Homework 4
due: March, 7, 2013, the day of the Midterm
Problems marked with ∗ will be graded. Please practice on the rest of them to the extent you find useful.
1. ∗Divide with the remainder. Find m, r such that a = m ⋅ b + r, 0 ≤ r < ∣b∣, for
(a) ∗a = 10, b = 3
(b) ∗a = −153, b = 12
(c) a = 15398473, b = 543
2. ∗What is Λ12,9 ? What is the simplest way you could describe all the elements of this set?
3. Find m, n such that m ⋅ 12 + n ⋅ 9 =
(a) ∗-24
(b) 9
(c) 0
(d) ∗6
(e) ∗3
4. Follow the GCD algorithm to find gcd(5, 12),
5. ∗Find the prime factorizations of 30 and 4 use them to find gcd(30, 4) and lcm(30, 4).
6. ∗Prove that a∣b, a∣c ⇒ a∣(b + c)
7. Prove that a∣b, k ∈ Z ⇒ a∣bk
8. Prove that gcd(a, b) = gcd(a, −b)
9. ∗Prove that a∣b ⇒ gcd(a, b) = ∣a∣
10. Let p be a prime. Prove that
p∣a ⇐⇒ p2 ∣a2
11. ∗Let a, b ≠ 0, c = gcd(a, b), and a = a′ ⋅ c, b = b′ ⋅ c. Prove that gcd(a′ , b′ ) = 1. Hint: you may use
Pr6).
12. ∗Let p, q be prime numbers, p ≠ q. Prove by induction on n that gcd(p, q n ) = 1. Hint: use Lemma
Pr2.
13. Find three numbers a, b, c such that gcd(a, b, c) = 1, but gcd(a, b) ≠ 1 and gcd(b, c) ≠ 1. Remember
that gcd(a, b, c) = gcd(gcd(a, b), c). On the other hand show that
gcd(a, b) = 1 ⇒ gcd(a, b, c) = 1
This problem shows that ’pairwise relatively prime’ is a stronger condition that ’jointly relatively
prime’.
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