAT 115B HOMEWORK 3 SOLUTION
GK 3.1 Assume that x Zp has a square root y , i.e. x = y 2 , such that y Qp but
y Zp . Consider the rightmost nonzero digit of y , it must be on the right of the p-adic
point. Then the rightmost nonzero digit of y 2 must also
MAT 115B HOMEWORK 2 SOLUTION
GK 2.1 (a) Suppose that 0 is friends with b in Z/p. We have by definition, 0 = -b or 0 = ib. In all the cases, b = 0. Hence 0 is friends with itself. Now suppose that a = 0 and p > 3 is a prime. Observe that elements in cfw_a,
Math 115b: Number theory
Homework 2
This problem set is due Friday, January 25.
GK2.1. The question arises of when i is present in Z/ p for different primes p. In other words,
when there is a solution to the equation x2 = 1, which you can then call i.
(a)
Math 115b: Number theory
Homework 1
This problem set is due Friday, January 18.
If a problem is starred, that means that it could be harder and it will be graded as extra credit. A
particular starred problem might be within reach for you, or it might be e
MAT 115B HOMEWORK 1 SOLUTION GK 1.1 Suppose that n = a2 + b2 . Since a2 , b2 0, 1 (mod 4), their sum a2 + b2 can only be 0, 1, 2 (mod 4). GK 1.2 Use the previous problem and the fact that n (mod 4) is determined by the last two digits of n. GK 1.3 First s
Math 115b: Number theory
Homework 4
This problem set is due Friday, February 8.
GK4.1. Using the Euclidean algorithm nd the gcd of x3 + 2x2 + 2 and x5 + x4 in (Z/3)[x], and
express it as a combination of the original two polynomials.
GK4.2. Prove that in
MAT 115B HOMEWORK 9 SOLUTION
GK 7.3.30 (b)(c): By the L-L test, M7 = 127 is prime, and M11 = 2047 is composite.
GK 9.1 (a): The chance that gcd(N, A) > 1 is (N (N )/N 0.15%. Now assume that
gcd(N, A) = 1. The group (Z/N ) is isomorphic to (Z/853) (Z/3517)
Math 115b: Number Theory
Solutions to the Second Midterm
1. How many elements in (Z/131) have order 10? How many have order 30? (You can accept
without proof that 131 is prime.)
Solution: From Jiahui Guan:
2. Identify all combinations that are true statem
MAT 115B HOMEWORK 8 SOLUTION
2.3.2: 2n4 + 3n3 + 17 2n4 + 3n4 + 17n4 22n4 . It follows 2n4 + 3n3 + 17 is O(n4 ).
2.3.10 Let r R, r 1. If f is O(log2 n), then K R+ such that f (n) K log2 n for
all n. We have f (n) K log2 n = (K log2 r)(logr n), where K log2
MAT 115B HOMEWORK 7 SOLUTION
14.2.16: (a) Observing that N (9 + i) = 82 = 2(42 + 52 ), we try factors 1 + i, 5 + 4i and
4 + 5i. Have 9 + i = (1 + i)(4 + 5i)(i).
14.2.28: (a) Using Euclidean algorithm to get 4 i = (1 i)(2 + i) + 1. So x 3(1 + i)
1 + 2i (m
MAT 115B HOMEWORK 5 SOLUTION
9.1.12: Let r = ord(a), s = ord(b), and t = ord(ab). We want to show that t = rs.
Since ar 1 (mod n) and bs 1 (mod n), we have (ab)rs 1 (mod n). Hence t | rs. On
the other hand, since (ab)t 1 (mod n), we have brt (ab)rt 1 (mod
Math 115b: Number Theory Solutions to the First Midterm
1. Prove that in the field Z/1009, at most 505 elements are squares of other elements. (You can take it on faith that 1009 is a prime number.) Solution: From Amber Nelson:
If I could give extra credi