MAT315H5S, SPRING 2005
Solutions to Assignment 4
119/40 By Lemma 3.5, it follows that p|a|. But then p|a.
149/2c 7 | 99 2 = 97 so 99 2 (mod 7)
2e 7|(9 5) = 14, so 9 5 (mod 7)
149/5 Solution in book.
149/7a n! 0 (mod 2) for n 2. Thus, by repeatedly using T
UNIVERSITY OF TORONTO
The Faculty of Arts and Science
APRIL 2013 EXAMINATION
MAT315HIS
Duration 3 hours
Instructor: Professor James Arthur
NO AIDS ALLOWED
Family Name:
First Name:
Student Number
Student Signature:
Instruction:
0 There are 7 questions on t
MAT315H5S, SPRING 2005
Solutions to Assignment 2
40/16 a = qb + r = (tc + s)b + r = tcb + bs + r
74/3 The solution is in the book.
74/10 The Conclude that . . . follows immediately from the previous sentence. Euclids proof
tells us that if we take the pro
UNIVERSITY OF TORONTO AT MISSISSAUGA
Test 2 Solutions
MAT315H5S
1. See text.
2. For denition, see text. 1, 2, 4, 8, 9, 13, 15, 16.
3. For Euler, see text.
2
7
2
71
2
8 1 (mod 7), so
2
7
= 1.
4. 65 = 5 13. The solutions of x2 + 4 0 (mod 5) are x 1 (mod 5)
Assignment 1, due January 13
Problem 1
Find GCD(2352, 45) and LCM(2352, 45).
Problem 2
Prove that if a|b and a|(b + c), then a| GCD(b, c).
Problem 3
What are the possible remainders when a perfect square is divided by 7?
Problem 4
Find a pair of integers
Assignment 2, due January 20
Problem 1
Prove that there are innitely many primes of the form 6q+5 for some integer
q.
Problem 2
Prove that if a = qb + r, where a, b, q, r are integers, show that
gcd(2a 1, 2b 1) = gcd(2b 1, 2r 1).
Problem 3
Show that if a|
Problem 1. Let m, n, k be natural numbers, k | mn, and (m, n) = 1. Show that there are
natural numbers k1 , k2 , such that k = k1 k2 , k1 | m and k2 | n.
k
Solution. Let k1 = (m, k) and k2 = k1 . Then k1 | m, k2 N and k = k1 k2 . It remains to
m
show k2 |
UNIVERSITY OF TORONTO AT MISSISSAUGA
Test 1, Solutions
MAT315H5S
1. See text.
2. See text for rst two parts. Since 43 and 59 are prime, (43 59) = (43) (59) =
42 58.
3. See text.
4. 3 4 2 4 (mod 4), but 3 2 (mod 4).
5. By the Chinese Remainder Theorem, the